The problem asks us to solve two systems of linear equations without graphing. System a: $-5x + 3y = -8$ $3x - 7y = -3$ System b: $-8x - 2y = 24$ $5x - 3y = 2$

1. Problem Description

The problem asks us to solve two systems of linear equations without graphing.

System a:

$-5x + 3y = -8$

$3x - 7y = -3$

System b:

$-8x - 2y = 24$

$5x - 3y = 2$

2. Solution Steps

System a:

$-5x + 3y = -8$ (1)

$3x - 7y = -3$ (2)

Multiply equation (1) by 3 and equation (2) by 5:

$3(-5x + 3y) = 3(-8)$

$5(3x - 7y) = 5(-3)$

$-15x + 9y = -24$ (3)

$15x - 35y = -15$ (4)

Add equation (3) and equation (4):

$(-15x + 9y) + (15x - 35y) = -24 + (-15)$

$-26y = -39$

$y = \frac{-39}{-26} = \frac{3}{2}$

Substitute $y = \frac{3}{2}$ into equation (1):

$-5x + 3(\frac{3}{2}) = -8$

$-5x + \frac{9}{2} = -8$

$-5x = -8 - \frac{9}{2} = -\frac{16}{2} - \frac{9}{2} = -\frac{25}{2}$

$x = \frac{-\frac{25}{2}}{-5} = \frac{25}{10} = \frac{5}{2}$

System b:

$-8x - 2y = 24$ (1)

$5x - 3y = 2$ (2)

Multiply equation (1) by -3 and equation (2) by -2:

$-3(-8x - 2y) = -3(24)$

$-2(5x - 3y) = -2(2)$

$24x + 6y = -72$ (3)

$-10x + 6y = -4$ (4)

Subtract equation (4) from equation (3):

$(24x + 6y) - (-10x + 6y) = -72 - (-4)$

$24x + 6y + 10x - 6y = -72 + 4$

$34x = -68$

$x = \frac{-68}{34} = -2$

Substitute $x = -2$ into equation (1):

$-8(-2) - 2y = 24$

$16 - 2y = 24$

$-2y = 24 - 16 = 8$

$y = \frac{8}{-2} = -4$

3. Final Answer

System a: $x = \frac{5}{2}, y = \frac{3}{2}$

System b: $x = -2, y = -4$