Pair of Linear Equations in Two Variables

32 Lifeskills' Complete NCERT Solutions Class-X Mathematics 3 Pair of Linear Equations in Two Variables Chapter In the Chapter In this chapter, you will be studying the following points: Two linear equations in the same two variables are called a pair of linear equations in two variables. The most general form of a pair of linear equations is a1 x + b1 y + c 1 = 0 a2 x + b2 y + c 2 = 0 where a1, a2, b1, b2, c1, c2 are real numbers, such that a12 + b12 ≠ 0, a22 + b22 ≠ 0. O A pair of linear equations in two variables can be represented, and solved, by the: (i) graphical method (ii) algebraic method O Graphical Method : The graph of a pair of linear equations in two variables is represented by two lines. (i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent. (ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent). (iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent. O Algebraic Methods : We have discussed the following methods for finding the solution(s) of a pair of linear equations : (i) Substitution Method (ii) Elimination Method (iii) Cross-multiplication Method O If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the following situations can arise : O a1 b1 (i) b ≠ b : In this case, the pair of linear equations is consistent. 2 2 a1 b1 c1 (ii) b = b ≠ c : In this case, the pair of linear equations is inconsistent. 2 2 2 a1 b1 c1 (iii) b = b = c : In this case, the pair of linear equations is dependent and consistent. 2 2 2 O There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations. TEXT BOOK QUESTION (SOLVED) EXERCISE 3.1 Q.1. Aftab 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." (Isn't this interesting?) Represent this situation algebraically and graphically. Ans. Let present age of Aftab be x years and present age of his daughter by y years. Case I. Seven years ago Age of Aftab = (x – 7) years Age of his daughter = (y – 7) years According to question : (x – 7) = (y – 7) ⇒ x – 7 = 7y – 49 ⇒ x – 7y = – 42 Case II. Three years later, Age of Aftab = (x + 3) years Age of his daughter = (y + 3) years According to question : x + 3 = 3(y + 3) ⇒ x + 3 = 3y – 9 ⇒ x – 3y = 6 So, algebraic expression be x – 7y = – 42 x – 3y = 6 Graphical representation For eq. (i), we have x – 7y = – 42 ⇒ x = 7y – 42 Thus, we have following table : x 0 42 y 6 12 From eqn. (ii), we have x – 3y = 6 ⇒ x = 3y + 6 Thus, we have following table x y 0 –2 6 0 42 12 18 4 When we plot the graph of equation. We find that both the lines intersect at the point (42, 12). Therefore, x = 42, y = 12 is the solution of the given system of equations. Q.2. The coach of a cricket of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and graphically. Ans. Let the cost of 1 bat be Rs x and cost of 1 ball be Rs. y Case I. Cost of 3 bats = 3x Cost of 3 balls = 3y According to question, 3x + 6y = 3900 Case II. Cost 1 bat = x Cost of 3 more balls = 3y According to question, x + 3y = 1300 So, algebraically representation be 3x + 6y = 3900 x + 3y = 1300 Graphical representation We have, 3x + 6y = 3900 ⇒ 3(x + 2y) = 3900 ⇒ x + 2y = 1300 ⇒ x = 1300 – 2y Thus, we have following table : x y We have, ⇒ 1300 –2 500 0 700 12 100 4 x + 3y = 1300 x = 1300 – 3y Thus, we have following table : x y 1300 0 100 400 4x + 2y = 300 ⇒ 2x + y = 150 Graphical representation, we have 2x + y = 160 ⇒ y = 160 – 2x 400 1000 300 100 When we plot the graph of equations, we find that both the lines intersect at the point (1300, 0). Therefore, x = 1300, y = 0 is the solution of the given system of equations. x y 50 60 40 80 Graphical representation, we have 2x + y = 150 ⇒ y = 150 – 2x x y 50 50 30 80 When we plot the graph of the equations we find that two lines do not intersect i.e. they are parallel. Q.3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically. Ans. Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y. So, algebraic representation 2x + y = 160 EXERCISE 3.2 Q.1. Form a pair of linear equations in the following problems and find their solutions graphically. (i) 10 students of Class X took part in 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. (ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of pencil and that of one pen. Ans. (i) Let the number of boys be x and number of girls be y. Case I. x + y = 10 ....(i) Case II. y = x+4 x – y = –4 ...(ii) We have, x + y = 10 x = 10 – y Thus, we have following table : 6 4 x y Scale X-axis : 1 cm = 1 unit Y-axis : 1 cm = 1 unit 0 3 7 5 5 We have, x–y = –4 ⇒ x = y–4 Thus, we have following table : 0 4 x y 3 7 x = y–2 Thus, we have following table : x y 1 5 0 2 3 5 2 4 Scale X-axis : 1 cm = 1 Re. When we plot the graph of the given equation, we find that both lines intersect at the point (3, 7). So x = 3, y = 7 is the required solution of the pair of linear equation. Y-axis : 1 cm = 1 Re, Hence, the number of boys be 3 and the number of girls be 7, who took part in quiz. (ii) Let cost of 1 pencil be Rs. x and cost of 1 pen be Rs. y. Case I. Cost of pencils = 5x Cost of 7 pens = 7y According to question, 5x + 7y = 50 Case II. Cost of pencils = 7x Cost of 5 pens = 5y According to question, 7x + 5y = 46 Thus, we have 5x + 7y = 50 7x + 5y = 46 12x + 12y 12(x + y) x+y 5x + 7y 7x + 5y – – ⇒ ⇒ Again, = = = = = – 96 0 8 50 46 –2x + 2y = 4 ⇒ –2(x – y) = ⇒ x–y = We have, x+y = ⇒ x = Thus, we have following table : x y We have, 4 4 3 5 4 –2 8 8–y 6 2 x – y = –2 3 2 1 When we plot the graph of the given equation, we find that both the lines intersect at the point (3, 5). So x = 3, y = 5 is the required solution of the pair of linear eqation Hence, the cost of 1 pencil be Rs. 3, cost of 1 pen be Rs. 5. a1 b1 c1 Q.2. On comparing the ratios a , b and c 2 2 2 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 Ans. Comparing the given equations with standard form 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 , b1 = – 4 ⇒ a1 ≠ b1 a2 = 7 b2 6 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 c1 12 1 9 1 b1 3 1 ∴ a = 18 = 2 , b = 6 = 2 and c = 24 = 2 2 2 2 a1 b1 c1 ⇒ a = b = c 2 2 2 Thus, the lines representing the pair of linear equations are intersecting. (iii) a1 = 6, b1 = –3, c1 = 10 a2 = 2, b2 = –1, c2 = 9 ∴ (iii) 3 x + 5 y = 7; 9x – 10y = 14 2 3 Here, a2 a1 b1 c 6 = – 3 = 3, 1 = 10 = = 3, a2 b2 –1 c2 9 2 a1 b1 c1 ⇒ a = b ≠ c 2 2 2 Thus, the lines representing the pair of linear equations are parallel. a1 b1 c1 Q.3. On comparing the rations a , b and c , 2 2 2 find out whether the following pair of linear equations are consistent or inconsistent. (i) 3x + 2y = 5; 2x – 3y = 7 (ii) 2x – 3y = 8; 4x – 6y = 9 3 x + 5 y = 7; (iii) 9x – 10y = 14 2 3 (iv) 5x – 3y = 11; – 10x + 6y = –22 4 (v) x + 2y = 8; 2x + 3y = 12 3 Ans. (i) 3x + 2y = 5; 2x – 3y = 7 Here, and a1 b1 Clearly, a ≠ b 2 2 Hence, the given lines are intersecting. So, the given pair of linear equations has exactly one solution and therefore it is consistent. (ii) 2x – 3y = 8; 4x – 6y = 9 Here, a 1 = 2, b1 = –3, c1= 8 a 2 = 4, b2 = –6, c2 = 9 ∴ a1 2 1 b1 – 3 1 c = = , = = , and 1 = 8 a2 4 2 b2 – 6 2 c2 9 a1 b1 c1 Clearly, a = b ≠ c 2 2 2 Hence, the given lines are parallel. So, the given pair of linear equations has no solution and therefore it is inconsistent. = 3/ 2 = 1 9 6 b1 b2 = 5/3 = 1 – 10 6 c1 c2 = 7 =1 14 2 Hence, the given lines are intersecting. So the given pair of linear equations has exactly one solution and therefore it is consistent. (iv) 5x – 3y = 11; – 10x + 6y = –22 Here, a 1 = 5, b1 = – 3, c1 = 11 a 2 = –10, b2 = 6, c2 = –22 a1 a2 = b1 b2 3 1 = – =– 6 2 and c1 c2 = Clearly, a1 a2 b1 c1 = b = c 2 2 a 2 = 2, b2 = –3, c2 = 7 ∴ a1 a2 a1 b1 Clearly, a ≠ b 2 2 a 1 = 3, b1 = 2, c1 = 5 a1 3 b1 – 2 c1 5 = , = , = a2 2, b2 3 c2 7 3 5 ,b = ,c =7 2 1 3 1 = 9, b2 = –10, c2 = 14 a1 = 5 ≠–1 – 10 2 11 = – 1 – 22 2 Hence, the given lines are consistent. So, the given pair of linear equations has infinitely many solutions and therefore it is consistent. (v) 4 x + 2y = 8; 2x + 3y = 12 3 Hence, a1 = 4 , b = 2, c1 = 8 3 1 a1 a2 = 4/3 = 2 2 3 b1 b2 = 2 3 and c1 c2 8 =2 = 12 3 (ii) Given pair of lines are x – y = 8, 3x – 3y = 16 Here, a1 = 1, b1 = –1, c1 = – 8; a2 = 3, b = – 1, c2 = –16 Clearly, a1 a2 b1 c1 = b = c 2 2 Here, a1 1 b1 −1 1 = , = = a2 3 b2 − 3 3 and c1 −8 1 = = c2 − 16 2 ∴ a1 b1 c1 ≠ ≠ a2 b2 c2 Hence, the given lines are consistent. So the given pair of linear equations has infinitely many solutions and therefore it is consistent. Q.4. Which of the following pairs of linear equations are consistent/inconsistent ? If Consistent, obtain the solution graphically : (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8; 3x – 3y = 16 (iii) 2x + 2y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 Ans. (i) Wehave x+y = 5 ⇒ y = 5–x Hence, the lines represented by the given equations are parallel. Therefore, it has no solution. Hence, the given pair of lines is inconsistent. (iii) We have, 2x + y – 6 = 0 ⇒ y = 6 – 2x x y Thus, we have following table : 0 5 x y 5 0 and 2x + 2y = 10 Thus, we have following table : 0 5 x y 2 3 5 0 When we plot the graph of the equations we find that both the lines are coincident. Hence, pair of linear equations has infinitely many solutions y = 2x – 2 Scale X-axis : 1 cm = 1 Unit Y-axis : 1 cm = 1 Unit 4 3 2 1 1 2 3 4 5 3 0 4x – 2y – 4 = 0 y = 2x – 0 Thus we have following table : 10 – 2 x y = 2 ⇒ 0 6 6 x y 0 –2 1 0 When we plot graph of the equations, we find that both the lines intersect at point (2, 2). Hence the solution of the given equation is x = 2, y = 2. Therefore, the pair of equation is consistent at point (2, 2). Thus, we have following table : x y (iv) 2x – 2y – 2 = 0 y = 1 Thus we have following table : x y 0 –1 10 26 20 16 16 20 Scale X-axis : 1 cm = 4 m 1 0 Y-axis : 1 cm = 4 m 4x – 4y – 5 = 0 ⇒ 5 y = x– – 4 when 5 x = 0, y = – 4 when x = x 0 y –5 4 5 ,y=0 4 5 4 0 When we plot graph of the given equations, we find that both the lines never meet. Hence lines are parallel and equations have no solution. Q.5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. Ans. Let the length of the garden be x m and width be y m. Case I. x = y+4 ⇒ x–y = 4 Case II. Half perimeter = 36 ⇒ x + y = 36 So algebraic representation be x–y = 4 x + y = 36 Graphical representation : We have, x–y = 0 x = 0 ⇒ Thus, we have following table x 0 4 12 y –4 0 8 ⇒ x+y = 0 x = 36 – y 0 If we plot the graph of both the equations, we find that the two lines intersect at the point (20, 16). So, x = 20, y = 16 is the required solution of the given equation i.e., the length of the garden is 20 m and breadth is 16 m. Q.6. Given the linear equations 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is : (i) intersecting lines (ii) parallel lines (iii) coincident lines. Ans. 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 Q.7. Draw the graph 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. Ans. We have, x + y +1 = 0 ⇒ y = x+1 Ans. We have, Thus, we have following table : 2 3 x y –1 0 3 4 We have, 3x +2 y +12 = 0 2y = 12 – x ⇒ ⇒ y = 12 – 3x 2 Thus, we have following table : x y 2 3 4 0 0 6 When we plot the graph of the given equations, we find that both the lines intersect at the pont (2, 3), therefore x = 2, y = 3 is the solution of the given system of equations. Vertices of triangle are (2, 3), (–1, 0) and (4, 0). EXERCISE 3.3 ⇒ ⇒ Q.1. Solve the following pairs of linear equations by the substitution method. (i) x + y = 14 (ii) s – t = 3 x–y = 4 (iii) 3x + y = 3 9x – 3y = 9 (v) 2x + 3y = 0 3x + 8 y = 0 Ans. (i) s+t =6 3 2 (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 (vi) 3x – 5 y = – 2 2 3 x – y = 13 3 2 6 x + y = 14 x–y = 4 The given pair of linear equations is x + y = 14 ...(i) x–y = 4 ...(ii) From equation (i), we have y = 14 – x ...(iii) Substitute this value of y in equation (ii), we get x – (14 – x) = 4 ⇒ x – 14 + x = 4 ⇒ 2x – 14 = 4 2x = 4 + 14 2x = 18 18 =9 2 Substitute this value of x in equation (iii), we ⇒ x = get y = 14 – 9 = 5 Therefore the solution is x = 9, y = 5 Verification. Substituting x = 9 and y = 5, we find that both the equations (i) and (ii) are satisfied as shown below x + y = 9 + 5 = 14 x–y = 9–5=4 This verifies the solution. Ans. (ii) We have s–t = 3 s+t = 6 3 2 and s+t = 6 3 2 2s + 3t = 6 6 ⇒ 2s – 3t = 36 The given pair of linear equations is s–t = 3 ...(i) 2s – 3t = 36 ...(ii) From equation (i), we have s = 3+t ...(iii) Substitute this value of s in equation (ii), we get 2s + 3t = 36 ⇒ 2(3 + t) + 3t = 36 ⇒ 6 + 2t + 3t = 36 ⇒ 5t + 6 = 36 ⇒ 5t = 30 ⇒ t = 6 Therefore, the solution is s = 9, t = 6 ⇒ ⇒ ⇒ s+t 9+6 = =3+3=6 3 2 3 2 This verifies the solution. (iii) 3x – y = 3 9x – 3y = 9 The given pair of linear equation is 3x – y = 3 ...(i) 9x – 3y = 9 ...(ii) From eqn. (i), we have y = 3x – 3 Substitute this value of y in eq. (ii), we get 9x – (3x – 3) = 9 ⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9 which is true. Therefore equn. (i) and (ii) have infinitely many solutions. Ans. (iv) 0.2x + 0.3y = 1.3 0.4x – 0.5y = 2.3 The given system of linear equations 0.2x + 0.3y = 1.3 ...(i) 0.4x – 0.5y = 2.3 ...(ii) From equation (i), we have 0.3y = 1.3 – 0.2x ⇒ y = 1.3 – 0.2 x 0 .3 ...(iii) Substituting this value of y in eqn. (ii), we get ⎛ 1.3 – 0.2 x ⎞ ⎟ 0.4x + 0.5 ⎜ 0.3 ⎠ = 2.3 ⎝ ⇒ 0.21 + 0.65 – 0.1x = 0.69 0.12x – 0.1x = 0.69 – 0.65 0.02x = 0.04 0.04 =2 0.02 Substituting this vaue of x in eqn. (iii), we get ⇒ x = y = 1.3 = 0.2(2) 0.3 = = 0 .9 =3 0.3 Therefore, the solution is x = 2, y = 3, Verification, Substituting x = 2 and y = 3, we find that both the equations (i) and (ii) are satisfied as shown below : 0.2x + 0.3y = (0.2) (2) + (0.3) (3) = 0.4 + 0.9 = 1.3 0.4 + 0.5y = (0.4) (2) + (0.5) (3) = 0.8 + 1.5 = 2.3 This verifies the solution. 2x + 3y = 0 (v) 3x – 8 y = 0 The given pair of linear eqn. is 2x + 3y = 0 ...(i) 3x – 8 y = 0 From eq. (ii), we have ...(ii) 3x = ⇒ x = 8y 8 y 3 ...(iii) Substituting this value of x in equation (i), we get 2. ⇒ 8 + 3y = 0 3 4 y + 3y = 0 3 ⎛ 4 ⎞ ⎜⎜ + 3 ⎟⎟ y = 0 ⎝ 3 ⎠ ⇒ y = 0 Substituting this value of y in eqn. (iii), we get ⇒ x = 8 (0) = 0 3 Therefore, the solution is x = 0, y = 0 Verification. Substituting x = 0 and y = 0, we find that both equations (i) and (ii) are satisfied as shown below : 2x + 3y = 2 ( 0) + 3 ( 0) = 0 3 x – 8 y = 3 (0) – 8 (0) = 0 This verifies the solution. 3x – 5 y 2 3 (vi) We have = –2 9 x – 10 y = –2 6 9x – 10y = –12 13 x+ y = 6 3 2 2x + 3y = 13 The given pair of linear equations is 9x – 10y = –12 2x – 3y = 13 From (ii), we have x = Therefore, the solutions is x = 2, y = 3 Verification. Substituting x = 2 and y = 3, we find that both the equations (i) and (ii) are satisfied as shown below : 5y 3 3 5 x– = (2) – (3) = 3 – 5 = –2 2 2 2 2 11 – 3 y ...(iii) 2 Putting the value of 'x' in (ii), we get x = and 141 ⎛ 13 + 3 y ⎞ 9⎜ ⎟ 47 ⎝ 2 ⎠ 13 + 3 y 2 ⇒ ...(i) ...(ii) ...(iii) 22 – 6 y – 8 y – 4y = – 24 2 22 – 14y = – 48 –14y = – 70 y =5 Putting the value of y in (iii), we get ⇒ ⇒ ⇒ ⇒ 9(13 + 3 y ) – 20 y = – 12 2 117 + 27y – 20y = – 24 – 47y = – 24 – 117 y = x = =3 Substituting the value of y in (iii), we get 13 – 3 y x = 2 13 – 3 × 3 x = 2 11 – 3 y 11 – 15 = 2 2 –4 =–2 2 Hence, the solution is x = – 2, y = 5 It is given that y = mx + 3 Putting the values of x and y in given condition we get 5 = m(–2) + 3 ⇒ 5 = – 2m + 3 ⇒ –2m = 2 ⇒ m = –1. Q.3. From the pair of linear equations for the following problems and find their solutions by substitution method. (i) The difference between two numbers is 26 ⇒ –10y = – 12 ⎛ 11 – 3 y ⎞ 2⎜ ⎟ – 4y = – 24 ⎝ 2 ⎠ ⇒ Substituting the value of x in (i), we get 9x – 10y = – 12 ⇒ 13 – 9 = 4 2 2 3 3 13 x+ y = + = 2 2 6 3 2 This verifies the solution Q.2. Solve 2x + 3y = 11 and 2x – 4y = –24 and hence find the value of 'm' for which y = mx + 3. Ans. The given equations are 2x + 3y = 11 ...(i) 2x – 4y = –24 ...(ii) From (i), we have = –2 13 x+ y = 6 3 2 3x – 5 y 2 3 = x = and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degree. Find them. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball. (iv) 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 Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km ? How much does a person has to pay for travelling a distance of 25 km ? 9 (v) Afraction becomes , if 2 is added to both 11 the numerator and the denominator. If, 3 is added to both the numerator and the 5 demoniator it becomes . Find the fraction. 6 (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages ? Ans. (i) Let the two numbers be x and y ∴ By the given conditions, x – y = 26 and x = 3y Sinstituting x = 3y in (i), we get 3y – y = 26 ⇒ 2y = 26 ⇒ y = 13 From (ii), x = 3y = 3 × 13 = 39 Thus the numbers are 39 and 13. (ii) Let the larger angle be x and smaller angle be y. According to given conditions, x + y = 180 ...(i) ⇒ (y + 18) + y = 180 ...(ii) ⇒ y + 18 + y = 180 ⇒ 2y = 180 – 18 ⇒ 2y = 162 ⇒ y = 81º ...(iii) Putting the value of (iii) in (ii) x = y + 18 = 81 + 18 = 90º Thus, the angles are 81º and 99º. (iii) Let the cost of each bat and each ball be x Rs. x and Rs. y respectively. According to given conditions, 7x + 6y = 3800 ...(i) 3x + 5y = 1750 ...(ii) From eqn. (ii) we get 3x + 5y = 1750 5y = 1750 – 3x 1750 – 3x ...(iii) 5 Substitue the value of eqn. (iii) in (i), we get 7x + 6y = 3800 y = ⎛ 1750 – 3x ⎞ ⎟ x = 3800 7x + 6 ⎜ 5 ⎝ ⎠ ⇒ ⇒ ⇒ ⇒ ⇒ 35 x + 6 (1750 – 3 x ) 5 35x + 10500 – 18x 17x + 10500 17x 17x = 3800 = = = = 19000 19000 19000 – 10500 8500 8500 = 500 17 Substituting the value of x in (iii) we get, ⇒ x = y = 1750 – 3x 5 = 1750 – 3 × 500 5 = 1750 – 1500 = 5 250 = 50 5 Hence, the cost of one bat = Rs. 500 and cost of one ball = Rs. 50. (iv) Let fixed charge be Rs. x and charge per km be Rs. y According to the given conditions, x + 10y = 105 ...(i) and x + 15y = 155 ...(ii) Subtracting (i) from (ii), we get (x + 5y) – (x + 10y) = 155 – 105 ⇒ x + 15y – x – 10y = 50 = ⇒ 5y = 50 ⇒ y = 10 Putting the value of y in (i), we get x + 10y = 105 ⇒ x + 10 (10) = 105 ⇒ x + 100 = 105 ⇒ x = 5 ∴ Fixed charge (x) = Rs. 5 and charge per km (y) = Rs. 10 Thus, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = Rs. 255 (v) Let the numerator of the fraction be x and denominator be y. Then, Case I. ⇒ ⇒ ⇒ x y Fraction = x+2 y+2 11(x + 2) 11x + 22 11x – 9y 9 11 = 9(y + 2) = 9y + 18 = –4 = x+3 5 y+3 = 6 ⇒ 6(x + 3) = 5(y + 3) ⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = – 3 Thus, we have following equations ⇒ 11x – 9y = – 4 ⇒ 6x – 5y = – 3 From (ii), we have ⇒ 6x – 5y = – 3 ⇒ 6x = 5y – 3 ...(i) Case II. ...(ii) 5y – 3 ...(iii) 6 Substituting the value of (iii) in (i), we get ⇒ 11x – 9y = – 4 ⇒ ⇒ x = ⎛ 5y – 3 ⎞ 11 ⎜ 6 ⎟ –9y = – 4 ⎝ ⎠ 11(5 y – 3) – 54 y = –4 6 ⇒ 55y – 33 – 54y = – 24 ⇒ y – 33 = – 24 ⇒ y = – 24 + 33 ⇒ y = 9 Now, substituting the value of y in (iii), we get ⇒ 5y – 3 6 5× 9 – 3 x = 6 45 – 3 42 = = =7 6 6 x = ⇒ Hence, the required fraction is 7 . 9 (vi) Let the present ago of Jacob = x years and present age of his son = y years. Case I. 5 years hence, Age of Jacob = (x + 5) years and age of his son = (y + 5) years According to given conditions, x + 5 = 3 (y + 5) ⇒ x + 5 = 3y + 15 ⇒ x – 3y = 10 Case II. 5 years ago, Age of Jacob = (x – 5) years and age of his son = (y – 5) years According to given conditions, x – 5 = 7 (y – 5) ⇒ x – 5 = 7y – 35 ⇒ x – 7y = – 35 + 5 ⇒ x – 7y = – 30 Thus, we have following equations x – 3y = 10 ...(i) x – 7y = – 30 ...(ii) From (i), we have x – 3y = 10 ⇒ x = 3y + 10 ...(iii) Substituting the value of x in (ii), we get x – 7y = – 30 ⇒ 3y + 10 – 7y = – 30 ⇒ – 4y + 10 = – 30 ⇒ – 4y = – 40 ⇒ y = 10 Now, substituting the value of y in (iii), we get x = 3y + 10 = 3(10) + 10 = 30 + 10 = 40 Hence, Age of Jacob = 40 years and Age of his son = 10 years EXERCISE 3.4 Q.1. Solve the following pair of linear equations by the elimination method and the substitution method. (i) x + y = 5 and 2x – 3 = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 y x + 2y = – 1 and x = = 3 2 3 3 Ans. Elimination Method : (i) x+y = 5 ...(i) 2x – 3y = 4 ...(ii) For making the coefficeint of y in (i) and (ii) equal, we multiply (i) by 3 and adding, we get 3x + 3y = 15 2x – 3y = 4 (iv) 5x = 19 19 5 Now, putting the value of x in equation (i), we ⇒ x = get x+y = 5 ⇒ 19 +y = 5 5 ⇒ y = 5– 19 25 – 19 6 = = 5 5 5 19 6 ,y= 5 5 Substitution method : We have following equations : x+y = 5 2x – 3y = 4 From (i), we have x+5 = 5 ⇒ x = 5–y Substituting the value of x in (ii), we get 2x – 3y = 4 ⇒ 2(5 – y) –3y = 4 ⇒ 10 –2y – 3y = 4 ⇒ 10 – 5y = 4 ⇒ –5y = 4 – 10 ⇒ –5y = – 6 Hence, x = 6 5 Now, substituting the value of y in (iii), we get x = 5–y ⇒ y = ⇒ x = 5– ⇒ x = 6 5 25 – 6 19 = 5 5 19 6 ,y= 5 5 (ii) 3x + 4y = 10 2x – 2y = 2 For making the coefficient of y in (i) and (ii) equal, we multiply (ii) by 2 and adding, we get 3x + 4y = 10 4x – 4y = 4 Hence, x = 7x = 14 14 =2 7 Now, putting the value of in (i), we get 3x + 4y = 10 ⇒ 3(2) + 4y = 10 ⇒ 6 + 4y = 10 ⇒ 4y = 4 ⇒ y = 1 Hence, x = 2, y = 1 Substitution method : We have, 3x + 4y = 10 2x – 2y = 2 From (i), we have 3x + 4y = 10 ⇒ 3x = 10 – 4y ⇒ ⇒ x = x = 10 – 4 y 3 ...(i) ...(ii) ...(iii) Substituting the value of (iii) in (ii), we get 2x – 2y = 2 ⇒ ⎛ 10 – 4 y ⎞ 2 ⎜ 3 ⎟ – 2y = 2 ⎝ ⎠ 2(10 – 4 y ) – 6 y = 2 3 ⇒ 20 – 8y – 6y = 6 ⇒ –14y = 6 – 20 ⇒ – 14y = –14 ⇒ y = 1 Now, substituting the value of y in (iii), we get ⇒ 10 – 4 y 10 – 4(1) 10 – 4 6 = = = =2 3 3 3 3 Hence, x = 2, y = 1. (iii) 3x – 5y = 4 ...(i) 9x – 2y = 7 ...(ii) For making the coefficient of x in (i) and (ii) equal, we multiply eqn. (i) by 3 and subtracting, we get 9x – 15y = 12 9x – 2y = 7 – + – –13y = 5 x= ⇒ ⇒ y4 + 5 y ⎞ ⎛15 ⎜39 ⎟ 3 ⎝ 3 ⎠ 3x + 3x = 4 – 25 1 13 ⇒ 3x = 52 – 25 13 ⇒ 3x = 27 13 ⇒ x = 9 13 ⎛ –5⎞ 9 , y = ⎜ 13 ⎟ 13 ⎝ ⎠ Substitution method : We have, 3x – 5y = 9x – 2y = From (i), we have 3x – 5y = ⇒ 3x = x = 9 = 27 × 1 = 9 13 3 13 9 –5 ,y= . 13 13 x + 2y y 3 = –1 y 3 = 3. Considering equation x + 2y = –1 2 3 3x + 4 y ⇒ = –1 6 ⇒ 3x + 4y = –6 x– = 3 3x – y = 3 3 ⇒ 3x – y = 9 Now, we have following pairs of equations 3x + 4y = – 6 ...(i) 3x – y = 9 ...(ii) Since the coefficients of 'x' (i) and (ii) are equal. So simply by subtracting we can eliminate the variable i.e., x. 3x + 4y = – 6 3x – y = 9 – + = – 5y = – 15 ⇒ ...(i) ...(ii) 4 4 + 5y 4 + 5y 3 ...(iii) Substituting the value of (iii) in (ii), we get 9x – 2y = 7 ⇒ 4 – 25 13 3 x– and 4 7 – 2y = 7 –5 13 52 – 25 13 = 3 1 4 + 5⎛⎜ – 5 ⎞⎟ ⎝ 13 ⎠ x = 3 (iv) Hence, x = ⇒ = Now, substituting the value of y in (iii), we get Hence, x = 25 =4 13 ⇒ y = = ⎛ –5⎞ 3x – 5 ⎜ 13 ⎟ = 4 ⎝ ⎠ ⇒ 36 + 39y = 21 ⇒ 39y = – 5 ⇒ ⎛ –5⎞ y = ⎜ 13 ⎟ ⎝ ⎠ Now putting the value of y in (i), we get 3x – 5y = 4 ⇒ 9(4 + 5 y ) – 6 y =7 3 36 + 45y – 6y = 21 ⇒ ⇒ ⇒ –15 5 y = –3 y = Now putting the value of y in eqn. (i), we get 3x + 4y = – 6 3x + 4(– 3) = – 6 ⇒ 3x – 12 = – 6 ⇒ 3x = 6 ⇒ x = 2 Hence, x = 2, y = – 3. Substitution method : ⇒ x + 2y = –1 2 3 3x + 4y = – 6 y = 3 3 ⇒ 3x – y = 9 Now, we have following pairs of equations 3x + 4y = – 6 ...(i) 3x – y = 9 ...(ii) From (ii), we have 3x – y = 9 y = 3x – 9 ...(iii) Substituting the value of (iii) in (i), we get 3x + 4y = – 6 3x + 4(3x – 9) = – 6 3I + 12x – 36 = – 6 15x = – 6 + 36 15x = 30 x = 2 Now, substituting the value of x in (iii) we get y = 3x – 9 = 3(2) – 9 = 6 – 9 = 3 Hence, x = 2, y = – 3 Q.2. From the pair of linear equation in the following problems and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to and x– 1 if we only add 1 to the 2 denominator. What is the fraction ? (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ? (iii) 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 digits. Find the number. (iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and 1. It become Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received. (v) A leading library has a fixed charge for the first three days and an additional charge for each day thereafter. Sarita paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra days. Ans. (i) Let the numerator of the fraction be x and denominator be y. Then x Fraction = y Then, according to given conditions, Case I. ⇒ x+1 ⇒ Case II. x +1 y –1 = 1 = y–1 x–y = –2 x 1 y +1 = 2 ⇒ 2x = y + 1 ⇒ 2x – y = 1 Thus, we have following equations x–y = –2 ...(i) ⇒ 2x – y = 1 ...(ii) From (i), we have x – y = –2 ⇒ x = y–2 ...(iii) Substituting the value of x in (ii), we get 2x – y = 1 ⇒ 2(y – 2) – y = 1 ⇒ 2y – 4 – y = 1 ⇒ y–4 = 1 ⇒ y = 5 Now, substituting the value of y in (iii), we get x = y–2 = 5–2=3 3 . 5 (ii) Let the present ago of Nuri be x years and present age of Sonu be y years. Case I. 5 years ago, Age of Nuri = (x – 5) years and Age of Sonu = (y – 5) years Hence, the required fraction = According to the given conditions, x – 5 = 3 (y – 5) ⇒ x – 5 = 3y – 15 ⇒ x – 3y = – 10 Case II. Ten yeras later, Age of Nuri = (x + 10) years Age of Sonu = (y + 10) years According to the given conditions, x + 10 = 2(y + 10) x + 10 = 2y + 20 x – 2y = 10 Thus, we have following equations : x – 3y = –10 x – 2y = 10 From (i), we have x – 3y = – 10 x = 3y – 10 9 ⇒ y = =1 9 Substituting this value of y in equation (i), we get ...(i) ...(ii) ...(iii) Substituting the value of (iii) in (ii), we get x – 2y = 10 3y – 10 – 2y = 10 y = 20 Now, substituting the value of y in (iii), we get x = 3y – 10 = 3(30) – 10 = 60 – 10 = 50 Hence, Age of Nuri = 50 years and Age of Sonu = 20 years (iii) Let the digit at Unit's place be x and digit at ten's place be y Then, number = 10y + x Also, the number obtained by reversing the order of the digits = 10x + y According to the given condition, x+y = 9 And 9(10y + x) = 2(10x + y) ...(i) ⇒ 90y + 9x = 20x + 2y ⇒ 11x – 88y = 0 ⇒ x – 8y = 0 ...(ii) Subtracting question (ii) from equation (i) from equation (i), we get 9y = 9 x+1 = 9 x = 9–1=8 Hence, the required number = 10y + x = 10(1) + 8 = 10 + 8 = 18 (iv) Suppose that Meena received x notes of Rs. 50 and y notes Rs. 100. Then, according to the given conditions x + y = 25 ...(i) And 50x + 100y = 2000 ⇒ x + 2y = 40 ..(ii) Subtracting this value of y in equation (i), we get y = 15 Substituting this value of y in equation (i), we get x + 15 = 25 ⇒ x = 25 – 15 = 0 Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100. (v) Let the fixed charge be Rs. a and the charge for each extra day be Rs. b. Then, according to the given conditions, a + 4b = 27 ...(i) [Extra days = 7 – 3 = 4] a + 2b = 21 [Extra days = 5 – 3 = 21] Subtracting equation (ii) from equation (i), we get 2b = 6 6 b = =3 2 Substituting it is value of b in equation (ii), we get a + 2(3) = 21 ⇒ a + b = 21 ⇒ a = 21 – 6 = 15 Hence, the fixed charges are Rs. 15 and the charge for each extra days is Rs. 3. EXERCISE 3.5 Q.1. Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it : (i) x – 3y – 3 = 0 3x – 9y – 2 = 0 (ii) 2x + y = 5 3x + 9y =8 (iii) 3x – 5y = 20 6x – 10y = 40 (iv) x – 3y – 7 = 0 3x – 9y – 2 = 0 Ans. x – 3y – 3 = 0 3x – 9y – 2 = 0 The given pair of linear equation is x – 3y – 3 = 0 3x – 9y – 2 = 0 Here, a 1 = 1, b1 = –3, c1 = –3 a 2 = 3, b2 = –9, c2 = –2 We see that a1 b1 c1 ≠ a2 = b2 c 2 Hence, the given pair of linear equations has no solutions. (ii) 2x + y = 5 3x + 2y = 8 The given pair of linear equation is 2x + y = 5 3x + 2y = 8 ⇒ 2x + y – 5 = 0 3x + 2y – 8 = 0 Here, a1 = 2, b1 = 1, c1 = –5 a2 = 3, b2 = 2, c2 = –8 We see that, a1 b1 ≠ a 2 b2 Hence, the given pair of linear equations has a unique solution. To solve the given equation by cross multiplication method, we draw the diagram below : 1 x –5 y 2 1 1 2 Then, –8 3 x (1)(– 8) – (2)(– 5) ⇒ ⇒ x – 8 + 10 x 2 2 y = (– 5)(3) – (– 8)(2) = (2)(2) – (3)(1) 1 y 1 = 4–3 – 15 + 16 y 1 = = 1 1 = ⇒ x = 2, y = 1 Hence, the required solution of the given pair of linear equation is x = 2, y = 1. (iii) 3x – 5y = 20 6x – 10y = 40 The given pair of linear equations is 3x – 5y – 20 = 0 ⇒ 6x – 10y – 40 = 0 Here, a 1 = 3, b1 = –5, c1 = –20 a 2 = 6, b2 = –10, c2 = –40 What see that a1 b1 c1 = = a2 b2 c2 Hence, the given pair of linear equation has infinitely many solution. (iv) x – 3y – 7 = 0 3x – 3y – 15 = 0 The given pair of linear equations is x – 3y – 7 = 0 3x – 3y – 15 = 0 Here, a1 = 1, b1 = –3, c1 = –7 a2 = 3, b2 = –3, c2 = – 15 We see that b1 a1 a2 ≠ b 2 Hence, the given pair of linear equqtion has a unique solution. To solve the given equations by cross multiplication method, we draw the diagram below : –3 x –7 y 1 1 –3 –3 Then, –15 3 –3 x (– 3)(– 15) – (– 7)(– 3) = y (– 7 )(3) – (– 15)(1) = y (1)(– 3) – (3)(– 3) y x 1 = = 45 – 21 –3+9 – 21 + 15 y x =1 = 24 –6 6 24 = 4, y = – 6 = – 1 x = 6 6 Hence, the required method of the given pair of linear equatin is x = 4, y = –1, Q.2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ? 2x + 3y = 7 (a – b) x + (a + b) y = 3a + b – 2 = ⇒ (ii) For which value of k will the following pair of linear equations have no solution ? 3x + y = 1 (2x – 1)x + (k + 1) y = 2k + 1 4 –1 3x k –21 ⇒ From first two, 2 3 = 7 = a–b a + b 3a + b – 2 2 a–b 2(a + b) 2a + 2b a – 5b ⇒ ⇒ ⇒ and, 3 a+b = 3(a – b) = 3a – 3b =0 = 3 7 = a+b 3a + b – 2 ⇒ 3(3a + b – 2) = 7(a + b) ⇒ 9a + 3b – 6 = 7a + 7b ⇒ 2a – 4b – 6 = 0 ⇒ a – 2b – 3 = 0 Thus, we have following equations a – 5b = 0 a – 2b – 3 = 0 –5 a 0 b 1 1 –5 a b =1 = 15 3 3 ⇒ a= 15 =5 3 and b= 3 =1 3 Hence, a = 5; b = 1. (ii) The given pair of linear equations 3x + y = 1 (2k – 1) x + (k – 1) y = 2k + 1 ⇒ 3x + y – 1 = 0 Here, (2x – 1) x + (k – 1) y – (2k + 1) = 0 a1 = 3, b1 = 1, c1 = –1 a2 = 2k –1, b2 = k –1, c2 = –2k –1 For having no solution, we must have Ans. (i) We have following equations 2x + 3y = 7 (a – b)x + (a + b)y = 3a + b – 2 Here, a 1 = 2, b1 = 3, c1 = 7 and a 2 = a – b, b2 = a + b, c2 = 3a + b – 2 For having an infinite number of solutions, we must have a1 b1 c1 = = a2 b2 c2 ⇒ a1 b1 c ≠ 1 a2 = b2 c 2 3 = 2k – 1 ⇒ 3(k – 1) = 2k – 1 ⇒ 3k – 3 = 2k – 1 ⇒ 3k – 2k = 3 – 1 ⇒ 2k = 2 Hence, k = 2. Q.3. Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x – 5y – 3 = 9 ...(i) 3x + 2y = 4 ...(ii) Ans. The given pair of linear equtions is 8x + 5y = 9 3x + 2y = 4 (i) By substitution method From equation (ii), we have 2x = 4 – 3x ⇒ y = –3 a 1 (– 5)(– 3) – (– 2)(0 ) 2 = b (0 )(1) – (– 3)(1) ...(iii) Substituting this value of y in equation (i), we get –2 Then 1 (1)(– 2) – (1)(– 5) ⎛ 4 – 3x ⎞ 8x + 5 ⎜ 2 ⎟ = 9 ⎝ ⎠ 16x + 20 – 15x = 18 x + 20 = 18 x = 18 – 20 x = –2 Substituting this value of x in equation (iii), we get 4 – 3(– 2 ) 2 4 + 6 = 10 = =5 2 2 So, the solution of the given pair of linear equations is x = –2, y = 5. (ii) By cross-multiplication method Let us write the given pair of linear equations in 8x + 5y – 9 = 0 3x + 2y – 4 = 0 Solving the equations, we get Then, y = x (5)(– 4 ) – (2)(– 9) = y (– 9)(3) – (– 4)(8) 1 = (8 )(2 ) – (3 )(5 ) = x y 1 = = – 20 + 18 – 27 + 32 16 – 15 y 1 x = = –2 5 1 ⇒ x = –2 and y = 5 Hence, the required solution of the given pair of linear equatins is x = –2, y = 5. Q.4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (i) A part of monthly charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day. (ii) A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction. (iii) Yash scored 40 markes in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 markes 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 ? (iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ? (v) 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. Ans. (i) Let the fixed hostel charges be Rs. x and the cost of food per day be Rs. y (Running charges) Case I. Hostel charges of 'A' Fixed charges be Rs x and cost of food for 20 days = 20y According to question, x + 20y = 1000 Thus, we have following equations x + 20y = 1000 x + 26y = 1180 or x + 20y – 1000 = 0 x + 26y – 1180 = 0 Now, by using cross multiplication method, x y 1 20 –1000 1 20 26 ⇒ = ⇒ –1180 1 x (20 )(– 1180 ) – (26 )(– 1000 ) y 1 = (– 1000)(1) – (– 1180) (26)(1) – (20 )(1) y x = – 23600 + 26000 – 1000 + 1180 = ⇒ 26 1 26 – 20 y x 1 = = 2400 6 180 y x 1 1 = and = 2400 6 6 180 x = 400 and y = 300 Hence, fixed charges is Rs. 400 and cost of food per day is Rs. 30. (ii) Let the numerator of the fraction be x and denominator be y. Then ⇒ x y Fraction = x –1 y 3(x – 1) 3x – 3 3x – y 3x – y – 3 Case I. ⇒ ⇒ ⇒ ⇒ Case II. = = = = = 1 3 y y 3 0 x 1 y +8 = 4 ⇒ 4x = y + 8 ⇒ 4x – y = 8 ⇒ 4x – y – 8 = 0 Thus, we have following equation 3x – y – 3 = 0 4x – y – 8 = 0 –1 x –3 y 3 1 –1 –1 Then, –8 x 4 (– 1)(– 8) – (– 3)(– 1) = = ⇒ –1 y (– 3)(4) – (– 8)(– 3) 1 (3)(– 1) – (– 1)(4) y x 1 = = 8–3 –3+ 4 – 12 + 24 y 1 x = = 5 12 1 ⇒ y x 1 = and =1 5 1 12 ⇒ x = 5 and y = 12 Hence, numerator (x) of the fraction be 5 and denominator (y) be 12. (iii) Let the number of correct answers of Yash be x and number of wrong answres 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 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 Thus, we have following equations 3x – y = 0 4x – 2y – 50 = 0 Now, using cross mulitiplication method, –1 x –40 y 3 1 –1 –2 –50 4 x (– 1)(– 50) – (– 2)(– 40) 1 = (3)(– 2) – (4)(– 1) ⇒ ⇒ ⇒ x 50 – 80 x – 30 = x – 30 = y 1 = –6+4 – 160 + 150 y = 1 – 10 – 2 = 1 and y = 1 –2 – 10 – 2 –30 and y = –10 –2 –2 ⇒ x = 15 and y = 5 Hence, total number of questions = Number of correct answer + Number of incorrect answers = 15 + 5 = 20. (iv) ⇒ x = ⇒ So fraction = 5 12 –2 1 2 Let the speed of the car starting from point A be x km/hr and speed of the car starting from point B be y km/hr. Case I. When they travel in the same direction, let they meet at point P (Fig. 1). So, AP = distance covered by first car from point A to P = speed × time = x × 5 = 5x and BP = distance covered by second car from point B to P = speed × time = y × 5 = 5y Therefore, AB = AP – BP ⇒ 100 = 5x – 5y ⇒ 5x – 5y = 100 ⇒ x – y = 20 Case II. (Fig. 2) When they travel in opposite direction, let they meet at point Q. So. AQ = Distance covered by first car car from point B to Q = speed × time = x×1=x and B Q = Distance covered by second car from point B to Q = speed × time = y×1=y Threrefore AB = AQ + BQ 100 = x + y x + y = 100 Thus, we have following eqn. x – y = 20 x + y = 100 ⇒ x – y – 20 = 0 x + y – 100 = 0 –1 x –20 y 1 1 –1 1 +1 Now, +1 –100 x (– 1)(– 100 ) – (– 20)(+ 1) = y (– 20)(– 1) – (– 100)(1) = 1 (1)(+ 1) – (1)(– 1) ⇒ y x 1 = = 100 + 20 +1+1 – 20 + 100 ⇒ x 120 = y =1 80 2 ⇒ x 120 = y 1 =1 and 2 80 2 ⇒ x = 60 and y = 40 Hence, speed of the car that starts from point A = 60km/hr Speed of the car that starts from point B = 40 km/hr. (v) Let the dimension (i.e., the length and the breadth) of the rectangle be x units and y units respectively. Then, area of the rectangle = length × bredth = xy square units According to the question, xy – 9 = (x – 5) (y + 3) xy – 9 = xy + 3x – 5y – 15 3x + 3y – 61 = 0 and xy + 67 = xy + 2x + 3y + 6 2x + 3y – 61 = 0 Thus, we have following equation 3x – 5y – 6 = 0 2x + 3y – 61 = 0 Then, –5 x –6 y 3 1 –5 3 –61 x (– 5)(– 61) – (3)(– 6) 2 = 3 y (– 6 )(2 ) – (61)(3) 1 = (3)(3) – (2)(– 5) y 1 = 9 + 10 – 12 + 183 x = y = 1 = 323 171 19 323 x = 17 and y = 9 19 Hence, the dimensions (i.e, the length and the breadth) of the rectangle are 17units and 9 units respectively. x 305 + 18 = EXERCISE 3.6 Q.1. Solve the following pair of equations by reducing them to a pair of linear equations : 1 1 (i) 2 x + 3 y = 0 2 + 3 x y =2 (ii) 4 – 9 x y =0 1 + 1 13 3x 2 y = 6 (iii) 5 1 4 + 3y = 14 (iv) x – 1 + y – 2 = 2 x 3 – 4 y = 23 x (v) ⇒ Now 12 – 2v 12 – 6 = 6 = 3 3 3 u = 2 u = 2 1 x ⇒ (vi) 6x + 3y = 6xy 2x + 4y = 5xy =4 x = 1 3 1 1 ,y= is the solution of the 2 3 given equations. 1 1 + = –1 2(3x + y ) 2(3 x – y ) 8 (ii) Let 1 1 = u and y = v x Then, the given system of equation becomes Ans. (i) Let ...(i) 1 = u and x ...(ii) 1 y =v Then the given system of equation becomes 2u + 3v = 2 ...(i) 4u – 6v = –1 ...(ii) For making the co-efficient of 'u' in (i) and (ii) equal, we multiply (i) by 2 and then subtracting 4u + 6v = 4 4u – 9v = –1 – + + 15v = 5 5 = 1 1 =v= 15 3 3 Putting the value of 'v' in (i), we get 2u + 3v = 2 ⇒ 12 – 2v 3 Putting the value of 'u ' in (ii), we get u = = 3 y = Hence, x = 1 1 3 (viii) 3 x + y + 3x – y = 4 1 1 u+ v = 2 2 3 ⇒ 3u + 2v = 12 1u+ 1 v 13 and = 3 2 6 2u + 3v 13 ⇒ = 6 6 ⇒ 2u + 3v = 13 Thus, we have two equations 2u + 2v = 12 2u + 3v = 13 From (i), wehave 1 y ⇒ ⇒ 15 – 5 x + y x – y = –2 = 2 1 2 v = 3 ⇒ and 8x + 7 y = 15 xy 10 6 12 –+ x+ – 1y yx –– 2y u= =1 7x – 2 y =5 xy (vii) ⎛ 12 – 2v ⎞ 2 ⎜ 3 ⎟ + 3v = 13 ⎝ ⎠ ⇒ 24 – 6u + 9v = 39 ⇒ 5v = 39 – 24 ⇒ 5v = 15 ⇒ v = 3 Putting the value of 'v ' in (ii), we get ⇒ ⇒ ⇒ ⇒ v = 2u + 3 1 = 2 2 2u = 2 – 1 2u = 1 u = 1 2 Now u = 1 2 = 1 2 1 x ⇒ ⇒ ⇒ x = 2 x = 4 (squaring both side) and v = 1 3 = 1 3 1 y ⇒ ⇒ = 3 (by squaring both side) y ⇒ y = 9 Hence, x = 4, y = 9 is the solution of the given equation. 1 =a x Thus, the given equation becomes 4a + 3y = 14 ...(i) 3a – 3y = 23 ...(ii) For making the coefficient of 'y' in (i) and (ii), we multiply (i) by 4 and (ii) by 3 and then adding, we get 16a + 12y = 56 9a – 12y = 6 9 25aa - 12y = 125 (iii) Let 125 25 ⇒ a = 5 Putting the value of 'a' in (i), we get 4a + 3y = 14 Þ 4 (5) + 3y = 14 Þ 20 + 3y = 14 Þ 3y = –6 Þ y = –2 Now a = 5 ⇒ Þ Þ Hence, x = equation. a = 1 x = 5 x = 1 5 1 and y = – 2 is the solution of the given 5 (iv) Let 1 1 = u and y − 2 = v x −1 Thus, the given equaion becomes 5u + v = 2 ...(i) 6u – 3v = 1 (iii) For making the coefficient of 'v' in (i) and (ii) equal, we multiply (i) by 3 and adding, we get 15u + 3v = 6 6u – 3v = 1 21ua - 12y = 7 7 =1 21 3 Putting the value of u in (i), we get 5u + v = 2 ⇒ u = ⇒ ⎛1⎞ 5⎜ 3 ⎟ +v = 2 ⎝ ⎠ ⇒ 5 +v = 2 3 Now, u = 1 3 1 = x –1 ⇒ x –1 = 3 ⇒ x = 4 1 and v = 3 1 1 ⇒ y–2 = 3 ⇒ y–2 = 3 ⇒ y = 5 Hence, x = 4, y =5. (v) Considering equation ⇒ 7 x – 2y = 5 xy ⇒ 7x – 2y = 5xy Dividing bot sides by xy, we get 2y 5 xy 7x xy – xy = xy 7 – 2 y x = 5 Considering equation ⇒ 8x + 7 y xy = 15 ⇒ 8x + 7y = 15xy Dividing both sides by xy, we get = ⇒ 8 + 7 y x 8 x + 7y xy xy 15 xy xy = 15 ⇒ ...(ii) 1 1 = u, y = v. Then the given system of x equation bcomes. 7v – 2u = 5 ...(iii) 8v + 7u = 15 ...(iv) For making the coefficient of 'u' in (iii) and (iv) equal, we multiply (iii) by '7' and (iv) by '2' and then adding 49v – 14u = 35 16v + 14u = 30 65v + 0 = 65 ⇒ v = 1 Putting the value of 'v' in (iii), we get 7v – 2u = 5 ⇒ 7(1) – 2u = 5 ⇒ –2u = –2 ⇒ u = 1 Now, u = 1 Let 8 x 7y + xy xy ⇒ and 1 = 1 x x = 1 v = 1 ⇒ 1 y ⇒ 3y 6x xy + xy ⇒ 6 + 3 y x = 6 xy xy = 6 Considering equation 2x + 4y = 5xy Dividing both sides by xy, we get ...(i) = 15 ...(ii) 1 1 = u, y = v. Then the given system of x equation bcomes. 6v + 3u = 6 ...(iii) 2v + 4u = 5 ...(iv) For making coefficient of u in (iii) and (iv) equal, we multiply (iii) by 4 and (iv) by 3 and by subtracting, we get 24v + 12u = 24 6v + 12u = 15 – – – 18v =9 9 1 = 18 2 Putting the value of v in (iii), we get 6v + 3u = 6 ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ y = 1 Hence, x = 1, y = 1 is the solution of the given equation. (vi) Considering equation 6x – 5y = 6xy Dividing both sides by xy, we get 15 xy xy Let Now, = 1 8 + 7 y x = ⇒ v = ⎛1⎞ 6 ⎜ 2 ⎟ + 3u = 6 ⎝ ⎠ 3u = 6 u = 1 u =1 and 1 =1 x x =1 ⇒ ⇒ v = 1 y 1 2 1 2 y =2 = Hence, x = 1, y = 2 1 1 (vii) Let x + y = u and x – y = v Then the given system of equation becomes 10u + 2v = 4 ...(i) 15u – 5v = –2 ...(ii) For making the coefficient of 'v' in (i) and (ii), equal we multiply (i) by 5 and (ii) by 2 and then adding, we get 50u + 10v = 20 30u – 10v = –4 80ua - 12y = 16 16 = 1 ⇒ u = 80 5 Putting the value of u in (i), we get 10v + 2v = 4 ⇒ ⇒ ⇒ ⇒ Now, ⇒ ⇒ ⎛1⎞ 10 ⎜ 5 ⎟ + 2v ⎝ ⎠ 2 + 2v 2v v = 4 = 2 = 1 1 5 u = and 1 1 x + y = 5 x+y = 5 ⇒ ⇒ v =1 1 x–y = 1 x–y = 1 ..(iv) Thus, we have x+y = 5 ⇒ x = 5–y ...(v) Substituting the value of x in (iv), we get x–y = 1 ⇒ 5–y–y = 1 ⇒ 5 – 2y = 1 ⇒ –2y = 1 – 5 ⇒ –2y = –4 ⇒ y = 2 Now, substituting the value of y in (v), we get x = 5–y = 5–2=3 Hence, x = 3, y = 2 1 1 (viii) Let (3x + y ) = u and (3 x – y ) = v Then the given system of equations become u+v = 3 4 –1 8 –1 ⇒ u–v = 4 Thus, we have following equations and u – v 2 2 = 3 4 –1 u–v = 4 u+v = Adding equations (i) and (ii), we get 3 ⎛ –1 ⎞ 2u = 4 + ⎜ 4 ⎟ ⎝ ⎠ ⇒ 2u = 1 4 Putting the value of 'u' in eqn. (i), we get 3 u+v = 4 1 3 ⇒ +v = 4 4 3 1 ⇒ v = – 4 4 2 ⇒ v = 4 1 ⇒ v = 2 1 Now, u = 4 1 1 ⇒ 3x + y = 4 ⇒ 3x + y = 4 ...(iii) 1 and v = 2 1 1 ⇒ 3x + y = 2 ⇒ 3x – y = 2 ...(iv) Adding (iii) and (iv), we get 6x = 6 ⇒ x = 1 Putting the value of x in (iii), we get 3x + y = 4 ⇒ 3(1) + y = 4 ⇒ y = 1 Hence, x = 1, y = 1 is the solution of the given equation. Q.2. Formulate the following problems as a pair of linear equations and hence find their solutions : (i) Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. (ii) 2 women and 5 men can together finish a piece of embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work and also that taken by 1 man alone. (iii) Rohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining ⇒ = 4 3 –1 2 = 4 4 ...(i) ...(ii) u = by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. Ans. Let the speed of the boat in still water = x km/hr and speed of the current = y km/hr = (x – y) km/hr now, the speed of the boat downstream = (x + y) km/hr Case I. Time taken to cover 20 km downstream 20 = x+ y 20 2 = x+ y ⇒ x + y = 10 ...(i) Case II. Time taken to cover 4 km upstream ⇒ 4 = x– y ⇒ ⇒ From (i), we have 4 2 = x– y x – y =2 ...(ii) x = 10 – y Putting the value of 'x' in (ii), we get 10 – y – y = 2 ⇒ 10 – 2y = 2 ⇒ –2y = –8 ⇒ y =4 Now, substituting the value of y in (i), we get x + y = 10 ⇒ x + y = 10 ⇒ x =6 Hence, speed of rowing her boat in still water = 6 km/hr and speed of the current = 4km/hr. (ii) Let 1 woman can finish embroidery work in x days and 1 man can finish embroidery work in y days ∴ and 1 woman's 1 day's work = 1 x 1 1 man's 1 day's work = y Case I. 2 women's 1 day's work = 2 x and 5 5 men's 1 day's work = y According to question Taking 3+ 6 x y = 1 3 1 1 = u, y = v, we have x 1 4 1 3u + 6v = 3 For making the coefficient of 'u' we multiply equation (i) by 3 and equation (ii) by '2' and then subracting, we get 2u + 5v = 3 4 2 6u + 12v = 3 – – – 3 2 3v = – 4 3 9–8 ⇒ 3v = 12 1 ⇒ 3v = 12 1 ⇒ v = 36 Putting the value of 'v'in (i), we get 1 2u + 5v = 4 5 1 ⇒ 2u + = 36 4 1 – 5 ⇒ 2u = 4 36 9–5 ⇒ 2u = 30 4 ⇒ 2u = 36 1 ⇒ 2u = 9 1 ⇒ u = 18 1 Now, u = 18 1 1 ⇒ = x x ⇒ x = 18 6u + 15v = 1 36 1 1 y = 36 and v = ⇒ ⇒ y = 36 Hence, I woman alone can finish the work in 18 days and 1 man alone can finish the work in 36 days. (iii) Let the speed of the train and the bus be x km/hour and y km/hour respectively. Case I. When she travels 100 km by train and the remaining (300 – 100)km, i.e., 200 km by bus, the time 60 + 240 x y =4 Time = Distance Speed 1 4 1 ⇒ x + y = ...(i) [Dividing by 60] 15 Case II. When she travels 100 km by train and the remaining (300 – 100) km, i.e., 200 km by bus, the 25 time taken is 4 hours 10 minutes, i.e., hours. 6 100 + 200 x y = ⇒ 4+ 8 x y = 25 6 1 6 ...(ii) [Dividing by 25] Multiplying equation (i) by 2, we get get 2+ 8 2 ...(iii) x y = 15 Subtracting equation (iii) from equation (ii), we 1 – 2 = 1 6 15 30 ⇒ x = 60 Substituting this value of x in equation (iii), we x = 25 taken is 4 hours 10 minutes, i.e, hours. 6 ∴ ∴ get ⇒ 2 + 8 60 y = 2 15 1 – 8 30 y = 2 15 8 2 – 1 = 1 y = 15 30 10 ⇒ y = 80 EXERCISE 3.7 (Optional) So, the soltion of the equations (i) and (ii) is x = 60 and y =II. 80.When x – y = – 3 Q.1. The ages of two friends Ani and Biju differ Case ...(iv) Hence, the speed thefrom trainEq. is 60 km/hour by 3 years. Ani's father Dharam is twice as old as On subtracting Eq.of(iv) (ii), we get and the speed of the bus is 80 km/hour. Ani and Biju is twice as old as his sister Cathy. The ∴ 3x = 60 + 3 ages of Cathy and Dharam differ by 30 years. Find ⇒ 3x = 63 the ages of Ani and Biju. ⇒ x = 21 Ans. Let the ages of Ani and Biju be x years and On putting x = 21 in Eq. (iv), we get y years respectively. Then, according to the question, 21 – y = – 3 ⇒ y = 24 yr. Hence, age of Ani is 19 yr and ageof Biju is 16 yr x – y = ±3 ...(i) or age of Ani is 21 yr and age of Biju is 24 yr. Age of Ani's father Dharam = 2x years Q.2. One says, "Give me a hundred, friend! I y shall then become twice as rich as you." The other Age of Biju's sister = years. 2 replies, "If you give me ten. I shall be six as rich as According to the question, you." Tell me what is the amount of their (respective) capital? y 2x – = 30 Ans. Let the amounts of thier respective capitals 2 be Rs. x and Rs. y respectively. ⇒ 4x – y = 60 ...(ii) Then, according to the question, Case I. When x – y = 3 ...(iii) x + 100 = 2(y – 100) On substracting Eq. (iii) from Eq. (i), we get ⇒ x – 2y = –300 ...(ii) 3x = 57 ⇒ x = 19 yr and 6(x – 10) = y + 10 On putting x = 19 in Eq. (iii), we get ⇒ 6x – y = 70 ...(ii) 19 – y = 3 ⇒ y = 16 yrs From eqution (i), we have ∴ x = 19 yrs and y = 16 yrs x = 2y – 300 ...(iii) ⇒ Substitute the value of x in equation (ii), we get 6(2y – 300) – y = 70 ⇒ ⇒ ⇒ 12y – 1800 – y = 70 11y = 1870 y = 1870 = 170 11 Substituting the value of y in equation (iii), we get x = 2(170) – 300 x = 340 – 300 = 40 So, the solution of the equtions (i) and (ii) is x = 40 and y = 170. Hence, the amounts of their respective capitals are Rs. 40 and Rs. 170 respectively. Q.3. A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. Ans. Let the actual speed of the train be x km/hr and the actual time taken be y hours. Then, Distance covered = (x y) km ...(i) Case I. If the speed is increased by 10 km/hr then time of journey is reduced by 2 hrs. i.e. when speed is (x + 10)km/hr, time of journey is (y – 2) hrs. ∴ ⇒ Distance covered = (x + 10) (y – 2) xy = xy – 2x + 10y – 20 [Using (i)] ⇒ 2x – 10y = –20 Case II. When the speed is reduced by 10 km/hr then time of journey is by 3 hrs. i.e. when speed is (x – 10) km/hr, time of journey is (y +3) hrs. ∴ Distance covered = (x – 10) (y + 3) xy = (x – 10) (y +3) [Using (i)] ⇒ ⇒ xy = xy + 3x – 10y – 30 –3x + 10y = – 30 Thus, we have following eqn. 2x – 10y = – 20 ...(i) –3x + 10y = – 30 ...(ii) Since the coefficient of y in both the equations are same, so we can eliminate it directly by adding i.e. 2x – 10y = – 20 –3x + 10y = – 30 –x = –50 x = 50 Putting the value of x in (i), we get 2x – 10y = –20 ⇒ 2(50) – 10y = –20 ⇒ 100 – 10y = –20 ⇒ –10y = –120 ⇒ y = 12 Hence, the distance covered by the train = (xy) km = (50 × 12) km = 600 km. Q.4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in row, there would be 2 rows more. Find the number of students in the class. Ans. Let the total no. students in each row be x and the total no. of rows be y ∴ Total no. of sutdents in the class = xy ...(i) Case I. If 3 students are extra in a row, there would be 1 row less, i.e. total no. of students in each row = (x + 3) and the total no. of rows = (y – 1) ∴ Total no. students in the class = (x + 3) (y – 1) ...(ii) Comparing (i) and (ii), we get xy = (x + 3) (y – 1) ⇒ xy = xy – x + 3y – 3 ⇒ x – 3y = –3 Case II. If 3 students are less in a row, there would be 2 rows more, i.e. Total no. of students in each row = (x – 3) and the total no. of rows = (y + 2) ∴ Total no. of students in the class = (x – 3) (y + 2) ...(iii) Comparing (i) and (iii), we get xy = (x – 3) (y + 2) ⇒ xy = xy + 2x – 3y – 6 ⇒ –2x + 3y = –6 Thus, we have following equations x – 3y = –3 ...(iv) –2x + 3y = –6 ...(v) From (iv), we have x – 3y = –3 ⇒ x = 3y – 3 ...(vi) Substituting the value of (vi) in (υ), we get –2x + 3y = –6 ⇒ –2(3y – 3) + 3y = –6 ⇒ –6y + 6 + 3y = –6 ⇒ –3y = –12 ⇒ y =4 Now, substituting the value of 'y' in (vi), we get x = 3y – 3 = 3(4) – 3 = 12 – 3 = 9 Hence, total no. of students in the class = xy = 9 × 4 = 36. ∠A + ∠B). Q.5. In a Δ ΔABC, ∠C = 3 ∠B = 2 (∠ Find the three angles. Ans. Let ∠A = xº, ∠B = yº. Then, ∠C = 3∠ ∠B ⇒ C = 3yº ∠C = ∠3 B = 2 (∠ ∠A + ∠B). We have,∠ ...(i) ∠B = 2(∠ ∠A + ∠B) From (i) Now, 0 3∠ ⇒ 3y = 2(x + y) ⇒ y = 2x Since A, B and C are angles of a triangle ∠A + ∠B + ∠C = 180º ∴ ⇒ x + y + 3y = 180 ⇒ x + 4y = 180 Putting y = 2x in equation (ii), we get x + 4(2x) = 180 ⇒ x + 8x = 180 ⇒ 9x = 180 ⇒ x = 20º Putting the value of x equation (i), we get in y = 40º ∠ A = 20º, ∠B = 40º Hence, ∠C = 3yº = (3 × 40º) = 120º and Q.6. Draw the graph of the equtions 5x – y = 5 and 3x – y = 3. Determine the coordiantes of the vertices of the triangle formed by these lines and the y axis. Ans. The given equations are 5x – y = 5 ...(i) 3x – y = 3 ...(ii) For equation (i), we have y = 5x – 5 Thus, we have following table : For equation (ii), we have y = 3x – 3 Thus, we have following table : x 0 1 y –3 0 Q.7. Solve the following pair of linear equations: (i) px + qy = p – q qx – py = p + q (ii) ax + by = c bx + ay = 1 + c x – y =0 a b ax + by = a2 + b2 (iv) (a – b)x + (a + b) y = a2 – 2ab – b2 (a + b) (x + y) = a2 + b2 (v) 152x – 378y = – 74 –378x + 152y = –604 Ans. We have px – qy = p – q ...(i) qx – py = p + q ...(ii) From (i), we get (iii) x = p – q – qy p Putting the value of 'x' in (ii), we get ⎛ p – q – qy ⎞ py ⎟– ⇒ q⎜ = p+q p ⎝ ⎠ 1 x 0 1 ⇒ y –5 0 ⇒ q( p – q – qy ) – p 2 y = p+q p qp – q2 –q2y – p2y = p(p + q) ...(iii) ⇒ ⇒ –q2y – p2y = p(p + q) – dp + q2 –y (q2 + y2) = p2 + pq – dp + q2 p2 + q2 ⇒ –y = =1 p2 + q 2 ⇒ y = –1 Putting thte value of 'y' in (iii), we get p – q + q(– 1) x = p p ⇒ x = = p ⇒ x =1 Hence, the solution is x = 1, y = –1 (ii) We have, ax + by = c ...(i) bx + ay = 1 + c ...(ii) From (i), we have c – by x = a Putting the value of (iii) in (ii), we get ...(iii) ⎛ c – by ⎞ b⎜ ⎟ + ay = (1 + c) ⎝ a ⎠ p –q+q p ⇒ ⇒ ⇒ ⇒ bc – b 2 y + ay = (1 + c) a 2 2 bc – b y + a y a bc – b2y + a2y –b2y + a2y y(–b2 + a2) = (1 + c) = a(1 + c) = a(1 + c) – bc = a + ac – bc a + ac – bc a2 – b2 Putting the value of 'y' in (iii), we get ⇒ y = x = c – b⎛ a + ac – bc ⎞ ⎜ ⎟ 1 ⎝ a2 – b2 ⎠ a ( ) c a 2 – b 2 – b(a + ac – bc ) a 2 – b2 = a 1 = a 2 c – b 2 c – ab – abc + b 2 c 1 × a a2 – b2 = a 2 c – ab – abc a a 2 – b2 ( ) a(ac – b – bc ) a2 – b2 ac – b – bc ⇒ x = a2 – b2 Hence, the solution is ac – b – bc x = a2 – b2 a + ac – bc and y = a2 – b2 y x (iii) – =0 ...(i) a b ax + by = a2 + b2 ...(ii) The given pair of their equations is y x = a b b x ⇒ y = ...(iii) a Substituting the value of y in equation (ii), we = get ⎛b ⎞ ax + b ⎜ a x ⎟ = a2 + b2 ⎝ ⎠ b2 x = a2 + b2 a a2x + b2x = a(a2 + b2) x(a2 + b2) = a(a2 + b2) ax + ( ) a a2 + b2 =a a2 + b2 Substituting the value of x in (iii), we get b y = x a b y= ×a–b a Hence, x = a, y = b. (iv) The given system of equations may be written as (a – b) x + (a + b) y = a2 – 2ab – b2 (a + b) x + (a + b) y = a2 + b2 (a – b) x – (a + b) x = – 2ab – 2b2 ⇒ x[(a – b) – (a + b)] = –2ab –2b2 ⇒ x[(a – b – a – b)] = –2ab –2b2 ⇒ x(– 2b) = –2ab –2b2 ⇒ x = a+b Putting the value of x in (ii), we get (a + b)x + (a + b)y = a2 + b2 ⇒ (a + b) (a + b) + (a + b)y = a2 + b2 ⇒ (a + b)y = (a2 + b2) – (a + b)2 ⇒ (a + b)y = (a2 + b2) – (a2 + b2 + 2ab) ⇒ (a + b)y = a2 + b2 – (a2 + b2 + 2ab) x = ⇒ (a + b)y = –2ab ⇒ y= ∠A + ∠C = 180º –2ab a+b Hence, x = a b and y = –2ab a+b (v) 152x – 378y = –74 –378x + 152y = –604 The given pair of linear equations is 152x – 378y = –74 ...(i) –378x + 152y = –604 ...(ii) Adding equation (i) and equation (ii), we get –226x – 226y = –678 x+y = 3 ...(iii) [Dividing throughout by –226] Subtracting equation (ii) from equation (i), we ⇒ ⇒ ⇒ 4y + 20 + 4x = 180º 4x + 4y = 160º x + y = 40º ...(i) [Dividing throughout y 4] ∠A + ∠C = 180º and ⇒ 3y – 5 + 7x + 5 = 180º ⇒ 7x + 3y = 180º ...(ii) From equation (i), we have y = 40 – x ...(iii) Substituting this value of y in equation (ii), we get 530x – 530y = 530 x–y = 1 [Dividing throughout by –226] Adding equation (iii) and equation (iv), we get 2x = 4 ⇒ 4 ⇒ x = =2 2 Subtracting equation (iv) from equation (iii), we get ⇒ ⇒ get 60 = 15º 4 Substituting x = 15 in the equation (iii), we get y = 40 – x = 40 – 15 = 20º Hence, required angles be ∠A = 4y + 20 = 4 × 25 + 20 = 120º ∠B = 3y – 5 = 3 × 25 – 5 = 75 – 5 = 70º ∠C = 4x = 4 × 15 = 60º ∠D = 7x + 5 = 7 × 15 + 5 = 105 + 5 = 110º ⇒ 2y = 2 2 =1 2 Hence, the solution of the given pair of linear equations is x = 2, y = 1. Q.8. ABCD is cyclic quadrilateral (See Fig.). Find the angles of the cyclic quadrilateral. Ans. We know that the opposite angles of a cyclic quadrilateral are supplementary, therefore, ⇒ 7x + 3(40 – x) = 180º 7x + 120 – 3x = 180º 4x = 60 y = x = Additional Questions Q.1. For all real values of c, the pair of equations x – 2y = 8 5x – 10y = c have a unique solution. Justify whether it is true or false. Ans. Given equations are x – 2y = 8 and 5x – 10y = c Here a1 1 = a2 5 b1 1 = b2 5 ∴ a1 b1 = a2 b2 which is not the case of unique solution. ∴ Given statement is false. Q.2. The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not. Ans. The line represented by x = 7 is of the form x = k, which is the form of a line parallel to y - axis. ∴ Given statement is false. Q.3. Two straight paths are represented by the equations x – 3y = 2 and – 2x + 60y = 5. Check whether the paths cross each other or not. Ans. Two straight paths are represented by : x – 3y =2 and –2x + 6y = 5 or, x – 3y – 2 = 0 and – 2x + 5y – 5 = 0 The paths will cross each other if the equations intersect. Now, a1 1 1 = =− a2 − 2 2 b1 −3 1 = =− b2 6 2 ∴ a1 b1 = a2 b2 ∴ The paths will not intersect i.e., paths will not cross each other. Q.4. Write a pair of linear equations which has the unique solution x = –1, y = 3. How many such pairs can you write ? Ans. One of such pairs can be x – y = – 4 and 2x + 3y = 7 We can write infinitely many pairs Q.5. Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How any such lines can we find ? Ans. Given equations are : x+y =2 ...(i) and 2x – y = 1 ...(ii) ∴ x+y =2 (i) × 1 2x – y = 1 (ii) × 1 Adding 3x = 3 ⇒ x =1 Substituting the value of x in (i), we get y = 1. ∴ x = 1 and y - 1 is the solution of the given pair of linear equations. Now, to write equation of a line we must get relationship between x and y for the point (1, 1) which is x = y. Likewise we can write many such equations by multiplying both sides of the above equations by any non-zero constant. Q.6. The angles of a triangle are x, y and 40°. The difference between the two angles x and y is 30°. Find x and y. Ans. The angles of a triangle are x , y and 40°. therefore, x + y = 40° = 180° by angle sum properly of a Δ. Hence, we obtain corresponding equation as x + y = 140 ...(i) Also, the difference between the two angles x and y is 30°. Therefore, we obtain the following equation : x – y = 30 ... (ii) Adding (i) and (ii), we get 2x = 170 170 2 = 85 Subtracting (ii) from (i), we get 2y = 110 ⇒ y = 55 ∴ x = 85° and y = 55°. Q.7. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. Ans. Let the age of the father is x years and the sum of the ages of his two children is y years. ∴ According to the questions : x = 2y ... (i) After 20 years, the age of the father will be x + 20 and the sum of ages of his two children will be y = 40. ∴ According to the questions : x + 20 = y + 40 ⇒ x – y = 20 ... (ii) Substituting the value of x from (i) and (ii), we get 2y – y = 20 ⇒ y = 20 Substituting the value of y in (i), we get x = 2 × 20 = 40 ∴ The age of the father is 40 years. Q.8. There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A and B. But if 20 students are sent from B to A, the ⇒ x = number of students in A become double the number of students in B. Find the number of students in the two halls. Ans. Let the number of students in hall A be x and the number of students in hall B be y. ∴ According to the question, x – 10 = y + 10 ⇒ x = 20 + y ...(i) and 2(y – 20) = x + 20 ⇒ 2y – 60 = x ...(ii) Equating the two values of x, we get 20 + y = 2y – 60 ⇒ y = 80 Substituting the value of y in (i), we get x = 100 ∴ Number of students in hall A is 100 and the number of students in hall B is 80. Q.9. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs. 22 for a book kept for six days, while Anand paid Rs. 16 for the book kept for four days. Find the fixed charges and the charge for each extra day. Ans. Let the fixed charge be Rs. x for two days. Let the additional charge for each day thereafter be Rs. y per day. Now, Latika paid Rs., 22 for a book kept for six day. ∴ x + 4y = 22 ... (i) and Anand paid Rs. 16 for a book kept for four days. ∴ x + 2y = 16 ...(ii) Subtracting (ii) from (i), we get 2y = 6 ⇒ y=3 Substituting the value of y in (i), we get x + 4 × 3 = 22 ⇒ x = 10 ∴ Fixed charges = Rs. 10 and charge per day after two days = Rs. 3. Q.10. For which values of a and b, will the following pair of linear equations have infinitely many solutions ? x + 2y = 1 (a – b) x + (a + b) y = a + b – 2 Ans. Given equations are x + 2y= 1 or x + 2y – 1 = 0 ....(i) and (a – b) x + (a + b)y = a + b – 2 or (a – b)x + (a + b)y – a –b + 2 = 0 ...(ii) ∴ a1 = 1, b1 = 2, c1 = –1 a2 = a – b2 = a + b, c2 = –a –b + 2 Now, for infinitely many solutions, a1 b1 c2 = = a2 b2 c2 1 2 −1 = = a −b a +b −a −b + 2 ⇒ a + b = 2a – 2b; –2a – 2b + 4 = –a –b; –a – b + 2 = –a +b ⇒ a – 3b = 0; –a – b + 4 = 0; –2b + 2 = 0 ⇒ a – 3b = 0; a + b = 4; b = 1 ⇒ a = 3; b = 1 ∴ For infinitely many solutions a = 3 and b = 1. ⇒ Multiple Choice Questions Q.1. Two lines is given to be parallel. The equation of one of the lines is 4x + 3y = 14. The eqution of the second line can be (a) 3x + 4y = 14 (b) 8x + 6y = 28 (c) 12x + 9y = 42 (d) – 12x = 9y Ans. (d) Q.2. If x = a, y = b is the solution of th pair of equation x – b = 2 and x + y = 4, then the respective values of a and b are : (a) 3, 5 (b) 5, 3 (c) 3, 1 (d) –1, – 3 Ans. (c) Q.3. The pair of equations : 3x + 4y = 18 16 y = 25 3 (a) No solution (b) Unique solution (c) Infinitely many solution (d) Can not say anything. Ans. (c) Q.4. If 31x + 43y = 117 and 43x + 31y = 105, then the value of (x + y) is : 4x + (a) –3 (b) 1 3 Lifeskills' Complete NCERT Solutions Class-X Mathematics 65 (b) − 1 3 (d) 3 Ans. (d). Q.5. If 19x – 17y = 55 and 17x – 19y = 53, then the value of (x – y) is : (a) –3 1 3 (d) 5 (b) (b) 3 Ans. (d) Q.6. If (6, k) is a solution of the equation 3x + y – 22 = 0, then the value of k is : (a) –4 (b) 4 (c) 3 (d0 –3 Ans. (b) Q.7. Rs. 4,900 were divided among 150 children. If each girl get Rs. 50 and a boy get Rs. 25, then the number of boys is : (a) 100 (b) 102 (c) 104 (d) 105 Ans. (c) Q.8. If the pair of equations 2x + 3y = 5 and 5x + 15 y = k represent two coincident lines, then 2 the value of k is : (a) -25 2 (b) – 5 (c) 25 2 (d) –5 2 Ans. (c) 2x Q.9. If 3x – 5y = 1, x − y = 4, then the value of (x + y) is (a) 3 (c) 1 3 (b) –3 (d) − 1 3 Ans. (a) Q.10. If 2x + 3y = 0 and 4x – 3y = 0, then x + y equal: (a) 0 (b) –1 (c) 1 (d) 2 Ans. (a)