Question

solve the system of equations,
$3x + 4y = -23$
$x = 3y + 1$
$x = \square$
$y = \square$
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Explanation:

Step1: Substitute \( x = 3y + 1 \) into \( 3x + 4y = -23 \)

Substitute \( x \) in the first equation: \( 3(3y + 1) + 4y = -23 \)

Step2: Simplify and solve for \( y \)

Expand: \( 9y + 3 + 4y = -23 \)
Combine like terms: \( 13y + 3 = -23 \)
Subtract 3: \( 13y = -26 \)
Divide by 13: \( y = -2 \)

Step3: Substitute \( y = -2 \) into \( x = 3y + 1 \)

Calculate \( x \): \( x = 3(-2) + 1 = -6 + 1 = -5 \)

Answer:

\( x = -5 \), \( y = -2 \)