Question

use substitution to solve the system.
$5x + 2y = -13$
$3y - 6 = x$
$x = \square$
$y = \square$

Explanation:

Step1: Substitute \( x = 3y - 6 \) into the first equation

We have the first equation \( 5x + 2y = -13 \) and \( x = 3y - 6 \). Substitute \( x \) in the first equation:
\( 5(3y - 6) + 2y = -13 \)

Step2: Simplify and solve for \( y \)

Expand the left - hand side:
\( 15y-30 + 2y=-13 \)
Combine like terms:
\( 17y-30=-13 \)
Add 30 to both sides:
\( 17y=-13 + 30=17 \)
Divide both sides by 17:
\( y = 1 \)

Step3: Substitute \( y = 1 \) back into \( x = 3y - 6 \) to find \( x \)

Substitute \( y = 1 \) into \( x = 3y-6 \):
\( x=3\times1 - 6=3 - 6=-3 \)

Answer:

\( x=-3 \), \( y = 1 \)