Questions and Answers #2 Systems of Equations

Questions and Answers Sheet 2 Systems of Equations Question #1 Solve the system of equations. 0.8 x + 0.3 y = 1.3  0.3x + 0.2 y = 1.1 Answer: 0.8x + 0.3 y = 1.3   0.3x + 0.2 y = 1.1 8 x + 3 y = 13 {Multiply both equations by ′′′′10′′′′ }  3x + 2 y = 11 In matrix form. 8 3  x  13 3 2  y  = 11      𝑅1 → 2𝑟1 and 𝑅2 → 3𝑅2 16 6  x   26 1 0  x   −1  9 6  y  =  33 R1 → R1 − R2 9 6  y  = 33           ⇒ 𝑥 = −1 and 9𝑥 − 6𝑦 = 33 ⇒ 3𝑥 + 2𝑦 = 11 ⇒ 2𝑦 = 11 − 3𝑥 ⇒ 2𝑦 = 11 − 3(−1) ⇒ 2𝑦 = 14 ⇒ 𝑦 = 7 So, 𝑥 = −1 and 𝑦 = 7 Question #2 Consider the following system of equations 2 x + 3 y = 9  3 x + 5 y = 14 Answer: 2𝑥 + 3𝑦 = 9 3𝑥 + 5𝑦 = 14 2 3  x  9 (−1) 3 5 x  y  = b 14 Inverse of the coefficient matrix = 𝐴       1  d −b  A( −1) = ad-bc −c a  A = 1  5 −3  5 −3 = 10-9 −3 2  −3 2  So, 𝑥 = 𝐴(−1) 𝑏  5 −3  9   45 − 42  3 =   =   =    x = 3; y = 1 −3 2  14  −27 + 28 1 Question #3 Solve the system of equations given below:  y = 9 + 3z   y = 51 − 3 z Answer:  y = 9 + 3z (i)   y = 51 − 3z → Put 𝑦 = 51 − 3𝑧 in equation (i) 51 − 3𝑧 = 9 + 3𝑧 51 − 9 = 6𝑧 42 = 6𝑧 𝑧=7 𝑦 = 51 − 3 ∗ 7 = 51 − 21 𝑦 = 30 Question #4 Solve the following system of equations for a and b: 4a + 2b = 20  15a + 5b = 60 Answer: 4𝑎 + 2𝑏 = 20 → 𝑎 = 20-26 4 = 5 − 0.5𝑏 15𝑎 + 5𝑏 = 60(𝑖𝑖) Put value of a in equation (ii) 15(5 − 0.5𝑏) + 5𝑏 = 60 75 − 7 ∗ 5𝑏 + 5𝑏 = 60 75 − 2.5𝑏 = 60 2.5𝑏 = 15 15 𝑏= ⇒6⇒𝑏=6 2*5 𝑎 = 5 − 0.5 ∗ 6 𝑎=2 Question #5 Solve the system of equations by the method of substitution 4 x + 5 y = 7  9 y = 31 + 2 x Answer: 4 x + 5 y = 7 − ( I )  9 y = 31 + 2 x − ( II ) Now put the value of y in equation(I) 4𝑥 + 5 ( 31+2x 9 )=7 36𝑥 + 10𝑥 + 155 = 63 46𝑥 = 63 − 155 𝑥 = −92/46 𝑥 = −2 Now put the value of x in equation (II) to get the value of y 𝑦= 31+2(-2) 9 = 27/9 𝑦 = 3 ⇒ 𝑥 = −2; 𝑦 = 3 Question #6 Solve the system of equations for z and y:  y = 8 + 2z   y = 56 − 6 z Answer: 𝑦 = 8 + 2𝑧 𝑦 = 56 − 6𝑧 Now we will equate the two equations equally. 8 + 2𝑧 = 56 − 6𝑧 2𝑧 + 6𝑧 = 56 − 8 8𝑧 = 48 𝑧 = 48 / 8 = 6 𝑦 = 8 + 2𝑧 = 8 + 2 ∗ 6 = 8 + 12 = 20 𝑦 = 20 and 𝑧 = 6 Question #7 Solve the system of equations for a and b: 9a + 3b = 30  8a + 4b = 24 Answer: 9𝑎 + 3𝑏 = 30(1) 8𝑎 + 4𝑏 = 24(2) We will multiply equation 1 by 4 and equation 2 by 3. 36𝑎 + 12𝑏 = 120(3) 24𝑎 + 12𝑏 = 72(4) Now we will subtract 4 from 3 12𝑎 = 48 𝑎 = 48 / 12 = 4 Now put the value of a in equation 1 to get b. 9(4) + 3𝑏 = 30 36 + 3𝑏 = 30 36 − 30 = −3𝑏 6 = −3𝑏 𝑏 = −6/3 = −2 𝑎 = 4 and 𝑏 = −2 Question #8 Solve the following system of equations for z and y :  y = 10 + 2 z   y = 38 − 5 z Answer:  y = 10 + 2 z 10 + 2 z = 38 − 5z  z = 4; y = 18   y = 38 − 5z 7 z = 28 Question #9 Solve the following system of equations for a and b: 4a + 2b = 20  15a + 5b = 60 Answer: 4a + 2b = 20 20-2b  4a = 20 − 2b = a =  4 15a + 5b = 60 15 [ 20-2b 4 ] + 5𝑏 = 60 300 − 30𝑏 + 20𝑏 = 60 ∗ 4 = 240 60 = 10𝑏 ⇒ 𝑏 = 6 and 𝑎 = 2 Question #10 Is (3,3) a solution to this system of equations? 5 x + y = 18  2 x + 4 y = 18 Answer: 5 x + y = 18  2 x + 4 y = 18 5𝑥 + 𝑦 = 18 -(i) 2x 2 + 4y 2 = 18 2 (divide by 2)) 𝑥 + 2𝑦 = 9-(ii) + 10𝑥 + 2𝑦 = 36 (multiply (i)by 2) 9𝑥 = 27 𝑥 = 27/9 = 3 𝑥=3 from equation(i) 5 ∗ 3 + 𝑦 = 18 𝑦 = 18 − 15 = 3 𝑦=3 Hence, (3,3) is a solution of this system of equations. Question #11 Is(-10,5) a solution to this system of equations? 8 x + 15 y = −5  7 x − 11y = −13 Answer: 8 x + 15 y = −5 − (i)  7 x − 11y = −13 − (ii) Multiply (i) by 7 and (ii) by 8 than substract: 56𝑥 + 105𝑦 = −35 56𝑥 − 88𝑦 = −104 Substraction: 193𝑦 = 69 𝑦 = 69/193 Since 𝑦 = 69/193, than (-10,5) is not a solution. Question #12 Solve the system of equations:  x = −2   6 x − y = −6 Answer:  x = −2   6 x − y = −6 Put 𝑥 = −2 in 6𝑥 − 𝑦 = −6 6(−2) − 𝑦 = −6 −12 − 𝑦 = −6 −12 + 6 = 𝑦 −6 = 𝑦 𝑦 = −6 (𝑥, 𝑦) = (−2, −6) Question #13 Solve the system of equations:  y = 4x + 9   y = 2x + 7 Answer:  y = 4 x + 9 − (i)   y = 2 x + 7 − (ii) equate y from (i) and (ii), 4𝑥 + 9 = 2𝑥 + 7 4𝑥 − 2𝑥 = 7 − 9 2𝑥 = −2 𝑥 = −1 from(i), 𝑦 = 4(−1) + 9 𝑦−4+9 𝑦=5 (𝑥, 𝑦) = (−1,5) Question #14 Solve this system, please. 2 x = 10 − y  3x − y = 5 Answer: 2 x = 10 − y  3x − y = 5 Solve equation for y. 𝑦 = 3𝑥 – 5 Substitute the equation (3) into (1) and solve for x. 2𝑥 = 10– (3𝑥– 5) 2𝑥 = 10 – 3𝑥 + 5 2𝑥 + 3𝑥 = 10 + 5 5𝑥 = 15 𝑥 = 3 Substitute 𝑥 = 3 into equation (3) and solve for y. 𝑦 = 3𝑥 – 5 𝑦 = 3 ∗ 3– 5 𝑦 = 9– 5 𝑦 = 4 Hence, the answer is (3; 4) Question #15 Solve this system, please. 4 x + 5 y = −4  3x + 8 y = −20 Answer: 4 x + 5 y = −4  3x + 8 y = −20 Multiply equation (1) by 3 and equation (2) by 4. 12 x + 15 y = −12  12 x + 32 y = −80 Subtract equation (3) from equation (4) 17𝑦 = −68 𝑦 = −68/17 𝑦 = −4 Substitute 𝑦 = −4 into equation (1) and solve for x. 4𝑥 + 5 ∗ (−4) = −4 4𝑥 – 20 = −4 4𝑥 = −4 + 20 4𝑥 = 16 𝑥 = 4 Hence, the solution is (4 ; -4) Question #16 Solve, please: 1  0.7 x − 2 y = 9.4  0.9 x + 7 y = 0 10  Answer: 1  0.7 x − 2 y = 9.4  0.9 x + 7 y = 0  10 Multiply equations (1) and (2) by 10. 7 x − 5 y = 94  9 x + 7 y = 0 Multiply equation (3) by 7 and equation (4) by 5. 49 x − 35 y = 658  45x + 35 y = 0 Add equations. 94𝑥 = 658 𝑥 = 658/94 𝑥 = 7 Substitute 𝑥 = 7 into equation (1) and solve for y. 0.7 × 7 – 1/2𝑦 = 9.4 4.9 – 1/2𝑦 = 9.4 𝑦 = 2(4.9– 9.4) 𝑦 = −9 So, the solution is (7;-9) Question #17 Solve the equation system: x + y = 6  2y x = Answer: x + y = 6  2y x = Substitute equation (2) into equation (1) and solve for y. 𝑥 + 𝑦 = 6 2𝑦 + 𝑦 = 6 3𝑦 = 6 𝑦 = 2 Substitute 𝑦 = 2 into equation (2) and solve for x. 𝑥 = 2𝑦 𝑥 = 2 ∗ 2 𝑥 = 4 The answer is (4; 2) Question #18 Solve the system of equations, please: = −2 x y  4 x + 2 y = 0 Answer: = −2 x y  4 x + 2 y = 0 Substitute equation (1) into equation (2) and solve for x. 4𝑥 + 2𝑦 = 0 4𝑥 + 2 ∗ (−2𝑥) = 0 4𝑥 – 4𝑥 = 0 0 = 0 The given system has infinite number of solutions that are all of the ordered pairs along 𝑦 = −2𝑥 or 4𝑥 + 2𝑦 = 0. Question #19 3x − 4 y = −14  5x + 7 y = 45 Solve the equation system: Answer: 3x − 4 y = −14  5x + 7 y = 45 Multiply equation (1) by 5 and equation (2) by 3. 15x − 20 y = −70  15x + 21y = 135 Subtract equation (3) from equation (4) 41𝑦 = 205 𝑦 = 205/41 𝑦 = 5 Substitute 𝑦 = 5 into equation (1) and solve for x. 3𝑥 – 4 × 5 = −14 3𝑥 – 20 = −14 3𝑥 = −14 + 20 3𝑥 = 6 𝑥 = 2 The solution is (2 ; 5) Question #20 Solve the system, please: 3x +  2 x − y = 9 y = 1 Answer: 3x +  2 x − y = 9 y = 1 Add equation (1) and equation (2) 5𝑥 = 10 𝑥 = 2 Substitute 𝑥 = 2 into the equation (1) and solve for y. 3 ×2 + 𝑦 = 9 6 + 𝑦 = 9 𝑦 = 9−6 𝑦 = 3 Then the solution is (2; 3)