Study Guide
Linear Equation in Two Variables
-
University:
Eastern Kentucky University -
Course:
MAT 114 | College Algebra Academic year:
2022
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Views:
492
Pages:
43
Author:
Kaiden Murphy
LINEAR EQUATION IN TWO VARIABLES
CONTENTS
Linear equation in two variables
General form of pair of linear equation
Graph of linear equation in two variables
Graphical representation of pair of
linear equation
Types of solutions
Methods of Solving system of linear
equation
Word problems
LINEAR EQUATIONS IN TWO VARIABLES
A statement of equality of two algebraic
expressions, which involve one or more unknown
quantities is known as an equation. If there are
two unknown quantities then equation is called
linear equation in two variables.
A linear equation is an equation which involves
linear polynomials.
A value of the variable which makes the two sides
of the equation equal is called the solution of the
equation.
Same quantity can be added/subtracted to/from
both the sides of an equation without changing the
equality.
Both the sides of an equation can be
multiplied/divided by the same non-zero number
without changing the equality.
Note :- To find value of variables in any equation
we required number of equation equal to number
of variables in equation.
GENERAL FORM OF PAIR OF LINEAR
EQUATION
a1 x b1 y c1 0
a 2 x b 2 y c 2 0
where a1, b1, c1 & a2, b2, c2 are constants.
GRAPH OF LINEAR EQUATION ax + by + c = 0
IN TWO VARIABLES, WHERE a 0, b 0
(i) Step I : Obtain the linear equation, let the
equation be ax + by + c = 0.
(ii) Step II : Express y in terms of x to obtain
ax c
b
y=–
(iii) Step III : Give any two values to x and calculate
the corresponding values of y from the expression
in step II to obtain two solutions, say (1, 1) and
(2, 2). If possible take values of x as integers in
such a manner that the corresponding values of y
are also integers.
(iv) Step IV : Plot points (1, 1) and (2, 2) on a
graph paper.
(v) Step V : Join the points marked in step IV to
obtain a line. The line obtained is the graph of the
equation ax + by + c = 0.
Ex.1
Sol.
EXAMPLES
Draw the graph of the equation y – x = 2.
We have,
y–x=2
y = x + 2
When x = 1, we have : y = 1 + 2 = 3
When x = 3, we have : y = 3 + 2 = 5
Thus, we have the following table exhibiting
the abscissa and ordinates of points on the
line represented by the given equation.
x
y
1
3
3
5 Plotting the points (1, 3) and (3, 5) on the
graph paper and drawing a line joining them,
we obtain the graph of the line represented by
the given equation as shown in Fig.
Y
X
To find the coordinates of the point when
x = –5, we draw a line parallel to y-axis and
passing through (–5, 0). This line meets the
graph of x – 2y = 3 at a point from which we
draw a line parallel to x-axis which crosses yaxis at y = –4. So, the coordinates of the
required point are (–5, –4).
Since y = 0 on x-axis. So, the required point is
the point where the line meets x-axis. From the
graph the coordinates of such point are (3, 0).
Hence, required points are (–5, –4) and (3, 0).
9
8
y – x= 2
7
6
5
(3,5)
4
3
2 (1, 3)
1
–7 –6 –5 –4 –3 –2 –1
1 2 3 4 5 6 7
–1
–2
–3
–4
–5
–6
–7
–8
–9
Y
X
PAIR OF LINEAR EQUATIONS
a2x + b2y = c2
Draw a graph of the line x – 2y = 3. From the
graph, find the coordinates of the point when
(i) x = – 5 (ii) y = 0.
Sol.
We have x – 2y = 3 y =
x3
2
Y
1 3
= –2
2
1
–1
1
2
Y
1 and 2
0
X
P()
Thus, we have the following table :
x
y
....(2)
We know that given two lines in a plane, only one
of the following three possibilities can happen (i) The two lines will intersect at one point.
(ii) The two lines will not intersect, however far
they are extended, i.e., they are parallel.
(iii) The two lines are coincident lines.
1 3
= –1
2
When x = –1, we have : y =
Let the system of pair of linear equations be
a1x + b1y = c1
....(1)
Ex.2
When x = 1, we have : y =
GRAPHICAL REPRESENTATION OF
0
–1
–2
X
Y
1 2
Plotting points (1, –1) & (–1, –2) on graph
paper & joining them, we get straight line as
shown in fig. This line is required graph of
equation x – 2y = 3.
Y
X
–7 –6 –5 –4 –3 –2 –1
(–1, –2)
Y
9
8
7
6
5
4
3
2
1
X' 0
X
Y'
EXAMPLES
Ex.3
x – 2y = 3
(3, 0)
1 2 3 4 5 6 7
–1 (1, –1)
–2
–3
–4
–5
–6
–7
–8
–9
X
Sol.
The path of highway number 1 is given by the
equation x + y = 7 and the highway number 2
is given by the equation 5x + 2y = 20.
Represent these equations geometrically.
We have, x + y = 7
y=7–x
....(1)
In tabular form
x
1
4
y
6 3
Po int s A B and 5x + 2y = 20
y=
x – 7y = –42
20 5 x
2
....(2)
In tabular form
x
y
2
4
y
14
x 42
7
6
7
8
A
B
C
x
6
12
18
x6
3
0
2
4
D
E
F
Po int s
Plot the points A(0, 6), B(7, 7), C(14, 8) and
join them to get a straight line ABC. Similarly
plot the points D(6, 0), E(12, 2) and F(18, 4)
and join them to get a straight line DEF.
X
y
Y'
Sol.
7
x – 3y = 6
Y
Ex.4
0
Point s
y
5 0
Po int s C D
7 A
6
(1,6)
5
(2,5) C
4
(4,3) B
3
2
1
(4,0) D
X'
0 1 2 3 4 5 6 7
x
Plot the points A (1, 6), B(4, 3) and join them
to form a line AB.
Similarly, plot the points C(2, 5). D (4, 0) and
join them to get a line CD. Clearly, the two
lines intersect at the point C. Now, every
point on the line AB gives us a solution of
equation (1). Every point on CD gives us a
solution of equation (2).
A father tells his daughter, “ Seven years ago,
I was seven times as old as you were then.
Also, three years from now, I shall be three
times as old as you will be.” Represent this
situation algebraically and graphically.
Let the present age of father be x-years and
that of daughter = y years
Seven years ago father’s age
= (x – 7) years
Seven years ago daughter’s age
= (y – 7) years
According to the problem
(x – 7) = 7(y – 7)
or x – 7y = – 42
....(1)
After 3 years father’s age = (x + 3) years
After 3 years daughter’s age = (y + 3) years
According to the condition given in the
question
x + 3 = 3(y + 3)
or x – 3y = 6
....(2)
15
1 G
10
B C
A
(7,7)
(0,6)
5
2
(42, 12)
F
E
0 5D10 15 20 25 30 35 40 45
Ex.5
Sol.
x
10 students of class X took part in a
Mathematics quiz. If the number of girls is 4
more than the number of boys, find the
number of boys and girls who took part in the
quiz.
Let the number of boys be x and the number
of girls be y.
Then the equations formed are
x + y = 10
....(1)
and
y=x+4
....(2)
Let us draw the graphs of equations (1) and
(2) by finding two solutions for each of the
equations. The solutions of the equations are
given.
x y 10
x
0
y x4
0
1
3
y 10 x 10 2
y x4 4
5
7
Po int s
Po int s
A
8
B
x
C D E
Plotting these points we draw the lines AB
and CE passing through them to represent the
equations. The two lines AB and Ce intersect
at the point E (3, 7). So, x = 3 and y = 7 is the
required solution of the pair of linear
equations. The two lines AB and CD intersect at the
point (20, 16), So, x = 20 and y = 16 is the
required solution of the pair of linear
equations i.e. length of the garden is 20 m and
width of the garden is 16 m.
Verification :
Putting x = 20 and y = 16 in (1).
We get
20 = 16 + 4 = 20, (1) is verified.
Putting x = 20 and y = 16 in (2). we get
20 + 16 = 36
y
(0,10) A
E
(3, 7)
D
(1, 5)
C
(0, 4)
(8, 2) B
x
x'
Ex.6
Sol.
i.e. Number of boys = 3
Number of girls = 7.
Verification :
Putting x = 3 and y = 7 in (1), we get
L.H.S. = 3 + 7 = 10 = R.H.S.,
(1) is verified.
Putting x = 3 and y = 7 in (2), we get
7 = 3 + 4 = 7, (2) is verified.
Hence, both the equations are satisfied.
Half the perimeter of a garden, whose length
is 4 more than its width is 36m. Find the
dimensions of the garden.
Let the length of the garden be x and width of
the garden be y.
Then the equation formed are
x=y+4
....(1)
Half perimeter = 36
x + y = 36
....(2)
x y 4
Ex.7
Sol.
36 = 36, (2) is verified.
Hence, both the equations are satisfied.
Draw the graphs of the equations x – y + 1 = 0
and 3x + 2y – 12 = 0. Determine the
coordinates of the vertices of the triangle
formed by these lines and the x-axis, and
shade the triangular region.
Pair of linear equations are :
x–y+1
=0
....(1)
3x + 2y – 12 = 0
....(2)
In tabular form
x
0
4
y
4
0
Po int s
A
B
x
y x 1 1
5
A B
In tabular form
x
0
2
y
6
3
12 3x
2
Po int s
C D
10 20
Plot the points A(0, 1), B(4, 5) and join them to
get a line AB. Similarly, plot the points C(0, 6),
D(2, 3) and join them to form a line CD.
y 36 x 26 16
Po int s
4
Po int s
x y 36
x
0
C
D
Plotting these points we draw the lines AB
and CD passing through them to represent the
equations.
y
(0,6)
6
y
B
5
4
28
24
C
20
3
(10, 26)
D(20, 16)
16
12
8
4
x'
0
–4
x'
B
4 8
A
y'
x
12 16 20 24
D
(2, 3)
2
A (0, 1)
1
(–1, 0)
F 0
1
(4, 5)
E
2
3
4
(4, 0)
x
y'
Clearly, the two lines intersect each other at
the point D(2, 3). Hence x = 2 and y = 3 is the
solution of the given pair of equations. The line CD cuts the x-axis at the point
E (4, 0) and the line AB cuts the x-axis at the
point F(–1, 0).
Hence, the coordinates of the vertices of the
triangle are ; D(2, 3), E(4, 0), F(–1, 0).
Verification :
Both the equations (1) and (2) are satisfied by
x = 2 and y = 3. Hence, Verified.
TYPES OF SOLUTIONS
There are three types of solutions :
1. Unique solution.
2. Infinitely many solutions
3. No solution.
(A) Consistent :
If a system of simultaneous linear equations
has at least one solution then the system is
said to be consistent.
(i) Consistent equations with unique solution :
The graphs of two equations intersect at a
unique point. For example. Consider
x + 2y = 4
7x + 4y = 18
The graphs of the above equations coincide.
Coordinates of every point on the lines are the
solutions of the equations. Hence, the given
equations are consistent with infinitely many
solutions.
(B) Inconsistent Equation :
If a system of simultaneous linear equations has
no solution, then the system is said to be inconsistent.
No Solution : The graph (lines) of the two
equations are parallel.
For example. Consider
4x + 2y = 10
6x + 3y = 6
Y
5
4
3
2
1
0
5
S.No
3
1
2
3
2
(2, 1)
1
0
1
2
3 4
Sol.
Show graphically that the system of equations
x – 4y + 14 = 0 ;
3x + 2y – 14 = 0
is consistent with unique solution.
The given system of equations is
x – 4y + 14 = 0
....(1)
y=
x 14
4
When x = 6,
y=
x = – 2, y =
6 14
=5
4
2 14
=3
4
In tabular form
3
2
1
0
27
2
Ex.8
When
4
X
1
2
3 4
X
EXAMPLES
The graphs (lines) of these equations intersect
each other at the point (2, 1) i.e., x = 2, y = 1.
Hence, the equations are consistent with
unique solution.
(ii) Consistent equations with infinitely many
solutions : The graphs (lines) of the two
equations will be coincident.
For example. Consider
Y
3 4
Graph of Two
Types of Equations
Equations
Intersecting lines Consistent, with unique solution
Coincident
Consistent with infinite solutions
Parallel lines
Inconsistent (No solution)
X
2x + 4y = 9 3x + 6y =
2
The graphs (lines) of the given equations are
parallel. They will never meet at a point. So, there
is no solution. Hence, the equations are
inconsistent.
Y
4
1
x
6 2
y
5 3
Po int s A B 3x + 2y – 14
=0
....(2)
When x =
3x 14
2
y=
0 14
When x = 0, y =
=7
2
When x = 4, y =
1
,
2
In tabular form
1 11
2 2
y
3 1
Po int s C D
x
In tabular form
x
0
4
y
7 1
Po int s C D
Y
Y
A
8
(–2, 4)
C(0, 7)
6
4
X'
D(4, 1)
–2 0
2
4
6
X'
X
The given equations representing two lines,
intersect each other at a unique point (2, 4).
Hence, the equations are consistent with
unique solution.
Show graphically that the system of equations
15
y = 24
2
2x + 5y = 16 ; 3x +
Sol.
has infinitely many solutions.
The given system of equations is
2x + 5y = 16
y=
When x = 3,
(1/2, 3)
2
B(3, 2)
1
D
–3 –2 –1O
(11/2, 1)
1 2 3 4 5
3y = 10 – 2x
....(1)
y=
16 6
=2
5
When x = 2, y =
16 2 (2)
=4
5
X
x
2
3
y
Po int s
4
A
2
B
15
y = 24
2
10 2 2
10 4
=
=2
3
3
x
4
2
y
Po int s
6
A
2
B
4x + 6y = 12
....(1)
....(2)
6y = 12 – 4x
6y = 12 – 4x
y=
12 4 x
6
10 2 x
3
10 2 (4)
10 8
=
=6
3
3
In tabular form
In tabular form
48 6 x
15
y =
When x = – 4, y =
When x = – 2, y =
y=
C
The lines of two equations are coincident.
Coordinates of every point on this line are the
solution.
Hence, the given equations are consistent
with infinitely many solutions.
Ex.10 Show graphically that the system of equations
2x + 3y = 10, 4x + 6y = 12 has no solution.
Sol.
The given equations are
2x + 3y = 10
16 2 x
5
3x +
4
Y'
Y'
Ex.9
5
3
2
11
48 6
2
y=
=1
15
11
,
When x =
2
3 4 14
=1
2
48 3
=3
15
y= When x = – 3, y =
When x = 3, y =
12 4 (3) 12 12
=
=4
6
6
12 4 (3) 12 12
=
=0
6
6
In tabular form
x
3
3
y
Po int s
4
C
0
D
Plot the points A (–4, 6), B(2, 2) and join
them to form a line AB. Similarly, plot the
points
C(–3, 4), D(3, 0) and join them to get a
line CD.
y
Ex.12 Solve the following system of linear
equations graphically;
3x + y – 11 = 0 ; x – y – 1 = 0
Shade the region bounded by these lines and
also y-axis. Then, determine the areas of the
region bounded by these lines and y-axis.
Sol.
We have ;
3x + y – 11 = 0 and x – y – 1 = 0
(a) Graph of the equation 3x + y – 11 = 0
We have, 3x + y – 11 = 0
y = – 3x + 11
When, x = 2,
y = –3 × 2 + 11 = 5
When, x = 3,
y = – 3 × 3 + 11 = 2
Then, we have the following table :
x 2
y 5
A(–4, 6) 6
5
4
C
3
2
B(2, 2)
,
(3
x'
0)
1
D
–4 –3 –2 –1 0
–1
1 2
3 4 5
x
–2
y'
Clearly, the graphs of the given equations are
parallel lines. As they have no common point,
there is no common solution. Hence, the
given system of equations has no solution.
Ex.11 Given the linear equation 2x + 3y – 8 = 0,
write another linear equation in two variables
such that the geometrical representing of the
pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Sol.
We have,
2x + 3y – 8 = 0
(i) Another linear equation in two variables such
that the geometrical representation of the pair
so formed is intersecting lines is
3x – 2y – 8 = 0
(ii) Another parallel lines to above line is
4x + 6y – 22 = 0
(iii) Another coincident line to above line is
6x + 9y – 24 = 0
3
2
Plotting the points P (2, 5) and Q(3, 2) on
the graph paper and drawing a line joining
between them, we get the graph of the
equation 3x + y – 11 = 0 as shown in fig.
(b) Graph of the equation x – y – 1 = 0
We have,
x–y–1=0
y=x–1
When, x = – 1, y = –2
When, x = 3, y = 2
Then, we have the following table :
x
y
1
2
3
2
Plotting the points R(–1, –2) and S(3, 2) on
the same graph paper and drawing a line
joining between them, we get the graph of the
equation x – y – 1 = 0 as shown in fig.
Y
12
T(0,11)
10
8
Scale :
For both axes 2 unit = 0.5 cm 6
4
P(2, 5)
Q(3, 2)
S(3, 2)
V(0, 2) 2
X'
–8 –6 –4 –2 O
R(–1, 2) –2
2 4 6
U(0,–1)
–4
Y'
8
X You can observe that two lines intersect at
Q(3, 2). So, x = 3 and y = 2. The area enclosed
by the lines represented by the given equations
and also the y-axis is shaded.
So, the enclosed area = Area of the shaded
portion
1
× base × height
2
1
1
× (TU × VQ) =
× (TO + OU) × VQ
=
2
2
= Area of QUT =
=
It is a clear fact that y = 2 is for every value
of x. We may take the points T (3, 2), U(6, 2)
or any other values.
Then, we get the following table :
x 3 6
y 2 2
Plotting the points T(3, 2) and U(6, 2) on the
same graph paper and drawing a line joining
between them, we get the graph of the
equation y = 2 as shown in fig.
1
1
(11 + 1) 3 = × 12 × 3 = 18 sq.units.
2
2
Hence, required area is 18 sq. units.
Ex.13 Draw the graphs of the following equations ;
2x – 3y = – 6 ; 2x + 3y = 18; y = 2
Find the vertices of the triangles formed and
also find the area of the triangle.
Sol. (a) Graph of the equation 2x – 3y = – 6;
We have, 2x – 3y = – 6 y =
When, x = 0, y =
20 6
=2
3
When, x = 3, y =
23 6
=4
3
2x 6
3
x 0 3
y 2 4
Plotting the points P(0, 2) and Q(3, 4) on
the graph paper and drawing a line joining
between them we get the graph of the
equation 2x – 3y = – 6 as shown in fig.
(b) Graph of the equation 2x + 3y = 18;
When, x = 0,
y
When, x = – 3, y
=
2 x 18
3
2 0 18
=6
3
2 (3) 18
=
=8
3
Then, we have the following table :
x
y
0
6
2x
y=2
9
S(–3, 8)8
+3
y=
18
7
6
R(0, 6)
–3 –2 –1
3
8
Plotting the points R(0, 6) and S(–3, 8) on the
same graph paper and drawing a line joining
between them, we get the graph of the
equation 2x + 3y = 18 as shown in fig.
(c) Graph of the equation y = 2
y=
–6
U(6, 2)
T(3, 2)
1
X'
3
x–
2
Q(3, 4)
5
4
3
P(0, 2)
2
1
–1
Y'
Then, we have the following table :
We have 2x + 3y = 18 y =
Y
2
3
4
5
6
X
Scale :
For both axes 2 unit = 0.5 cm
From the fig., we can observe that the lines
taken in pairs intersect each other at points
Q(3, 4), U (6, 2) and P(0, 2). These form the
three vertices of the triangle PQU.
To find area of the triangle so formed
The triangle is so formed is PQU (see fig.)
In the PQU
QT (altitude) = 2 units
and
PU (base) = 6 units
so,
=
area of PQU =
1
(base × height)
2
1
1
(PU × QT) =
× 6 × 2 sq. untis
2
2
= 6 sq. units.
IMPORTANT POINTS TO BE REMEMBERED
Pair of lines
a1 x b1 y c1 0
a2x b 2y c2 0
2 x 3y 4 0
5x 6 y 9 0
x 2y 5 0
3x 6 y 15 0
2 x 3y 4 0
4 x 6 y 10 0
a1
a2
b1
b2
c1
c2
Compare
the ratio
2
5
1
3
2
4
3
6
2
6
3
6
4
9
5
15
4
10
a 1 b1
a2 b2
a 1 b 1 c1
a 2 b2 c2
a 1 b1 c1
a 2 b2 c2 Thus, the lines representing the pair of linear
equation coincide.
(iii) a1 = 6, b1 = – 3, c1 = 10;
Graphical
Algebraic
representation
interpretation
Intersecting
Exactly one
lines
solution (unique)
Coincident
Infinitely many
lines
solutions
Parallel lines
No solution
a2 = 2, b2 = – 1, c2 = 9
From the table above you can observe that if the
line a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
are
a1
b
1
a2 b2
a
b
c
(ii) for the coincident lines then 1 1 1
a 2 b2 c2
a 1 b 1 c1
(iii) for the parallel lines then
a 2 b2 c2
(i )
for the int er sec ting lines then
Ex.14 On comparing the ratios
Sol.
a1
b
c
, 1 and 1
a2
b2
c2
and without drawing them, find out whether
the lines representing the following pairs of
linear equations intersect at a point, are
parallel or coincide.
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0
Comparing the given equations with standard
forms of equations a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0 we have,
(i) a1 = 5, b1 = – 4, c1 = 8;
a2 = 7, b2 = 6, c2 = – 9
a1
5 b
4
= , 1 =
a2
7 b2
6
a1
b
1
a2
b2
Thus, the lines representing the pair of linear
equations are intersecting.
(ii) a1 = 9, b1 = 3, c1 = 12;
a2 = 18, b2 = 6, c2 = 24
a1
1 b
1
9
3
=
= , 1 =
=
2 b2
2
a2
18
6
and
12
1
c1
=
=
c2
24
2
a1
a2
=
b1
c
= 1
b2
c2
a1
6
3
b
c
10
= 3, 1 =
= 3, 1 =
=
9
a2
b2
c2
2
1
a1
b
c
= 1 1
a2
b2
c2
Thus, the lines representing the pair of linear
equations are parallel.
ALGEBRAIC SOLUTION OF A SYSTEM OF
LINEAR EQUATIONS
Sometimes, graphical method does not give an
accurate answer. While reading the coordinates of
a point on a graph paper, we are likely to make an
error. So, we require some precise method to
obtain accurate result. Algebraic methods given
below yield accurate answers.
(i) Method of elimination by substitution.
(ii) Method of elimination by equating the
coefficients.
(iii) Method of cross multiplication.
SUBSTITUTION METHOD
In this method, we first find the value of one
variable (y) in terms of another variable (x) from
one equation. Substitute this value of y in the
second equation. Second equation becomes a linear
equation in x only and it can be solved for x.
Putting the value of x in the first equation, we can
find the value of y.
This method of solving a system of linear
equations is known as the method of elimination
by substitution.
‘Elimination’, because we get rid of y or
‘eliminate’ y from the second equation.
‘Substitution’, because we ‘substitute’ the value
of y in the second equation.
Working rule :
Let the two equations be
a1x + b1y + c1 = 0
....(1)
a2x + b2y + c2 = 0
....(2)
Step I : Find the value of one variable, say y, in
terms of the other i.e., x from any equation, say (1).
Step II : Substitute the value of y obtained in
step 1 in the other equation i.e., equation (2). This
equation becomes equation in one variable x only. Step III : Solve the equation obtained in step II to
get the value of x.
Step IV : Substitute the value of x from step II to
the equation obtained in step I. From this
equation, we get the value of y. In this way, we
get the solution i.e. values of x and y.
Remark : Verification is a must to check the
answer.
EXAMPLES
Ex.15 Solve each of the following system of
equations by eliminating x (by substitution) :
(i) x + y = 7
(ii) x + y = 7
2x – 3y = 11
12x + 5y = 7
(iii) 2x – 7y = 1
(iv) 3x – 5y = 1
4x + 3y = 15
5x + 2y = 19
(v) 5x + 8y = 9
2x + 3y = 4
Sol. (i) We have ;
x+y=7
....(1)
2x – 3y = 11
....(2)
We shall eliminate x by substituting its value
from one equation into the other. from
equaton (1), we get ;
x+y=7
x = 7 – y
Substituting the value of x in equation (2), we
get ;
2 × (7 – y) – 3y = 11
14 – 2y – 3y = 11
–5y = – 3 or, y = 3/5.
Now, substituting the value of y in equation
(1), we get;
x + 3/5 = 7
x = 32/5.
Hence, x = 32/5 and y = 3/5.
(ii) We have,
x+y=7
12x + 5y = 7
From equation (1), we have;
x+y=7
x + 11 = 7 x = – 4
Hence, x = – 4,
y = 11.
(iii) We have;
2x – 7y = 1
4x + 3y = 15
From equation (1), we get
2x – 7y = 1
7y 1
2
Substituting the value of x in equation (2), we
get ;
4×
7y 1
+ 3y = 15
2
28y 4
+ 3y = 15
2
28y + 4 + 6y = 30
34y = 26
y =
26
13
=
34
17
Now, substituting the value of y in equation
(1), we get;
13
=1
17
2x – 7 ×
2x = 1 +
Hence, x =
91
108
108
54
=
x=
=
17
17
34
17
54
13
,y=
17
17
(iv) We have ;
3x – 5y = 1
.... (1)
5x + 2y = 19
.... (2)
From equation (1), we get;
3x – 5y = 1
....(1)
....(2)
x =
....(1)
....(2)
x =
5y 1
3
Substituing the value of x in equation (2), we
get ;
5×
5y 1
+ 2y = 19
3
x = 7 – y
Substituting the value of y in equation (2), we
get ;
25y + 5 + 6y = 57 31y = 52
12(7 – y) + 5y = 7
Now, substituting the value of y in equation
(1), we get ;
84 – 12y + 5y = 7
–7y = – 77
Now, Substituting the value of y in equation
(1), we get ;
y = 11
Thus, y =
52
31
3x – 5 ×
52
=1
31 3x –
x=
(ii) We have;
7x + 11y – 3 = 0
8x + y – 15 = 0
From equation (1), we get;
7x + 11y = 3
260
291
= 1 3x =
31
31
291
97
=
31 3
31
Hence, x =
97
52
, y=
31
31
y =
(v) We have,
5x + 8y = 9
2x + 3y = 4
From equation (1), we get;
5x + 8y = 9
x =
2×
8x +
9 8y
5
88x + 3 – 7x = 165
18 – 16y + 15y = 20
7x + 2 × (3x – 3) = 20
7x + 6x – 6 = 20
13x = 26
x=2
Now, substituting x = 2 in equation (1), we
get;
3×2–y=3
y = 3
Hence, x = 2,
y = 3.
x=2
Now, substituting, x = 2 in the equation (2),
we get ;
8 × 2 + y = 15
y = – 1
Hence, x = 2, y = – 1.
(iii) We have,
2x + y = 17
17x – 11y = 8
From equation (1), we get;
–y = 2 or y = – 2
Now, substituting the value of y in equation
(1), we get ;
5x + 8 (–2) = 9
3x – y = 3
y = 3x – 3
Substituting the value of ‘y’ in equation (2),
we get ;
3 7x
= 15
11
81x = 162
9 8y
+ 3y = 4
5
5x = 25
x = 5
Hence, x = 5, y = – 2.
Ex.16 Solve the following systems of equations by
eliminating ‘y’ (by substitution) :
(i) 3x – y = 3
(ii) 7x + 11y – 3 = 0
7x + 2y = 20
8x + y – 15 = 0
(iii) 2x + y – 17 = 0
17x – 11y – 8 = 0
Sol. (i) We have;
3x – y = 3
....(1)
7x + 2y = 20
....(2)
From equation (1), we get ;
3 7x
11
Substituting the value of ‘y’ in equation (2),
we get;
....(1)
....(2)
Substituting the value of x in equation (2), we
get ;
....(1)
.....(2)
....(1)
....(2)
2x + y = 17
y = 17 – 2x
Substituting the value of ‘y’ in equation (2),
we get ;
17x – 11 (17 – 2x) = 8
17x – 187 + 22x = 8
39x = 195
x = 5
Now, substituting the value of ‘x’ in equation
(1), we get ;
2 × 5 + y = 17
y = 7
Hence, x = 5, y = 7.
Ex.17 Solve the following systems of equations,
(i)
2
15
+
= 17
u
1
1
36
+
=
u
5
(ii)
11
7
=1
–
u
9
4
–
=6
u Sol.
(i) The given system of equation is ;
15
2
+
= 17
v
u
....(1)
1
1
36
+
=
u
v
5
....(2)
u =
Considering 1/u = x, 1/v = y, the above
system of linear equations can be written as :
15x + 2y = 17
....(3)
x+y=
36
5
....(4)
Multiplying (4) by 15 and (iii) by 1, we get ;
15x + 2y = 17
....(5)
15x + 15y =
36
× 15 = 108
5
....(6)
Subtracting (6) form (5), we get;
–13y = – 91
y = 7
Substituting y = 7 in (4), we get ;
36
x+7=
5
But, y =
1
=7
and,
x=
36
1
x =
–7=
5
5
1
7
=
1
1
= u = 5
u
5
– 19x = –38 x = 2
Substituting the above value of x in (2), we
get;
Hence, the required solution of the given
system of the equation is ;
=
1
= 2
y=
u=
1
3
a2 b2
ab
–
x
ab
ab
Substituting y =
....(3)
a2 b2
ab
–
x in (1), we
ab
ab
get
a 2 b2 a b
x = a2 + b2
a b
a b
(a + b) x +
1
2
(a b)(a 2 b 2 )
(a b ) 2
–
x
ab
(a b )
= a2 + b2
a 2 2ab b 2
(a + b) x –
ab
= a2 + b2 –
x
(a b)(a 2 b 2 )
ab
a 2 2ab b 2
(a + b) x –
ab
x
a b
= (a2 + b2) 1
a b
2
2
2
2
(a 2ab b ) x (a 2ab b ) x
ab
a ba b
ab
–4y = – 12
=
1
,
2
Ex.18 Solve the following system of equations by
the method of elimination (substitution).
(a + b) x + (a – b) y = a2 + b2
(a – b) x + (a + b) y = a2 + b2
Sol.
The given system of equations is
(a + b) x + (a – b) y = a2 + b2
....(1)
2
2
(a – b) x + (a + b) y = a + b
....(2)
From (2), we get (a + b) y = a2 + b2 – (a – b) x
= (a2 + b2)
y=3
But, x =
1
3
(a + b) x + (a – b)
Hence, the required solution of the given
system is u = 5, v = 1/7.
(ii) The given system of equation is ;
11
9
7
4
–
–
= 1;
=6
u
u
Taking 1/ = x and 1/u = y, the above system
of equations can be written as ;
11x – 7y = 1
....(1)
9x – 4y = 6
....(2)
Multiplying (1) by 4 and (2) by 7, we get,
44x – 28y = 4
....(3)
63x – 28y = 42
....(4)
Subtracting (4) from (3) we get,
9 × 2 – 4y = 6
1
=3
u
and, y =
(a 2 b 2 )2ab
4ab
x=
ab
ab 4abx = 2b (a2 + b2)
x =
2
a b
2a
Putting x =
y=
y=
2
Step I : Let the two equations obtained be
a1x + b1y + c1 = 0
....(1)
a2 b2
in (3), we get
2a
2
a b
ab
2
2
–
(a 2 b 2 )
ab
a 2 b2
=
ab
a2x + b2y + c2 = 0
2
(a b ) ( a b )
ab
2a
a b
1 2a
2a a b
2a
a 2 b2
y =
ab
a b 2a
y=
a2 b2
2a
Hence, the solution is
x=
a 2 b2
,
2a
y=
a 2 b2
2a
Ex.19 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence
find the value of ‘m’ for which y = mx + 3.
Sol.
We have,
2x + 3y = 11
....(1)
2x – 4y = – 24
....(2)
From (1), we have 2x = 11 – 3y
Substituting 2x = 11 – 3y in (2), we get
11 – 3y – 4y = –24
–7y = – 24 – 11
–7y = – 35
y=5
Putting y = 5 in (1), we get
2x + 3 × 5 = 11
2x = 11 – 15
x = –
4
=–2
2
Hence, x = – 2 and y = 5
Again putting x = – 2 and y = 5 in y = mx +
3, we get
5x = m(–2) + 3
–2m = 5 – 3
m=
METHOD OF ELIMINATION BY
EQUATING THE COEFFICIENTS
2
=–1
2
....(2)
Step II : Multiplying the given equation so as to
make the co-efficients of the variable to be
eliminated equal.
Step III : Add or subtract the equations so
obtained in Step II, as the terms having the same
co-efficients may be either of opposite or the
same sign.
Step IV : Solve the equations in one varibale so
obtained in Step III.
Step V : Substitute the value found in Step IV in
any one of the given equations and then copmpute
the value of the other variable.
Type I : Solving simultaneous linear
equations in two variables
EXAMPLES
Ex.20 Solve the following system of linear
equations by applying the method of
elimination by equating the co-efficients :
(i) 4x – 3y = 4
(ii) 5x – 6y = 8
2x + 4y = 3
3x + 2y = 6
Sol. (i) We have,
4x – 3y = 4
....(1)
2x + 4y = 3
....(2)
Let us decide to eliminate x from the given
equation. Here, the co-efficients of x are 4
and 2 respectively. We find the L.C.M. of 4
and 2 is 4. Then, make the co-efficients of x
equal to 4 in the two equations.
Multiplying equation (1) with 1 and equation
(2) with 2, we get ;
4x – 3y = 4
....(3)
4x + 8y = 6
....(4)
Subtracting equation (4) from (3), we get ;
–11y = –2
y =
2
11
Substituting y = 2/11 in equation (1), we get;
4x – 3 ×
2
=4
11 4x –
4x = 4 +
4x =
y
y
2x
2x
+ 1= 9
–
– =8
5
7
2
2
6
=4
11
The equation (1) can be expressed as :
6
11
5x 4 y
=8
10
50
11
Hence, solution of the given system of
equation is :
25
x=
,
22
2
y=
11
(ii) We have;
5x – 6y = 8
....(1)
3x + 2y = 6
....(2)
Let us eliminate y from the given system of
equations. The co-efficients of y in the given
equations are 6 and 2 respectively. The
L.C.M. of 6 and 2 is 6. We have to make the
both coefficients equal to 6. So, multiplying
both sides of equation (1) with 1 and equation
(2) with 3, we get ;
5x – 6y = 8
....(3)
9x + 6y = 18
....(4)
Adding equation (3) and (4), we get ;
x=
26
13
=
14
7
Putting x = 13/7 in equation (1), we get ;
5×
13
65
– 6y = 8
– 6y = 8
7
7
9
3
y=
=
42
14
Hence, the solution of the system of equations
3
13
, y=
14
7
Ex.21 Solve the following system of equations by
using the method of elimination by equating
the co-efficients.
y
x
2y
2x
+
+ 2 = 10;
–
+1=9
2
5
2
7
Sol.
The given system of equation is
x
2y
x
2y
+
+ 2 = 10
+
=8
2
5
2
5
4x 7 y
= 8 4x – 7y = 112
14
....(3)
...(1)
....(4)
Now the new system of equations is
5x + 4y = 80
....(5)
4x – 7y = 112
....(6)
Now multiplying equation (5) by 4 and
equation (6) by 5, we get ;
20x – 16y = 320
....(7)
20x + 35y = 560
....(8)
Subtracting equation (7) from (8), we get ;
y=
240
51
Putting y =
240
in equation (5), we get ;
51
240
= 80
51
5x + 4 ×
5x = 80 +
x=
5x –
960
= 80
51
960
4080 960
5040
=
=
51
51
51
5040
1008
336
336
=
=
x =
255
51
17
17
Hence, the solution of the system of equations
is, x =
65
65 56
9
6y =
–8=
=
7
7
7
is x =
5x + 4y = 80
Similarly, the equation (2) can be expressed
as :
50
25
x =
=
44
22
14x = 26
....(2)
336
80
,y=
.
17
17
Ex.22 Solve the following system of linear
equations by usnig the method of elimination
by equating the coefficients :
3x + 4y = 25 ;
5x – 6y = – 9
Sol.
The given system of equations is
3x + 4y = 25
....(1)
5x – 6y = – 9
....(2)
Let us eliminate y. The coefficients of y are 4
and – 6. The LCM of 4 and 6 is 12.
So, we make the coefficients of y as 12 and
– 12.
Multiplying equation (1) by 3 and equation
(2) by 2, we get
9x + 12y = 75
....(3)
10x – 12y = – 18
....(4) Adding equation (3) and equation (4), we get
19x = 57
x = 3.
Putting x = 3 in (1), we get,
3 × 3 + 4y = 25
4y = 25 – 9 = 16 y = 4
Hence, the solution is x = 3, y = 4.
Verification : Both the equations are satisfied
by x = 3 and y = 4, which shows that the
solution is correct.
Ex.23 Solve the following system of equations :
15x + 4y = 61; 4x + 15y = 72
Sol.
The given system of equation is
15x + 4y = 61
....(1)
4x + 15y = 72
....(2)
Let us eliminate y. The coefficients of y are 4
and 15. The L.C.M. of 4 and 15 is 60. So, we
make the coefficients of y as 60. Multiplying
(1) by 15 and (2) by 4, we get
225x + 60y = 915
....(3)
16x + 60y = 288
....(4)
Substracting (4) from (3), we get
209x = 627
x =
15 × 3 + 4y = 61 45 + 4y = 61
16
4y = 61 – 45 = 16 y =
=4
4
Hence, the solution is x = 3, y = 4.
Verification : On putting x = 3 and y = 4 in
the given equations, they are satisfied. Hence,
the solution is correct.
Ex.24 Solve the following system of linear
equatoins by using the method of elimination
by equating the coefficients
Sol.
2y=
3;
5x+ 3y=
2
The given equations are
3x–
2y=
3
....(1)
5x+
3y=
2
....(2)
Let us eliminate y. To make the coefficients
of equal, we multiply the equation (1) by 3
and equation (2) by
3x –
6y=3
10 x +
6y=2
x=
=
10 3
×
10 3 10 3
5
5
=
3 10
5( 10 3)
= 5(– 3)
10 9
Putting x = 5( 10 – 3) in (1) we get
3 × 5( 10 – 3) –
5 30 – 15 3 –
2y=
3
2y=
3
2 y = 5 30 – 15 3 –
3
2 y = 5 30 – 16 3
y=
5 30
2
–
16 3
2
= 5 15 – 8 6
Hence, the solution is x = 5 ( 10 – 3) and
y = 5 15 – 8 6 .
Verification : After verifying, we find the
solution is correct.
Ex.25 Solve for x and y :
627
=3
209
Putting x = 3 in (1), we get
3x–
3x + 10 x = 5 (3 + 10 ) x = 5
ax
by
–
= a + b ; ax – by = 2ab
b
a
Sol.
The given system of equations is
ax
by
–
=a+b
b
a
....(1)
ax – by = 2ab
Dividing (2) by a, we get
x–
....(2)
by
= 2b
a
....(3)
On subtracting (3) from (1), we get
ax
a
– x = a – b x 1 = a – b
b
b
x=
(a b ) b
= b x = b
ab
On substituting the value of x in (3), we get
b–
1–
2 to get
y
a
by
= 2b
a
b 1 = 2b
y
=2
a
y
= –1
a
....(3)
....(4)
Hence, the solution of the equations is
x = b, y = – a
Adding equation (3) and equation (4), we get
y
=1–2
a
y = –a Ex.26 Solve the following system of linear
equations :
2(ax – by) + (a + 4b) = 0
2(bx + ay) + (b – 4a) = 0
Sol.
2ax – 2by + a + 4b = 0 .... (1)
2bx + 2ay + b – 4a = 0 .... (2)
Multiplyng (1) by b and (2) by a and
subtracting, we get
Sol.
We have ;
1
1
+
=8
2y
x
....(2)
u+
1
2
v
=8
2
....(4)
Let us eliminate v from the system of
equations. So, multiplying equation (3) with
1/2 and (4) with 1, we get ;
Hence x = –1/2, and y = 2
Ex.27 Solve (a – b) x + (a + b) y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2
Sol.
The given system of equation is
(a – b) x + (a + b) y = a2 – 2ab – b2 ....(1)
(a + b) (x + y) = a2 + b2
....(2)
2
2
(a + b) x + (a + b) y = a + b
....(3)
Subtracting equation (3) from equation (1),
we get
1
u
v
–
=–
4
2
2
v
=8
u+
2
5u
15
=
4
2
u=
x=
Hence, the solution is x = a + b,
y=
2ab
ab
Type II : Solving a system of equations
which is reducible to a system of
simultaneous linear equations
EXAMPLES
Ex.28 Solve the following system of equations
1
1
1
1
= – 1;
=8
–
+
y
2y
x
2x
(a + b) y = – 2ab
2ab
ab
15
4
×
2
5
(a – b) (a + b) + (a + b) y = a2 – 2ab – b2
(a + b) y = a2 – 2ab – b2 – (a2 – b2)
y =
....(6)
1
u
+u=
+8
4
2
u=6
We know,
1
1
= u = 6
x
x
Putting the value of x in (1), we get
....(5)
Adding equation (5) and (6), we get ;
(a – b) x – (a + b) x = (a2 – 2ab– b2) – (a2 + b2)
–2bx = – 2ab – 2b2
2b 2
2ab
–
x=
=a+b
2b
2b
....(1)
Let us consider 1/x = u and 1/y = v.
Putting 1/x = u and 1/y = v in the above
equations, we get;
u
–v =–1
....(3)
2
2(b2 + a2) y = 4 (a2 + b2) y = 2
Multiplying (1) by a and (2) by b and adding,
we get
2(a2 + b2) x + a2 + b2 = 0
2(a2 + b2) x = – (a2 + b2) x = –
1
1
–
=–1
y
2x
1
x
Putting 1/x = 6 in equation (2), we get ;
1
1
6+
=8
=2
2y
2y
1
=4
y
y =
1
4
Hence, the solution of the system is,
x=
1
1
,y=
4
6
Ex.29 Solve,
1
2
2
1 3
+
+
= ;
=2
3y
3y
x
5 x
and also find ‘a’ for which y = ax – 2. Sol.
Considering 1/x = u and 1/y = v, the given
system of equations becomes
Ex.30 Solve,
2
6
+
=4
x 2y
2x y
v
1
2u +
=
3
5
5
1
+
=1
2( x 2 y)
3(2 x y)
6u v
1
=
3
5
30u + 5v = 3
2v
3u +
= 2 9u + 2v = 6
3
....(2)
Multiplying equation (1) with 2 and equation
(2) with 5, we get ;
60u + 10v = 6
....(3)
45u + 10v = 30
....(4)
Subtracting equation (4) from equation (3),
we get ;
15u = – 24
u =
24
8
=
15
5
Putting u =
where, x + 2y 0 and 2x – y 0
....(1)
8
in equation (2), we get;
5
8
+ 2v = 6
5
Sol.
Taking
1
1
= u and
= v, the above
x 2y
2x y
system of equations becomes
2u + 6v = 4
5u
v
+
=1
3
2
72
+ 2v = 6
5
2v = 6 +
72
102
=
5
5
....(2)
Multiplying equation (2) by 18, we have;
45u + 6v = 18
....(3)
Now, subtracting equation (3) from equation
(1), we get ;
–43u = – 14
u =
14
43
Putting u = 14/43 in equation (1), we get
2×
9×
....(1)
6v = 4 –
Now, u =
14
+ 6v
43
=4
28
172 28
144
=
v =
43
43
43
1
14
=
x 2y
43
14x + 28y = 43
....(4)
1
144
=
2x y
43
v=
102
51
=
25
5
And, v =
Here
1
8
= u = –
x
5
288x – 144y = 43
....(5)
Multiplying equation (4) by 288 and (5) by
14, the system of equations becomes
288 × 14x + 28y × 288 = 43 × 288
288x × 14 – 144y × 14 = 43 × 4
4022x + 8064y = 12384
....(6)
4022x – 2016y = 602
....(7)
Subtracting equation (7) from (6), we get;
10080y = 11782 y = 1.6(approx)
Now, putting 1.6 in (4), we get,
14x + 28 × 1.6 = 63
14x + 44.8 = 63
14x = 18.2
x = –
And,
5
8
1
51
5
y=
=v=
y
5
51
Putting x =
5
5
and y =
in y = ax – 2, we
8
51
get;
5
5a
=
–2
51
8
5a
5
102 5
107
=–2–
=
=
8
51
51
51
a=
107
8
856
856
× =
a =
51
5
255
255
x=
18.2
= 1.3 (approx)
14
Thus, solution of the given system of
equation is x = 1.3 (approx), y = 1.6 (approx). Ex.31 Solve,
We may use the following method to solve
the above type of equations.
1
2
+
=2
xy
xy
Steps :
2
1
–
=3
xy
xy
Sol.
where x + y 0 and x – y 0
1
1
Taking
= u and
= v the above
xy
xy
system of equations becomes
u + 2v = 2
....(1)
2u – v = 3
....(2)
Multiplying equation (1) by 2, and (2) by 1,
we get;
2u + 4v = 4
....(3)
2u – v = 3
....(4)
Subtracting equation (4) from (3), we get;
5v = 1 v =
1
2
8
= 2 u = 2 –
=
5
5
5
Here, u =
1
8
=
8x + 8y = 5 ....(5)
xy
5
And, v =
1
1
=
x–y=5
xy
5
....(6)
Multiplying equation (5) with 1, and (6) with
8, we get;
8x + 8y = 5
....(7)
8x – 8y = 40
....(8)
Adding equation (7) and (8), we get;
16x = 45
x=
45
16
Now, putting the above value of x in equation
(6), we get;
45
–y=5
16
y =
45
35
–5=
16
16
Hence, solution of the system of the given
equations is ;
x=
45
,
16
y=
x+y=
cd
ab
....(1)
(a – b) x – (a – b)y = c – d
x–y=
cd
ab
....(2)
Step III : We get the values of x and y after
adding or subtracting the equations (1) and (2).
EXAMPLES
1
5
Putting v = 1/5 in equation (1), we get;
u+2×
Step I : Let us write the equations in the form
ax + by = c
bx + ay = d
Step II : Adding and subtracting the above
type of two equations, we find :
(a + b) x + (a + b) y = c + d
35
16
Type-III : Equation of the form,
ax + by = c and bx + ay = d where a b.
Ex.32 Solve the following equations.
156x + 112y = 580; 112x + 156y = 492
Sol.
The given system of equation is
156x + 112y = 580
....(1)
112x + 156y = 492
....(2)
Adding equation (1) and (2) we get ;
268x + 268y = 1072
268(x + y) = 1072
x+y=4
....(3)
Subtracting equation (2) from equation (1),
we get
44x – 44y = 88
x–y=2
....(4)
Adding equation (3) with equation (4), we get;
2x = 6 x = 3
Putting x = 3 in equation (3), we get;
y=1
Thus, solution of the system of equations is
x = 3, y = 1
Ex.33 Solve the following system of equations.
43x + 35y = 207; 35x + 43y = 183
Sol.
The given system of equations is ;
43x + 35y = 207
....(1)
35x + 43y = 183
....(2)
Adding equation (1) and (2), we get;
78x + 78y = 390 78(x + y) = 390
x+y=5
....(3)
Subtracting equation (2) from the equation
(1), we get ;
8x – 8y = 24
x–y=3
....(4)
Adding equation (3) and (4), we get;
2x = 8 x = 4
Putting x = 4 in equation (3), we get;
4+y=5 y=1
Hence, the solution of the system of equation
is ; x = 4, y = 1.
Type IV : Equation of the form,
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
We may use the following method to solve
the above type of equations.
Steps :
Step I : Consider any one of the three given
equations.
Step II : Find the value of one of the
variable, say z, from it.
Step III : Substitute the value of z found in
Step II in the other two equations to get two
linear equatons in x, y.
Step IV : Taking the help of elimination
method, solve the equations in x, y obtained
in Step III.
Step V : Substitute the values of x, y found in
Step IV and Step II to get the value of z.
EXAMPLES
Ex.34 Solve the following system of equations.
x–z =5
y+z =3
x–y =2
Sol.
The given system of equations to ;
x–z =5
....(1)
y+z =3
....(2)
x–y =2
.....(3)
From equation (1), we have;
z=x–5
Putting z = x – 5 in equation (2), we get ;
y+x–5=3
x+y=8
....(4)
Adding equations (3) and (4), we get;
2x = 10
x = 5
Again putting x = 5 in equation (1), we get;
5–z=5
z = 0
Hence, the solution of the given system of
equation is x = 5, y = 3, z = 0.
Other method :
adding (1), (2) & (3)
2x = 10
x=5
now put this value in equation (1) & (3), we
get z = 0, y = 3 respectively
Ex.35 Solve,
x + 2y + z = 12
2x – z = 4
x – 2y = 4
Sol.
We have,
x + 2y + z = 12
....(1)
2x – z = 4
....(2)
x – 2y = 4
....(3)
From equation (1), we have z = 12 – x – 2y.
Putting, z = 12 – x – 2y in the equation (2),
we get;
2x – (12 – x – 2y) = 4
2x – 12 + x + 2y = 4
3x + 2y = 16
....(4)
Adding equations (3) and (4), we get;
4x = 20
x=5
Putting the value of x = 5 in equation (2), we
get
2×5–z=4
z = 10 – 4 = 6
Again putting the value of x = 5 in equation
(3), we get
5 – 2y = 4 y = 1/2
Hence, the solution of the given system of
equations is ;
x = 5, y = 1/2 , z = 6 CROSS-MULTIPLICATION METHOD
By the method of elimination by substitution, only
those equations can be solved, which have unique
solution. But the method of cross multiplication
discussed below is applicable in all the cases;
whether the system has a unique solution, no
solution or infinitely many solutions.
Let us solve the following system of equations
a1x + b1y + c1 = 0
....(1)
a2x + b2y + c2 = 0
....(2)
Multiplying equation (1) by b2 and equation (2)
by b1, we get
a1b2x + b1b2y + b2c1
=0
....(3)
a2b1x + b1b2y + b1c2
=0
....(4)
Subtracting equation (4) from equation (3), we get
(a1b2 – a2b1) x + (b2c1 – b1c2) = 0
x =
b1c 2 b2c1
a1b 2 a 2b1
1
x
=–
2
2
and
y
x
1
=
=
b1c 2 b 2c1
a 1c 2 a 2 c1
a1b 2 a 2b1
EXAMPLES
Ex.36 Solve the following system of equations by
cross-multiplication method.
2x + 3y + 8 = 0
4x + 5y + 14 = 0
Sol.
The given system of equations is
2x + 3y + 8 = 0
4x + 5y + 14= 0
By cross-multiplication, we get
–y
1
8 = 2 3
14
4 5
3
5
8 = 2
14
4
y
x
1
=
=
3 14 5 8 2 14 4 8
25 43
y
x
1
=
=
42 40
28 32
10 12
y
1
= – y = – 2.
4
2
Hence, the solution is x = – 1, y = – 2
We can verify the solution.
Ex.37 Solve the follownig system of equations by
the method of cross-multiplication.
2x – 6y + 10 = 0
3x – 7y + 13 = 0
Sol.
The given system of equations is
2x – 6y + 10 = 0
....(1)
3x – 7y + 13 = 0
....(2)
By cross-multiplication, we have
–6
–7
c1a 2 c 2a1
a1b 2 a 2b1
These values of x and y can also be written as
x
x
y
1
=
=
2
4
2
x=–1
a1b 2 a 2 b1 0
a1 b1
a 2 b2
Similarly, y =
x
–y
1
10 = 2 –6
13
3 –7
x
y
=
6 13 (7) 10
2 13 3 10
=
10 = 2
3
13
1
2 (7) 3 (6)
y
x
1
=
=
– 78 70
26 30
14 18
y
x
1
=
=
8
4
4
x
1
=
8
4
x=–2
y
1
=
4
4
y=1
Hence, the solution is x = – 2, y = 1
Ex.38 Solve the following system of equations by
the method of cross-multiplication.
11x + 15y = – 23; 7x – 2y = 20
Sol.
The given system of equations is
11x + 15y + 23 = 0
7x – 2y – 20 = 0
Now, by cross-multiplication method, we
have x
15
–2
23 = 11
7
–20
–y
23 = 11
7
–20
15
–2
y
x
=
15 (20) ( 2) 23
11 (20) 7 23
=
1
1
11 (2) 7 15
y
x
1
=
=
300 46
220 161
22 105
y
x
1
=
=
254
381
127
x
1
=
x=2
254
127
Ex.40 Solve the following system of equations by
cross-multiplication method.
x + y = a – b; ax – by = a2 + b2
Sol.
The given system of equations can be
rewritten as :
x + y – (a – b) = 0
ax – by – (a2 + b2) = 0
By cross-multiplication method, we have
Hence, x = 2, y = – 3 is the required solution.
Ex.39 Solve the following system of equations by
cross-multiplication method.
ax + by = a – b; bx – ay = a + b
Sol.
Rewriting the given system of equations, we
get
ax + by – (a – b) = 0
bx – ay – (a + b) = 0
By cross-multiplication method, we have
b
–a
x
–(a – b) = a
–(a + b) b
–y
1
b
–(a – b) = a
–(a + b) b –a
and
2
2
(a b )
x
(a 2 b 2 )
y
2
2
(a b )
=
=
=
y
2
2
(a b )
1
(a 2 b 2 )
x
(a 2 b2 ) b(a b)
2
2
(a b )
1
(a b 2 )
y=–1
Hence, the solution is x = 1, y = –1.
2
=
y
2
a b 2 a 2 ab
1
(a b )
y
x
1
=
=
a (a b )
b (a b )
( a b)
x
1
=
x=a
a (a b )
( a b)
and
y
1
=
y=–b
b (a b )
( a b)
y
x
=a+b
+
a
b
x
a
Sol.
2
+
y
b2
=2
The given system of equations is rewritten as :
x=1
2
2
=
2
1
2
a b ab b
(a b )
=
1
y
=
(
b
a)
(a b ) a(a b)
2
x
2
1
2
y
1
=
(a b2) a {(a b)} b a
Hence, the solution is x = a, y = – b.
Ex.41 Solve the following system of equations by
the method of cross-multiplication :
y
x
=
ab b2 a 2 ab
a 2 ab ab b 2
x
=
x
b {(a b)} (a ) {(a b)}
=
y
1
=
=
a (a b) b(a b)
a2 b2
x
(a b2 ) (b) {(a b)}
=
y
1
=
y=–3
and
381
127
1
–y
–(a – b) = 1
1
–(a2 + b2) a –b
x
–(a – b) = 1
–(a2 + b2) a
1
–b
y
x
– (a + b) = 0
+
a
b
x
a
2
+
y
b2
–2=0
....(1)
....(2)
Multiplying equation (1) by ab, we get
bx + ay – ab (a + b) = 0
....(3) Multiplying equation (2) by a2 b2, we get
b2x + a2y – 2a2b2 = 0
....(4)
By cross multiplication method, we have
x
–ab(a + b) = b
b2
–2a2b2
a
a2
=
x
a ba b
x
=
3
a b (a b )
y
4
2 3
ab a b
y
3
ab (a b)
=
=
1
ab(a b)
1
ab(a b)
1
=
3
ab(a b)
a b (a b )
ab 3 (a b)
=
(a b )
bx + ay =
bx + ay –
–1
x
2ab
2
a b2
a b2
a
a 2 b2
3
2ab a ab
=0
b
a 2 b2
x
2
=
2a 2 b
=
2
x
2
2
a (a b )
x
2
2
a (a b )
....(2)
1
a 2 b2
y
2a b a 2 b b 3
2
a 2 b2
=
=
1
a2 b2
1
y
= 2
2
b( b a )
a b2
a 2 b2
2
1
2
a b
and
2
x =
a
2
a b2
1
y
= 2
b( b 2 a 2 )
a b2
a 2 b2
y=
b
2
a b2
y=
a 2 b2
2ab
y
=
2ab 2
Hence, the solution is x =
a 2 2ab b 2 a 2 b 2
2
a2 b2
a2 b2
2
–1
a 2 b2
The given system of equations can be written
as.
ax + by – 1 = 0
....(1)
b
a
y
1
=
a a b b
2ab
a 2
b (1)
a b2
a2 b2
1
b
=
ax + by = 1; bx + ay =
bx + ay =
–
=a
a2 b2
ab (a b)
= b2
ab(a b)
a 2 b2
a
b
–y
–1
2ab
x
2ab
b 2
a (1)
a b2
1
ab(a b)
(a b ) 2
=
a2 b2
3
bx + ay =
–
a 3 b (a b )
x=
= a2
ab(a b)
y
a b2
x
–1
2ab
=
1
ab(a b)
Hence, the solution x = a2, y = b2
Ex.42 Solve the following system of equations by
cross-multiplication method -
Sol.
x
And
=
3 2
b
a
1
a2b ab2
x
–y
=
4
3 2
2 3
2a b a b a b
2a b a 2b3 ab4
=0
2
Now, by cross-multiplication method, we
have
x
y
=
2a3b2 a 3b(a b)
2a 2b3 ab3 (a b)
3 2
2ab
bx + ay –
–y
1
a
–ab(a + b) = b
2
b2 a
–2a2b2
=
Rewritting the equations (1) and (2), we have
ax + by – 1 = 0
a
a2 b2
,
b
2
a b2
Ex.43 Solve the following system of equations in x
and y by cross-multiplication method
(a – b) x + (a + b) y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2 Sol.
The given system of equations can be
rewritten as :
(a – b) x + (a +b) y – (a2 – 2ab – b2) = 0
(a + b) x + (a + b) y – (a2 + b2) = 0
By cross-multiplication method, we have
=
=
(a – b)
(a + b)
(a + b)
(a + b)
x
(a b) {(a 2 b 2 )} (a b) {(a 2 2ab b 2 )}
y
2
ax + bx + ay – by – (a2 – ab + b2) = 0
(a + b) x + (a – b) y – (a2 – ab + b2) = 0 ....(1)
And ax – bx + ay + by – (a2 + ab + b2) = 0
(a – b) x + (a + b) y – (a2 + ab + b2) = 0 ...(2)
Now, by cross-multiplication method, we
have
2
1
( a b ) ( a b) ( a b ) ( a b )
(a b)(a b ) (a b)(a 2 2ab b 2 )
=
2
2
2
2
y
(a b)(a b) (a b) 2
(a b)[(a 2 b 2 ) (a b)(a 2 2ab b 2 )]
=
y
2
(a b)(a 2ab b 2 ) (a b)(a 2 b 2 )
1
(a b)(a b a b)
=
=
(a – b)
(a + b)
1
=
2
2
(a b) {(a ab b )} (a b) {(a 2 ab b 2 )}
2
=
3
a a b 3ab b a 3 ab 2 a 2 b b 3
x
1
=
2(a b)(a b)b
2 b( a b )
x=a+b
1
(a b) (a b) (a b)(a b)
x
(a b)(a 2 ab b 2 ) (a b)(a 2 ab b 2 )
=
1
=
(a b)(2b)
(a – b)
(a + b)
(a + b)
(a – b)
y
y
x
1
=
=
(a b)(2a 2b)b 4ab 2 2b(a b)
1
2
2
y
3
–(a2 – ab + b2)
–(a2 + ab + b2)
x
(a b) {(a ab b )} (a b) {(a 2 ab b 2 )}
2
x
(a b)(2ab 2b 2 )
–y
x
=
–(a2 – ab + b2) (a + b)
2
2
–(a +ab + b ) (a – b)
x
=
2
(a b)(a b ) (a b)(a 2ab b )
=
2ab
.
ab
Ex.44 Solve the following system of equations by
cross-multiplications method.
a(x + y) + b (x – y) = a2 – ab + b2
a(x + y) – b (x – y) = a2 + ab + b2
Sol.
The given system of equations can be
rewritten as
x
2ab
ab
x = a + b, y = –
(a b) {(a b )} (a b) {(a 2 2ab b 2 )}
=
4ab
1
2 b( a b )
=
2
Hence, the solution of the given system of
equations is
–(a2 – 2ab – b2)
–(a2 + b2)
1
y
y=–
–y
x
=
–(a2 – 2ab – b2) (a – b)
2
2
–(a + b )
(a + b)
(a + b)
(a + b)
and
y
2
2
(a b)(a ab b ) (a b)(a 2 ab b 2 )
=
=
1
2
(a b ) (a b ) 2
x
3
3
(a b ) (a 3 b 2 )
y
a 3 2a 2 b 2ab 2 b 3 a 3 2a 2 b 2ab 2 b 3 =
x
2b
=
3
x
2b
=
3
x
=
2b 3
And
1
2
2
a 2ab b a 2ab b
y
2
4a b 2 b
=
3
y
2
2
2b(2a b )
2
1
4ab
1
4ab
=
=
2b( 2a 2 b 2 )
....(2)
–v
= 1
0
2
2
–(a + b )
b2
= 1
b2
u
=
u
2
a b
2
=
–1
a2
v
v
2
1
(a 2 b 2 ) b 2 0
=
v = 1
....(2)
b
c
a1
= 1 = 1
b2
c2
a2
It shows that lines represented by equation
(1) and (2) are coincident.
(c) It is inconsistent, if
a
b
= u and
= v in equatinos (1)
y
x
a2 b2 a2 0
a b2
It shows that lines represented by equations
(1) and (2) are not parallel.
(b) It is consistent with infinitely many
solutions, if
and (2) the system of equations reduces to
u–v+0=0
b2u + a2v – (a2 + b2) = 0
By the method of cross-multiplication, we
have
1
2
b
a1
1
a2
b2
....(1)
a 2b
ab 2
– (a2 + b2) = 0
+
y
x
(a b )
=
u=1
(a) It is consistent with unique solution, if
where x 0, y 0
The given system of equations is
b
a
=0
–
y
x
0
–(a2 + b2)
2
a b2
a2x + b2y + c2 = 0
a
ab 2
b
a 2b
–
+
=0;
= a2 + b2 ;
y
x
x
y
–1
a2
v
2
1
2
The system of equations is given by
....(1)
a1x + b1y + c1 = 0
b2
2a 2 b 2
,y=
2a
2a
u
=
y=b
Hence, the solution of the given system of
equations is x = a, y = b.
1
4ab
2a 2 b 2
2a
Putting
a b
2
a
= 1 x = a
x
b
v=
=1
y
Ex.45 Solve the following system of equations by
the method of cross-multiplication.
Sol.
u
2
and u =
Hence, the solution is
x=
and
1
b2
x=
4ab
2a
y
y=
2
2
(a b )
=
1
2
a (b 2 )
b
c
a1
= 1 1
a2
b2
c2
It shows that lines represented by equation
(1) and (2) are parallel and non–coincident.
EXAMPLES
Ex.46 Show that the following system of equations
has unique solution
2x – 3y = 6; x + y = 1.
Sol.
The given system of equation can be written as
2x – 3y – 6 = 0
x+y–1=0
The given equations are of the form
a1x + b1y + c1 = 0
1
2
a b2
a2x + b2y + c2
=0
Where, a1 = 2, b1 = – 3, c1 = – 6
and
a2 = 1,
b2 = 1, c2 = – 1 a1
b
2
3
=
= 2, 1 =
=3
a2
b2
1
1
c1
6
=
=6
1
c2
a1
b
1
a2
b2
Clearly,
So, the given system of equations has a
unique solution. i.e., it is consistent.
Ex.47 Show that the following system of equations
has unique solution :
x – 2y = 2 ; 4x – 2y = 5
Sol.
The given system of equations can be written as
x – 2y – 2 = 0
4x – 2y – 5 = 0
The given equations are of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Where, a1= 1,
and
b1 = –2, c1 = –2
a2 = 4, b2 = –2, c2 = –5
a1
2
c
2
2
1 b
= 1, 1 =
=
= , 1 =
4 b2
2
a2
5
5
c2
a1
b
1
a2
b2
Clearly,
So, the given system of equations has a
unique solution i.e. It is consistent.
Ex.48 For what value of k the following system of
equations has a unique solution :
x – ky = 2 ; 3x + 2y = –5
Sol.
The given system of equation can be written
as
x – ky – 2 = 0
3x + 2y + 5 = 0
The given system of equations is of the form
a1x + b1y + c1 = 0
a2x + b2y + c2
where, a1 = 1, b1
and
a2 = 3,
=0
= –k, c1 = –2
b2 = 2,
c2 = 5
Clearly, for unique solution
k
1
2
3
k
a1
b
1
a2
b2
2
3
Ex.49 Show that the following system has infinitely
many solutions.
Sol.
x = 3y + 3 ; 9y = 3x – 9
The given system of equations can be written
as
x – 3y – 3 = 0
3x – 9y – 9 = 0
The given equations are of the form
a1x + b1y + c1 = 0
a2x + b2y + c2
=0
Where, a1 = 1, b1 = –3, c1 = –3
and
a2 = 3,
b2– = –9, c2 = –9
a1
1 b
3
1 c
3
1
= , 1 =
= , 1 =
=
a2
3 b2
9
3 c2
9
3
Clearly,
a1
b
c
= 1 = 1 so the given system
a2
b2
c2
of equations has infinitely many solutions.
Ex.50 Show that the following system has infinitely
many solutions :
2y = 4x – 6 ; 2x = y + 3
Sol.
The given system of equations can be written
as
4x – 2y – 6 = 0
2x – y – 3 = 0
The given equations are of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Where, a1 = 4, b1 = –2, c1 = –6
and
a2 = 2, b2 = –1, c2 = –3
a1
b
c
4
2
6
=
=2
= 2, 1 =
= 2, 1 =
a2
2
1
b2
c2
3
Clearly,
a1
a2
=
b1
b2
=
c1
, so the given
c2
system of equations has infinitely many
solutions.
Ex.51 Find the value of k for which the following
system of equations has infinitely many
solutions.
(k – 1) x + 3y = 7; (k + 1) x + 6y = (5k – 1)
Sol.
The given system of equations can be written
as
(k – 1)x + 3y – 7 = 0
(k + 1) x + 6y – (5k – 1) = 0
Here a1 = (k – 1), b1 = 3, c1 = –7
and a2 = (k + 1), b2 = 6, c2 = – (5k – 1) 2a – 3a = –3b – 2b and 9a – 7a = 7b – 3b + 6
–a = –5b and 2a = 4b + 6
For the system of equations to have infinite
number of solutions.
a1
b
c
= 1 = 1
a2
b2
c2
k 1
7
3
=
=
(5k 1)
k 1
6
k 1
1
7
=
=
k 1
2
5k 1
Taking I and II
k 1
1
=
k 1
2
2k – 2 = k + 1 k = 3
Taking II and III
1
7
=
5k – 1 = 14
2
5k 1
5k = 15 k = 3
Hence, k = 3.
Ex.52 For what values of a and b, the following
system of equations have an infinite number
of solutions:
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
Sol.
The given system of linear equations can be
written as
2x + 3y – 7 = 0
(a – b) x + (a + b) y – (3a + b – 2) = 0
The above system of equations is of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0,
where a1 = 2, b1 = 3, c1 = –7
a2 = (a – b), b2 = (a + b), c2 = – (3a + b – 2)
For the given system of equations to have an
infinite number of solutions
a1
b
c
= 1 = 1
a2
b2
c2
Here,
a1
b
2
3
=
, 1 =
and
a2
a b b2
ab
7
c1
7
=
=
c2
(3a b 2)
3a b 2
2
3
7
=
=
ab
ab
3a b 2
2
3
3
7
=
and
=
ab
ab
ab
3a b 2
2a + 2b = 3a – 3b and 9a +3b – 6 = 7a + 7b
a = 5b .... (3) and a = 2b + 3
Solving (3) and (4) we get
.... (4)
5b = 2b + 3 b = 1
Substituting b = 1 in (3), we get a = 5 × 1 = 5
Thus, a = 5 and b = 1
Hence, the given system of equations has
infinite number of solutions when
a = 5, b = 1
Ex.53 Show that the following system of equations
is inconsistent.
2x + 7y = 11; 5x +
Sol.
35
y = 25
2
The given system of equations can be written
as
2x + 7y – 11 = 0
5x +
35
y – 25 = 0
2
The given equations are of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Where, a1 = 2,
b1 = 7, c1 = –11
and
b2 =
a2 = 5,
35
, c = – 25
2 2
a1
2 b
7
2 c
11
11
= , 1 =
= , 1 =
=
35
5 b2
a2
5 c2
25
25
2
Clearly,
a1
b
c
= 1 1
a2
b2
c2
So, the given system of equations has no
solution, i.e. it is inconsistent. Proved.
Ex.54 Show that the following system of equations
has no solution :
2x + 4y = 10 ; 3x + 6y = 12
Sol.
The given system of equations can be written
as
2x + 4y – 10 = 0
3x + 6y – 12 = 0
The given equations are of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Where a1 = 2, b1 = 4, c1 = – 10
and
a2 = 3, b2 = 6, c2 = – 12 2x + 3y – 5 = 0
kx – 6y– 8 = 0
a1 = 2, b1 = 3, c1= 5,
a1
2 b
4
2 c
10
5
= , 1 =
= , 1 =
=
a2
3 b2
6
3 c2
6
12
Clearly,
Here,
a1
b
c
= 1 1
a2
b2
c2
a2 = k, b2 = –6, c2 = –8
So, the given system of equations has no
solution, i.e., it is inconsistent. Proved.
Ex.55 For what values of k will the following
system of liner equations has no solution.
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Sol.
The given system of equations may be written
as
3x + y – 1 = 0
(2k – 1) x + (k – 1) y – (2k + 1) = 0
The above system of equations is of the form
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
where a1 = 3,
b1 = 1, c1 = –1
and a2 = (2k – 1), b2 = (k – 1), c2 = –(2k+1)
a1
3
b
1
c
1
=
, 1 =
, 1 =
a2
(2k 1)
2k 1 b 2
k 1 c2
Clearly, for no solution
a1
b
c
= 1 1
a2
b2
c2
3
1
=
2k 1
k 1
3k – 3 = 2k – 1
k=2
1
1
and
( 2k 1)
k 1
and
1
b1
3
=
=
2
b2
6
a1
b
1
a2
b2
Here
1
2
k
2
k – 4
Thus the given system of equations have a
unique solution for all real values of k except –4.
(ii) The given system of equations may be written
as
2x + ky – 1 = 0
5x – 7y – 5 = 0
Here, a1 = 2, b1 = k, c1 = –1,
a2 = 5, b2 = –7, c2 = –5
a1
b
k
2
k
=
and 1 =
=
7
a2
5
b2
7
We have
a1
b
k
2
1
a2
7
b2
5
Here
2k + 1 k – 1
k–2
So,
3
1
2k 1
2k 1
6k + 3 2k – 1
a1
2
=
a2
k
It satisfies the condition that the system of
given solutions has a unique solution.
and
As the given equations have unique solution,
we get,
4k – 4 k – 1
Hence the given system of linear equations
has no solution, when
k = 2 and k –2 and k – 1.
Ex.56 Determine the value of k for each of the
following given system of equations having
unique/consistent solution.
(i) 2x + 3y – 5 = 0; kx – 6y = 8
(ii) 2x + ky = 1; 5x – 7y – 5 = 0
Sol. (i) The given system of equations may be written
as
k
2
7
5
k
14
5
Thus, the given system of equations has a
unique solution for all real values of k except
14
.
5
Ex.57 Determine the value of k for each of the
following given system of equations having
unique/consistent solution.
(i) x – ky – 2 = 0; 3x + 2y + 5 = 0
(ii) 2x – 3y – 1 = 0; kx + 5y – 7 = 0
Sol. (i) We have,
x – ky – 2 = 0 3x + 2y + 5 = 0
a1 = 1, b1 = –k, c1 = –2,
Here,
and
a2 = 3, b2 = 2, c2 = 5
Since, the given system of equations has a
unique solution, we have
a1
b
1
a2
b2
or
k
1
2
3
k
Thus, the given system of equations has a
2
3
(ii) We have
2x – 3y – 1 = 0
kx + 5y – 7 = 0
Here, a1 = 2, b1 = –3, c1 = –1,
a2 = k, b2 = 5, c2 = –7
Since, the given system of equations has a
unique solution, we get
2
3
1
=
=
k 1
k2
3
2
3
=
k 1
k2
or
3
1
=
k2
3
2k + 4 = 3k – 3 or k + 2 = 9
3k – 2k = 4 + 3 or k = 7
2
3
solution for all values of k except
a1
b
c
= 1 = 1
a1
b2
c2
k=7
or k = 7
k=7
It shows that the given system of equations
has infinitely many solutions at k = 7
(ii) We have
2x + 3y – 2 = 0
(k + 2) x + (2k + 1) y – 2 (k – 1) = 0
Here, a1 = 2, b1 = 3, c1 = –2,
a2 = k + 2, b2 = 2k + 1, c2 = –2 (k – 1)
Since, the given system of equations has
infinitely many solutions, we get
a1
b
3
c
2
2
, 1 =
=
, 1 =
a2
3k 1 c 2
2(k 1)
k 2 b2
a1
b
1
a2
b2
and
a1
b
c
= 1 = 1
a1
b2
c2
2
3
k
5
2
3
1
=
=
k2
2k 1
k 1
2
3
=
k2
2k 1
or
4k + 2 = 3k + 6
4k – 3k = 6 – 2
or
or
k
10
3
Thus, the given system of equation has a
10
unique solution for all value of k except
.
3
Ex.58 Find the value of k for each of the following
systems of equations having infinitely many
solutions.
(i) 2x + 3y = k; (k – 1) x + (k + 2) y = 3k
(ii) 2x + 3y = 2; (k + 2) x + (2k + 1) y = 2 (k – 1)
Sol. (i) We have
2x + 3y – k = 0
(k – 1) x + (k + 2) y – 3k = 0
Here a1 = 2, b1 = 3, c1 = –k,
a2 = k – 1, b2 = k + 2, c2 = –3k
Since, the given system of equations has
infinitely many solutions, we get
a1
2
b
3
c
k
=
, 1 =
, 1 =
a2
k 1 b2
k 2 c2
3k
3
1
=
2k 1
k 1
3k – 3 = 2k + 1
3k – 2k = 1 + 3
k=4
or k = 4
k=4
Ex.59 Determine the values of k for the following
system of equations having no solution.
x + 2y = 0; 2x + ky = 5
Sol.
The given system of equations may be written
as
x + 2y = 0
2x + ky – 5 = 0
Here, a1 = 1, b1 = 2, c1 = 0,
a2 = 2, b2 = k, c2 = –5
As the given system of equations has no
solution, we get a1
b
c
1
2
0
=
= 1 =
1 =
a2
k
5
b2
c2
2
We must write
a1
b
c
= 1 1
a2
b2
c2
2
2a 5 1
2b 1
9
3
2
1
2a 5 1
and
2b 1 3
9
3
2b + 1 = 6 or 6a + 15 = 9
1
2
=
k
2
b =
k=4
Here, for this value of k, we get
a1
b
c
= 1 1
a2
b2 c2
Ex.60 Find the value of k of the following system of
equations having infinitely many solutions.
2x – 3y = 7; (k + 2) x – (2k + 1) y = 3(2k –1)
Sol.
A given system of equations has infinitely
many solutions, if
a1
b
c
= 1 = 1
a1
b2
c2
Thus, the given system of equations has
infinitely many solutions at a = –1, b =
2
3
7
k 2 ( 2k 1) 3(2k 1)
2
3
=
or
k2
2k 1
4k + 2 = 3k + 6
3
7
=
2k 1
6k 3
or 18k – 9 = 14k + 7
k = 4 or k = 4
k=4
Thus, the given system of equations has
infinitely many solutions at k = 4.
Ex.61 Determine the values of a and b so that the
following given system of linear equations
has infinitely many solutions.
2x – (2a + 5) y = 5 ; (2b + 1) x – 9y = 15
Sol.
We have
2x – (2a + 5) y – 5 = 0
(2b + 1) x – 9y – 15 = 0
Hence, a1 = 2, b1 = –(2a + 5), c1 = –5,
a2 = 2b + 1, b2 = –9, c2 = – 15 :
The given system of equations has infinitely
many solutions, if
a1 b1 c1
a 2 b2 c2
such that
(2a 5)
2
5
=
=
2b 1
9
15
5
.
2
Ex.62 Find the value of c if the following system of
equation has no solution.
cx + 3y = 3; 12x + cy = 5
Sol.
We have
cx + 3y –3 = 0
12x + cy – 6 = 0
The given system of equations has no
solution, if
a1
b
c
c 3
3
= 1 1 such that
a2
b2
c2
12 c
6
So, we get
5
and a = –1
2
3
1
c 3
and
12 c
c
2
c2 = 36
c=±6
Thus, the given system of equation has no
solution at c = ± 6.
Ex.63 For what value of p, the system of equations
will have no solution ?
px – (p –3) = –3y; py = p – 12x
Sol.
The given system of equations may be written
as
px + 3y – (p – 3) = 0
12x + py – p = 0
Here, a1 = p, b1 = 3, c1 = –(p – 3),
a2 = 12, b2 = p, c2 = –p
The given system of equations will have no
solution, if
a1 b1 c1
a1 b2 c2
For it we get,
p 3
3 p3
and
12 p
p
p
p2 = 36 p = ± 6
When p = 6,
p3 63 1
3 3 1
and
p 6 2
p
6
2 So,
1
3
p3
= . Thus, p = 6 does not
=
p
p
2
satisfy the equation
When p = –6,
3 p3
p
p
3
3
1
p 6 2
and
p 3 6 3 9 3
p
6
6 2
So,
3 p3
p
p
Thus, p = –6 satisfy the equation
3 p3
.
p
p
Thus, the given system of equations will have
no solution, if p = – 6
Ex.64 Find the value of k for the following system
of equations has no solution.
(3k + 1) x + 3y = 2 ; (k2 + 1) x – 5 = – (k – 2)y
Sol.
The given system of equations may be written
as
(3k + 1) x + 3y – 2 = 0
(k2 + 1) x + (k – 2) y – 5 = 0
Here, a1 = 3k + 1, b1 = 3, c1 = –2,
a2 = k2 + 1, b2 = k – 2, c2 = – 5
Since the given system of equations has no
solution therefore, we can write ;
a1
b
c
= 1 1
a2
b2
c2
So,
3k 1
2
k 1
3k 1
2
k 1
3k 1
k 2 1
3
k 2
=
3
3
and
k2
k2
=
3
k 2
3k2 – 6k + k – 2 = 3k2 + 3
– 5k = 5 k = – 1
3
2
Putting k = – 1 in the equation
,
k2 5
we get
a2 = a + b, b2 = 2 (a – b), c2 = – (5a – 1)
Since, the given system of equations has
infinite number of solutions therefore, we get
a1
b
c
= 1 = 1
a1
b2
c2
3
4
12
=
=
ab
2(a b)
(5a 1)
3
4
4
12
and
a b 2( a b )
2(a b) (5a 1)
6a – 6b = 4a + 4b and 20a – 4 = 24a – 24b
6a – 4a – 6b – 4b = 0 and
20a – 24a + 24b = 4
2a – 10b = 0 and
24b – 4a = 4
a – 5b = 0 and 6b – a = 1
Adding the above two equations, we get
– 5b + 6b = 1
b = 1
Putting b = 1 in the equation 6b – a = 1, we
get
6×1–a=1 6–a=1a=5
Thus, the given system of equations has
infinitely many solutions at a = 5, b = 1.
HOMOGENEOUS EQUATIONS
The system of equations
a1x + b1y = 0
a2x + b2y = 0
called homogeneous equations has only
3
= –1
1 2
Thus, k = –1 satisfy
Thus, the given system of equation has no
solution at k = –1
Ex.65 Determine the values of a and b so that the
following system of equations has infinite
number of solutions.
3x + 4y – 12 = 0
2(a – b) y – (5a – 1) = – (a + b) x
Sol.
The given system of equations may be written
as
3x + 4y – 12 = 0
(a + b) x + 2 (a – b) y – (5a – 1) = 0
Here, a1 = 3, b1 = 4, c1 = – 12,
solution x = 0, y = 0, when
3
2
k2 5
(i) when
a 1 b
,
a2 b
a1
b
a2 b The system of equations has only one
solution, and the system is consistent.
(ii) When
a 1 b1
a 2 b2
3500 + 6y = 3800
The system of equations has infinitely
many solutions and the system is
consistent.
Ex.66 Find the value of k for which the system of
equations
4x + 5y = 0; kx + 10y = 0
has infinitely many solutions.
Sol.
The given system is of the form
a1x + b1y = 0
a2x + b2y = 0
a1 = 4, b1 = 5 and a2 = k, b2 = 10
If
a1
b
= 1 , the system has infinitely many
a2
b2
solutions.
5
4
=
k
10
k=8
WORD PROBLEMS ON SIMULTANEOUS
LINEAR EQUATION
Problems Based on Articles
EXAMPLES
Ex.67 The coach of a cricket team buys 7 bats and 6
balls for j 3800. Later, he buys 3 bats and 5
balls for j 1750. Find the cost of each bat and
each ball.
Sol.
Let the cost of one bat be j x and cost of one
ball be j y. Then
7x + 6y = 3800
....(1)
3x + 5y = 1750
....(2)
From (1) y =
3800 7 x
6
Putting y =
3800 7 x
in (2), we get
6
3800 7 x
= 1750
6
3x + 5
Multiplying (3) by 6, we get
18x + 5(3800 – 7x) = 10500
18x + 19000 – 35x = 10500
–17x = 10500 – 19000
–17x = –8500
x = 500
Putting x = 500 in (1), we get
7(500) + 6y = 3800
....(3)
6y = 3800 – 3500
6y = 300
y = 50
Hence, the cost of one bat = j 500
and the cost of one ball = j 50
Ex.68 Meena went to a bank to withdraw j 2000.
She asked the cashier to give j 50 and
j 100 notes only. Meena got 25 notes in all.
Find how many notes of j 50 and j 100 she
received ?
Sol.
Let the number of notes of j 50 be x,
and the number of notes of j 100 be y,
Then according to the question,
x + y = 25
....(1)
50x + 100y = 2000
....(2)
Multiplying (1) by 50, we get
50x + 50y = 1250
....(3)
Subtracting (3) from (2), we have
50y = 750 y = 15
Putting y = 15 in (1), we get
x + 15 = 25 x = 25 – 15 = 10
Hence, the number of notes of j 50 was 10
and that of j 100 was 15.
Ex.69 Yash scored 40 marks in a test, receiving 3
marks for each right answer and losing 1
mark for each wrong answer. Had 4 marks
been awarded for each correct answer and 2
marks been deducted for each incorrect
answer, then Yash would have scored 50
marks. How many questions were there in the
test ?
Sol.
Let the number of correct answers of Yash be
x and number of wrong answers be y. Then
according to question :
Case I. He gets 40 marks if 3 marks are given
for correct answer and 1 mark is deducted for
incorrect answers.
3x – y = 40
....(1)
Case II. He gets 50 marks if 4 marks are
given for correct answer and 2 marks are
deducted for incorrect answers.
4x – 2y = 50
....(2)
Multiplying (1) by 2, we get
6x – 2y = 80
....(3) Subtracting (2) from (3), we get
2x = 30
x =
30
= 15
2
Putting x = 15 in (1); we get
3 × 15 – y = 40
45 – y = 40 y = 5
Total number of questions = number of
correct answers + number of incorrect
answers.
= 15 + 5 = 20
Problems Based on Numbers
Ex.70 What number must be added to each of the
numbers, 5, 9, 17, 27 to make the numbers in
proportion ?
Sol.
Four numbers are in proportion if
First × Fourth = Second × Third.
Let x be added to each of the given numbers
to make the numbers in proportion. Then,
(5 + x) (27 + x) = (9 + x) (17 +x)
135 + 32x + x2 = 153 + 26x + x2
32x – 26x = 153 – 135
6x = 18 x = 3
Ex.71 The average score of boys in an examination
of a school is 71 and that of girls is 73. The
average score of the school in the
examination is 71.8. Find the ratio of the
number of boys to the number of girls that
appeared in the examination.
Sol.
Let the number of boys = x
Average score of boys = 71
Total score of boys = 71x
Let the number of girls = y
Average score of girls = 73
Total score of girls = 73 y
According to the question,
Average score
=
Total average
Total number of students
71.8 =
71x 73y
xy
71.8x + 71.8y = 71x + 73y 0.8x = 1.2y
3
x
1.2
=
=
y
2
0.8
Hence, the ratio of the number of boys to the
number of girls = 3 : 2.
Ex.72 The difference between two numbers is 26 and
one number is three times the other. Find them.
Sol.
Let the numbers be x and y.
Difference of two numbers is 26.
i.e.,
x – y = 26
....(1)
One number is three times the other.
i.e.,
x = 3y
....(2)
Putting x = 3y in (1), we get
3y – y = 26
2y = 26 y = 13
Putting y = 13 in (2), we get
x = 3 × 13 = 39
Hence, the numbers are x = 39 and y = 13.
Problems Based on Ages
Ex.73 Father’s age is three times the sum of ages of
his two children. After 5 years his age will be
twice the sum of ages of two children. Find
the age of father.
Sol.
Let the age of father = x years.
And the sum of the ages of his two children
= y years
According to the question
Father’s age = 3 × (sum of the ages of his two
children)
x = 3y
....(1)
After 5 years
Father’s age = (x + 5) years
sum of the ages of his two childrens
= y + 5 + 5 = y + 10
[Age of his each children increases by 5 years]
According to the question,
After 5 years
Father’s age = 2 × (sum of ages of his two
children)
x + 5 = 2 × (y + 10)
x + 5 = 2y + 20
x – 2y = 15
....(2)
Putting x = 3y from (1) in (2), we get
3y – 2y = 15
y = 15 years
And x = 3y x = 3 × 15 = 45
x = 45 years.
Hence, father’s age = 45 years
Ex.74 Five years hence, the age of Jacob will be
three times that of his son. Five years ago, Sol.
Jacob’s age was seven times that of his son.
What are their present ages.
Let the present age of Jacob and his son be x
and y respectively.
Case I. After five years age of Jacob = (x + 5),
After five years the age of his son = (y + 5).
According to question
x + 5 = 3(y + 5)
x=9–y=9–3=6
Hence, the number is x + 10y = 6 + 10 × 3 = 36.
Ex.76 The sum of a two-digit number and the
number obtained y reversing the order of its
digits is 165. If the digits differ by 3, find the
number.
Sol.
Let unit’s place digit = x
And ten’s place digit = y
x – 3y = 10
....(1)
Case II. Five years ago Jacob’s age = x – 5,
and his son’s age = y – 5. Then, according to
question,
x – 5 = 7 (y – 5)
Original number = x + 10y
The number obtained by reversing the digits
= 10 x + y
According to first condition.
The original number + Reversed number = 165
x = 7y – 30
....(2)
Putting x = 7y – 30 from (2) in (1), we get
7y – 30 – 3y = 10
x + 10y + 10x + y = 165
11x + 11y = 165
4y = 40 y = 10
Putting y = 10 in (1), we get
x – 3 × 10 = 10
x = 10 + 30 x = 40
Hence, age of Jacob is 40 years, and age of
his son is 10 years.
Problems Based on two digit numbers
Ex.75 The sum of a two digit number and the
number obtained by reversing the order of its
digits is 99. If the digits differ by 3, find the
number.
Sol.
Let the unit’s place digit be x and the ten’s
place digit be y.
Original number = x + 10 y
The number obtained by reversing the
digits = 10x + y
According to the question,
Original number + Reversed number = 99
(x + 10y) + (10x + y) = 99
11x + 11y = 99
x+y=9
x=9–y
....(1)
Given the difference of the digit = 3
x–y=3
....(2)
On putting the value of x = 9 – y from
equation (1) in equation (2), we get
(9 – y) – y = 3 9 – 2y = 3
2y = 6 y = 3
Substituting the the value of y = 3 in equation
(1), we get
x+y=
165
= 15
11
x = 15 – y
According to second condition.
The difference of the digits = 3
....(1)
x–y=3
....(2)
Substituting x = 15 – y from equation (1) in
equation (2), we get
(15 – y) – y = 3
15 – 2y = 3
2y = 12
y = 6
Putting y = 6 in equation (1), we have
x = 15 – 6 x = 9
Hence, the original number = x + 10y
= 9 + 10 × 6 = 69
Ex.77 The sum of the digits of a two-digit number is
9. Also, nine times this number is twice the
number obtained by reversing the order of the
number. Find the number.
Sol.
Let the ten’s and the unit’s digits in the
number be x and y, respectively. So, the
number may be written as 10x + y.
When the digits are reversed, x becomes the
unit’s digit and y becomes the ten’s digit.
The number can be written as 10y + x.
According to the given condition,
x+y=9
....(1)
We are also given that nine times the number
i.e., 9(10x + y) is twice the numbers obtained
by reversing the order of the number i.e.
2(10y + x). 9(10x + y) = 2 (10y + x)
90x + 9y = 20y + 2x
90x – 2x + 9y – 20y = 0
88x – 11y = 0
Sol.
8x – y = 0 ....(2)
Adding (1) and (2), we get
9x = 9
x=1
Putting x = 1 in (1), we get
y=9–1=8
Thus, the number is
10 × 1 + 8 = 10 + 8 = 18
Problems Based on Fraction
Ex.78 The sum of the numerator and denominator of
a fraction is 4 more than twice the numerator.
If the numerator and denominator are
increased by 3, they are in the ratio 2 : 3.
Determine the fraction.
Sol.
Let Numerator = x and Denominator = y
Fraction =
Then, fraction =
x + y = 2y – 3
2y – y = 3 + x
y=3+x
According to the second condition,
Decreased numerator by 1 =
denominator)
(x – 1) =
1
(y – 1)
2
2x – y = 1
....(2)
Substituting y = 3 + x in equation (2), we
have
2x – (3 + x) = 1
....(1)
2x – x = 1 + 3
x = 4
On putting x = 4 in equation (1), we get
y=3+4
x 3
2
=
y3
3
y=7
3x + 9 = 2y + 6
3x – 2y + 3 = 0
....(2)
Substituting the value of y form equation (1)
into equation (2), we get
3x – 2(x + 4) + 3 = 0
3x – 2x – 8 + 3 = 0
x=5
On putting x = 5 in equation (1), we get
y = 5 + 4
y=9
Hence, the fraction =
1
(decreased
2
2x – y = –1 + 2
Increased numerator by 3
2
=
Increased deno min ator by 3
3
....(1)
2(x – 1) = y – 1
x
y
x + y = 2x + 4
y=x+4
According to the second condition,
x
y
According to the first condition,
Numerator + denominator = twice of the
denominator – 3
According to the first condition,
Numerator + denominator = 2 × numerator + 4
denominator. If the numerator and
denominator are decreased by 1, the
numerator becomes half the denominator.
Determine the fraction.
Let Numerator = x and Denominator = y,
Hence, the fraction =
Ex.80 A fraction becomes
Ex.79 The sum of the numerator and denominator of
a fraction is 3 less than twice the
9
, if 2 is added to both
11
the numerator and the denominator. If 3 is
added to both the numerator and the
denominator it becomes
Sol.
5
. Find the fraction.
6
Let the numerator be x and denominator be y.
Then, according to the question,
Case 1 :
x
5
=
y
9
x
4
=
y
7
x2
9
=
y2
11
11(x + 2) = 9(y + 2)
11x + 22
= 9y + 18
11x – 9y = – 4
....(1) x3
5
=
y3
6
Case 2 :
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x – 5y = – 3
x=
....(2)
5y 3
6
Putting x =
5y 3
– 9y = – 4
6
11
....(3)
Hence, the required fraction is
7
.
9
4
if 1 is added to each of
5
the numerator and denominator. However, if
we subtract 5 from each, the fraction becomes
1
. Find the fraction.
2
Let the required fraction be
x
where x be the
y
numerator and y be the denominator.
First Case :
According to the question,
x 1
4
=
y 1
5
5x + 5 = 4y + 4
5x – 4y = – 1
Second Case : 5 is subtracted from x and y
So,
x
7
=
y
9
Hence, the required fraction is
Multiplying (3) by 6, we get
11(5y – 3) – 54y = –24
55y – 33 – 54y = – 24
y = 33 – 24 = 9
Putting y = 9 in (1), we get
11x – 9 × 9 = – 4
11x = – 4 + 81 = 77 x = 7
Sol.
y=9
So,
5y 3
in (1), we get
6
Ex.81 A fraction becomes
Subtracting (4) from (3), we get
–3x = – 21
x=7
Substituting the value of x in (2) we get
2×7–y=5
x 5
1
=
y5
2
2x – 10 = y – 5
2x – y = 5
....(2)
Multiplying equation (2) by 4 and equation
(1) by 1, we get
5x – 4y = – 1
....(3)
8x – 4y = 20
....(4)
7
.
9
Problem on Fixed Charges & Running Charges
Ex.82 A Taxi charges consist of fixed charges and
the remaining depending upon the distance
travelled in kilometers. If a persons travels 10
km, he pays j 68 and for travelling 15 km,
he pays j 98. Express the above statements
with the help of simultaneous equations and
hence, find the fixed charges and the rate per
km.
Sol.
Let fixed charges of taxi = j x.
And running charges of taxi = j y per km.
According to the question,
Expenses of travelling 10 km = j 68.
x + 10y = 68
....(1)
Again expenses of travelling 15 km = j 98.
x + 15y = 98
....(2)
Subtracting equation (1) from equation (2),
we get
5y = 30
y=6
On putting y = 6 in equation (1), we have
x + 10 × 6 = 68
x = 68 – 60
x = 8
Hence, fixed charges of taxi = x = j 8 and
running charges per km = y = j 6.
Ex.83 A lending library has a fixed charge for the
first three days and an addition charge for
each day thereafter. Sarika paid j 27 for a
book kept for seven days. While Susy paid
j 21 for the book the kept for five days. Find
the fixed charge and the charge for each extra
day.
Sol.
Let fixed charge be j x.
and the charge for each extra day be j y.
According to the question Case I. Sarika paid j 27 for 7 days i.e. 4
extra days.
x + 4y = 27
....(1)
Susy paid j 21 for 5 days i.e. 2 extra days
x + 2y = 21
Subtracting (2) from (1), we get
2y = 6
....(2)
y=3
Putting y = 3 in (1), we get
x + 4 × 3 = 27
x = 27 – 12 = 15
Hence, the fixed charge is j 15 and the
charge for each extra days is j 3.
Ex.84 The taxi charges in a city consist of a fixed
charge together with the charge for the
distance covered. For a distance of 10 km, the
charge paid is j 105 and for a journey of 15
km, the charge paid is 155. What are the fixed
charges and the charges per kilometer ? How
much does a person have to pay for travelling
a distance of 25 km ?
Sol.
Let fixed charges of taxi = j x
And running charges of taxi = j y per km.
According to the question,
Express of travelling 10 km = j 105
x + 10y = 105
....(1)
Again expenses of travelling 15 km = j 155
x + 15y = 155
x = 155 – 15y
Putting x = 155 – 15y in (1), we get
155 – 15y + 10y = 105
155 – 5y
....(2)
= 105
–5y
= 105 – 155
–5y = – 50
y = 10
Putting y = 10 in (2), we get
x + 15 × 10 = 155
x + 150 = 155
x = 155 – 150 = 5
Hence, fixed charges of taxi = x = j 5 and
running charges per km = y = j 10 A person
should pay for travelling 25 km = 5 + 25 × 10
= 5 + 250 = j 255
Problems Based on Speed & Time
Ex.85 Places A and B are 100 km apart on the
highway. One car stars from A and another
from B at the same time. If the cars travel in
Sol.
the same direction at a different speed, they
meet in 5 hours. If they travel towards each
other, they meet in 1 hour. What are the speed
of the two cars ?
Let the speed of the first car, starting from
A = x km/hr.
And the speed of second car, starting from
B = y km/hr.
Distance travelled by first car in 5 hours
= AC = 5x
Distance travelled by second car in 5 hours
= BC = 5y
According to the question,
Let they meet at C, when moving in the same
direction.
AC = AB + BC
5x = 100 + 5y
x = 20 + y
....(1)
5x km
B
A
100 km
C
5y km
Distance = Speed × Time
When moving in the opposite direction, let
they meet at D
Distance travelled by first car in 1 hour = AD = x.
Distance travelled by second car in 1 hour
= BD = y
A
x km
D
y km
B
100 km
AD + BD = AB
x + y = 100
....(2)
Substituting x = 20 + y from equation (1) in
equation (2), we have
(20 + y) + y = 100
20 + 2y = 100
2y = 100 – 20 = 80
y = 40 km/hour
On putting y = 40 in equation (1), we get
x = 20 + 40 = 60 km/hour
Hence, the speed of first car = 60 km/hour
and the speed of the second car = 40 km/hour.
Ex.86 Two places A and B are 120 km apart from
each other on a highway. One car starts from
A and another from B at the same time. If
they move in the same direction, they meet in
6 hours and if they move in opposite Sol.
directions, they meet in 1 hour and 12
minutes. Find the speed of the cars.
Let the speed of car starting from
A = x km/hr.
6x
B
A
C
6y
120 km
And the speed of car starting from B = y km/h.
While moving in the same-direction let they
meet at C.
Distance travelled by first car in 6 hours
= AC = 6x.
Distance travelled by second car in 6 hours
= BC = 6y.
According to the first condition.
AC = AB + BC
6x = 120 + 6y
2y = 80
y = 40 km/hour
Putting y = 40 in equation (1), we have
x = 20 + 40 = 60 km/hour
Hence, the speed of first car = 60 km/hour.
And the speed of second car = 40 km/hour.
Ex.87 A plane left 30 minutes later than the
scheduled time and in order to reach the
destination 1500 km away in time, it has to
increase the speed by 250 km/hr from the
usual speed. Find its usual speed.
Sol.
Let the usual speed of plane = x km/hr.
The increased speed of the plane = y km/hr.
(distance = Speed × Time)
x = 20 + y
According to the second condition,
Distance travelled by first car in
= AD =
6
hours
5
6
x
5
Distance travelled by second car in
= BD =
....(1)
6
hours
5
1500
1500
1
=
–
y
x
2
6
y
5
While moving in the opposite direction let
they meet at D.
AD + DB = AB
A
6
x
5
6
y
5
D
B
[1 hour 12 minutes =
....(2)
Dis tan ce
Time
Speed
1500x 375000 1500x
1
=
x ( x 250)
2
x(x + 250) = 750000
x2 + 250x – 750000 = 0
6
6
x + y = 120
5
5
x + y = 120 ×
1500
1
1500
–
=
x 250
x
2
120 km
y = (x + 250) km/hour.
....(1)
Distance = 1500 km.
According to the question,
(Scheduled time) – (time in increasing the
speed) = 30 minutes.
6
hours]
5
5
6
x + y = 100
....(2)
Substituting x = 20 + y from equation (1) in
equation (2), we get
(20 + y) + y = 100
x2 + 1000x – 750x – 750000 = 0
(x – 750) (x + 1000) = 0
x = 750 or x = – 1000
But speed can never be – ve
Hence, Usual speed = 750 km/hr.
Problems Based on Boat & Stream
Ex.88 A boat goes 16 km upstream and 24 km
downstream in 6 hours. It can go 12 km
upstream and 36 km downstream in the same Sol.
time. Find the speed of the boat in still water
and the speed of the stream.
Let the speed of stream = y km/hr ;
speed of boat in still water = x km/hr.
And the speed of boat in upstream = (x – y)
km/hr.
The speed of boat in downstream = (x + y)
km/hr.
16p = 6 – 2 = 4
p = 1/4
1
1
1
1
=
and
=
xy
xy
4
12
x–y=4
And, x + y = 12
By adding 2x = 16
....(7)
....(8)
x=8
Putting x = 8 in equation (7), we get
8 – y = 4
According to the question,
Time taken in going 16 km upstream + time
taken in going 24 km downstream = 6 hours.
16
24
+
=6
....(1)
xy
xy
Dis tan ce
Time
Speed
Again, according to the question,
Time taken in going 12 km upstream + time
taken in going 36 km downstream = 6 hours.
12
36
+
=6
xy
xy
Let
1
1
= p,
=q
xy
xy
1
6
=
12
72
Putting the value of q in equation (3), we get
16p + 24
16p + 2 = 6
1
= 6
12
y=8–4=4
Hence, speed of boat in still water = 8 km/hr.
and speed of stream = 4 km/hr.
Ex.89 A boat goes 30 km upstream and 44 km
downstream in 10 hours. In 13 hours, it can
go 40 km up stream and 55 km down stream.
Determine the speed of the stream and that of
the boat.
Sol.
Let the speed of the boat in still water be
x km/hr and speed of the stream be y km/hr.
Then the speed of the boat downstream =
(x + y) km/hr, and the speed of the boat upstream
= (x – y) km/hr. Also time = distance/speed.
In the first case, when the boat goes 30 km
upstream, let the time taken be t1. Then
t1 =
30
( x y)
....(2)
Equation (1) becomes 16p + 24q = 6 ....(3)
Equation (2) becomes 12p + 36q = 6 ....(4)
Multiplying equation (3) by 3 and equation
(4) by 4, we get
48p + 72q = 18
....(5)
48p + 144q = 24
....(6)
Subtracting equation (5) from equation (6),
we get
72q = 6 q =
Let t2 be the time taken by the boat to go 44
km downstream. Then t2 =
44
. The total
( x y)
time taken, t1 + t2, is 10 hours Therefore, we
get the equation
30
44
+
= 10
( x y)
( x y)
....(1) In the second case in 13 hours it can go 40
km upstream and 55 km downstream. We get
the equation
40
55
+
= 13
xy
xy
1
1
Let
= u and
=v
( x y)
( x y)
We know that speed =
time =
....(2)
....(3)
u=
x + y = 12
For the second case, we get
1=
1
11
1
5
On putting these values of u and v in equation
(3), we get
1
1
1
1
= and
=
( x y)
( x y)
11
5
i.e.,
(x – y) = 5 and (x + y) = 11
Adding these equations, we get
i.e.,
2x = 16 i.e.,
x=8
Subtracting the equations, we get
2y = 6 i.e., y = 3
Hence, the speed of the boat in still water is
8 km/hr and the speed of the stream is
3 km/hr.
Ex.90 A sailor goes 8km downstream in 40 minutes
and returns in 1 hour. Determine the speed of
the sailor in still water and speed of the
current.
Sol.
We know that
40 minutes =
40
2
hr =
hr
60
3
Let the speed of the sailor in still water be
x km./hr and the speed of the current be
y km/hr.
dis tan c
time
speed
2x + 2y = 24
....(1)
dis tan ce
time
speed
8
xy
x–y=8
....(2)
Adding equations (1) and (2), we get
2x = 20
x = 10 km/hr
Substituting x = 10 in equation (1), we get
10 + y = 12
30u = 6
dis tan ce
speed
8
2
=
xy
3
Substituting the value of v in equation (4), we
get
30u + 4 = 10
time
speed of upstream = (x – y) km/hr
and speed of downstream = (x + y) km/hr
For the first case, we get
On substituting these values in equations (1)
and (2), we get the linear pair
30u + 44v = 10
....(4)
40u + 55v = 13
....(5)
Multiplying equation (3) by 4 and equation
(5) by 3, we get
120u + 176 v = 40
120u + 165v = 39
On subtracting the two equations, we get
11v = 1, i.e., v =
dis tan ce
y = 2 km/hr
Hence, speed of the sailor in still water and
speed of the current are 10 km/hr and 2km/hr
respectively.
Ex.91 A person rows downstream 20 km in 2 hours
and upstream 4 km in 2 hours. Find man’s
speed of rowing in still water and the speed of
the current.
Sol.
Let man’s speed of rowing in still water and
the speed of the current be x km/hr and y
km/hr respectively.
Then, the upstream speed = (x – y) km/hr
and the downstream speed = (x + y) km/hr
we know that
time =
dis tan ce
speed
First case :
2=
20
xy
2(x + y) = 20
x + y = 10
2nd Case :
....(1) 2=
4
2 (x – y) = 4
xy
x–y=2
Adding (1) and (2), we get
....(2)
2x = 12
x = 6
Substituting the value of x in equation (1), we
get
6 + y = 10 y = 4
Hence, man’s speed of rowing in still water
and the speed of the current are 6 km/hr and
4 km/hr respectively.
Problems Based on Area
Ex.92 If in a rectangle, the length is increased and
breadth reduced each by 2 metres, the area is
reduced by 28 sq. metres. If the length is
reduced by 1 metre and breadth increased by
2 metres, the area increases by 33 sq. metres.
Find the length and breadth of the rectangle.
Sol.
Let length of the rectangle = x metres
And breadth of the rectangle = y metres
Area = length × breadth = xy sq. metres
Case 1: As per the question
x – 12
xy – 12
y–2
y+3
Reduced length = x – 5
Increased breadth = y + 3
Reduced area = (x – 5) (y + 3)
Reduction in area = 9
Original area – Reduced area = 9
xy – [(x – 5) (y + 3)] = 9
xy –[xy + 3x – 5y – 15] = 9
xy – xy + 2x – 2y + 4 = 28
2x – 2y = 28 – 4 = 24
....(1)
x–1
2x – y = 33 + 2 = 35
2x – y = 35
....(2)
Subtracting equation (1) from equation (2),
we get
x = 23
Substituting the value of x in equation (1), we
get
2x – y = 12
y = 23 – 12 = 11
Length = 23 metres.
Breadth = 11 metres.
Ex.93 The area of a rectangle gets reduced by 9
square units if its length is reduced by 5 units
and breadth is increased by 3 units. If we
increase the length by 3 units and the breadth
by 2 units, the area increases by 67 square
units. Find the dimensions of the rectangle.
Sol.
Let the length of rectangle be x units
and the breadth of the rectangle be y units
Area of the rectangle = xy
Case 1 : According to the first condition,
x–5
Increased length = x + 2
Reduced breadth = y – 2
Reduced area = (x + 2) (y – 2)
Reduction in area = 28
Original Area – Reduced area = 28
xy – [(x + 2) (y – 2)] = 28
xy – [xy – 2x + 2y – 4] = 28
x – y = 12
Case 2 :
Increase in area = 33
Increased area – original area = 33
(x – 1) (y + 2) – xy = 33
xy + 2x – y – 2 – xy = 33
xy – xy – 3x + 5y + 15 = 9
3x – 5y = 6
....(1)
Case 2. According to the second condition,
x+3
xy + 33
y+2
y+2
Reduced length = x – 1
Increased breadth = y + 2
Increased area = (x – 1) (y + 2)
Increased length = x + 3
Increased breadth = y + 2 A + 40º + 120º = 180º
A = 180º – 40º – 120º = 20º
Increased area = (x + 3) (y + 2)
Increase in area = 67
Increased area – Original area = 67
A = 20º, B = 40º, C = 120º
(x + 3) (y + 2) – xy = 67
xy + 2x + 3y + 6 – xy = 67
2x + 3y = 61
....(2)
On solving (1) and (2), we get
x = 17 units and and y = 9 units
Hence, length of rectangle = 17 units,
and breadth of rectangle = 9 units.
Problems Based on Geometry
Ex.94 The larger of two supplementary angles
exceeds the smaller by 18 degrees. Find them.
Sol.
Let the angles be x and y. Then according to
the question.
x + y = 180
....(1)
and x = y + 18
....(2)
Putting x = y + 18 from (2) in (1), we get
y + 18 + y = 180
2y = 180 – 18
2y = 162
....(1)
(given)
Adding 2C on both sides of (1), we get
C + 2C = 2 (A + B) + 2C
3C = 2 (A + B + C)
C =
2
× 180º = 120º
3
A
B
C
Again C = 3B
120º = 3B
120º
B =
= 40º
3
But A + B + C = 180º
A = (2x + 4)º, and C = (2y + 10);
D
C
A
B
But A + C = 180º
(Cyclic quadrilateral)
(2x + 4)º + (2y + 10)º = 180º
2x + 2y = 166º
Also B = (y + 3)º, D = (4x – 5)º
But B + D = 180º
(Cyclic
quadrilateral)
(y + 3)º + (4x – 5)º = 180º
4x + y = 182º
On solving (1) and (2), we get x = 33º, y = 50º
B = (y + 3)º = (50 + 3)º = 53º
C = (2y + 10)º = (100 + 10)º = 110º,
Ex.95 In a ABC, C = 3 B = 2 (A + B). Find
the three angles.
C = 2(A + B)
Sol.
A = (2x + 4)º = (66 + 4)º = 70º
y = 81
Putting y = 81 in (2), we get
x = 81 + 18 = 99
Hence, angles are x = 99º and y = 81º.
Sol.
Ex.96 Find a cyclic quadrilateral ABCD, A =
(2x + 4)º, B = (y + 3)º, C = (2y + 10)º and
D = (4x – 5)º. Find the four angles.
(given)
D = (4x – 5)º = (4 × 33 – 5)º = 127º
A = 70º, B = 53º, C = 110º, D = 127º
Ex.97 The area of a rectangle remains the same if
the length is decreased by 7 dm and breadth is
increased by 5 dm. The area remains
unchanged if its length is increased by 7 dm
and and breadth decreased by 3 dm. Find the
dimensions of the rectangle.
Sol.
Let the length and breadth of a rectangle be x
and y units respectively. So, area = (xy) sq.
units.
First Case : Length is decreased by 7 dm and
breadth is increased by 5 dm.
According to the question,
xy = (x – 7) (y + 5)
xy = xy + 5x – 7y – 35
5x – 7y – 35 = 0 ....(1)
Second Case : Length is increased by 7 dm
and breadth is decreased by 3 dm.
Here, area also remains same
so, we get xy = (x + 7) (y – 3) = xy – 3x + 7y – 21
3x – 7y + 21 = 0
....(2)
So, the system of equations becomes
5x – 7y – 35 = 0
....(3)
3x – 7y + 21 = 0
....(4)
Subtracting equation (4) from (3), we get
2x – 56 = 0
2x = 56
x = 28 dm
Substituting x = 28 in equation (3), we get
5 × 28 – 7y = 35
7y = 105
y = 15 dm
Hence, length and breadth of the rectangle are
28 dm and 15 dm respectively.
Ex.98 In a triangle PQR, P = xº, Q = (3x – 2)º,
R = yº, R – Q = 9º. Determine the three
angles.
Sol.
It is given that
P = xº, Q = (3x – 2)º,
R = yº and
R – Q = 9º
P
x°
(3x–2)°
Q
y°
R
We know that the sum of three angles in a
triangle is 180º.
So, P + Q + R = x + 3x – 2 + y = 180
4x + y = 182
It is also given that
R – Q = 9º
or y – (3x – 2) = 9
....(1)
y – 3x + 2 = 9
3x – y = –7
....(2)
Adding equation (1) with (2), we get
7x = 175
x = 25
Substituting x = 25 in equation (2), we get
3 × 25 – y = – 7
y = 75 + 7 = 82
Thus, P = xº = 25º
Q = (3x – 2)º = (3 × 25 – 2)º = 73º
and R = y = 82º. IMPORTANT POINTS TO BE REMEMBERED
1. An equation of the form ax + by + c = 0 is linear
in two variables x and y. For all a and b are the
coefficients of x and y respectively such that
a, b R and a 0 , b 0
2. The graph of a linear equation in two variables is
a straight line
3. A linear equation in two variables has infinitely
many solutions
4. Slope of the line ax + by + c = 0 is – a/b
5. Equation of x-axis is y = 0 and equation of y-axis
is x = 0
(ii) Algebraic method
12. To solve a pair of linear equations in two
variables by Graphical method, we first draw the
lines represented by them.
(i) If the pair of lines intersect at a point, then we
say that the pair is consistent and the
coordinates of the point provide us the unique
solution.
(ii) If the pair of lines are parallel, then the pair
has no solution and is called inconsistent pair
of equations.
8. Every point on the graph of a linear equation in
two variables is a solution of the equation.
(iii) If the pair of lines are coincident, then it has
infinitely many solutions each point on the
line being of solution. In this case, we say
that the pair of linear equations is consistent
with infinitely many solutions.
9. A pair of linear equations in two variables x and y
can be represented algebraically as follows :
13. To solve a pair of linear equation in two variables
algebraically, we have following methods :
6. The graph of the line x = a is parallel to y-axis
7. The graph of the line y = b is parallel to x-axis.
a1x + b1y + c1 = 0
(i) Substitution method
a2x + b2y + c2 = 0
(ii) Elimination method
where a1, a2, b1, b2, c1, c2 are real number such
that
a12 +
b12
0,
a22 +
b22 0.
10. Graphically or geometrically a pair of linear
equations
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
in two variables represents a pair of straight lines
which are
(i) intersecting, if
(ii) parallel, if
a1 b1
a 2 b2
a1
b
c
= 1 1
a2
b2
c2
a
b
c
(iii) coincident , if 1 = 1 = 1
a2
b2
c2
11. A pair of linear equations in two variables can be
solved by the :
(i) Graphical method
(iii) Cross-multiplication method
14. If
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
is a pair of linear equation in two variable x and y
such that :
(i)
a1
b
1 , then the pair of linear equations is
a2
b2
consistant with a unique solution.
(ii)
a1
a2
=
b1
b2
c1
, then the pair of linear
c2
equations is inconsistent.
(iii)
a 1 b1 c1
=
= ,
a 2 b 2 c2
then the pair of linear
equations is consistent with infinitely many
solutions.
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