Question

1. Solve each system of linear equations below, then check your work.

Answer 1

Answer:

$a) \: x = - 3 \: and \: y = 2 \\ b) \: x = 1 \: and \: y = - \frac{5}{4} \\ c) \: \: x = 25 \: and \: y = 0 \\ d ) \: x = 10 \: and \: y = - 12$

Step-by-step explanation:

$question (a) \: \: \: 3x - y = - 11 \\ \: \: \: - x + y = 5 \\ \\ 2x = - 6 \\ \frac{2}{2} x = \frac{ - 6}{2} \\ x = - 3$

$\\ since \: x = - 3 \: then \: y = \\ \\ - x + y = 5 \\ - ( - 3) + y = 5 \\ y = 5 - 3 \\ y = 2$

$\\ (check) \\ recall \: that \: x = - 3 \: and \: y = 2 \\ \\ - x + y = 5 \\ - ( - 3) + 2 = 5 \\ 3 + 2 = 5 \\ 5 = 5$

$question (b) \: \: \: -2y + 3 = 4x + 2 \\ 6x + 4y = 1 \\ \\ ( - 4x - 2y = - 1) \times 2 \\\: \: \: 6x + 4y = 1 \\ \\ - 8x - 4y = - 2 \\ 6x + 4y = 1 \\ \\ - 2x = - 2$

$\frac{ - 2}{ - 2} x = \frac{ - 2}{ - 2} \\ x = 1 \\ \\ since \: x = 1 \: then \: y = \\ \\ 6x + 4y = 1 \\ 6(1) + 4y = 1$

$4y = - 6 + 1 \\ \frac{4}{4} y = - \frac{ 5}{4} \\ y = - \frac{ 5}{4} \\ \\ (check) \\ - 2y + 3 = 4x + 2$

$- 2( - \frac{5}{4} ) + 3 = 4(1) + 2 \\ \frac{5}{2} + 3 = 6\\ 6 = 6$

$question (c) \: \: \: 32y - x = - 25 \\ \: \: \: 5x = 100 + x - 8y \\ \\ 32y - x = - 25 \\ 8y + 5x - x = 100 \\ \\ (32y - x = - 25) \times 4 \\ 8y + 4x = 100$

$128y - 4x = - 100 \\ 8y + 4x = 100 \\ 136y = 0 \\ y = 0 \\ \\ since \: y = 0 \: then \: x = \\ \\ 32y - x = - 25 \\ 32(0) - x = - 25 \\ - x = - 25 - 0 \\ \frac{ - 1}{ - 1} x = \frac{ - 25}{ - 1} \\ x = 25$

$(check) \\ 5x = 100 + x - 8y \\ 5(25) = 100 + (25) - 8(0) \\ 125 = 125 - 0 \\ 125 = 125$

$question (d) \: \: \: 2y + 3x = 6 \\ \: \: \: 4x + 5y + 20 = 0 \\ \\ ( 2y + 3x = 6) \times 4 \\ (5y + 4x = - 20) \times - 3 \\ \\ 8y + 12x = 24 \\ - 15y - 12x = 60$

$- 7y = 84 \\ \frac{ - 7}{ - 7} y = \frac{84}{ - 7} \\ y = - 12 \\ \\ since \: y = - 12 \: then \: x = \\ \\ 2y + 3x = 6 \\ 2( - 12) + 3x = 6 \\ - 24 + 3x = 6 \\ 3x = 6 + 24 \\ x = 10 \\ \\ (check) \\ 4x + 5y + 20 = 0$

$4(10) + 5( - 12) + 20 = 0 \\ 40 - 60 + 20 = 0 \\ - 20 + 20 = 0 \\ 0 = 0$