Question

solve the system of linear equations below.

x + y = 4
2x + 3y = 0

a. x = -6, y = 2
b. x = -1, y = 2
c. x = 11/5, y = 9/5
d. x = 12, y = -8

Explanation:

Step1: Solve for x from first equation

From \( x + y = 4 \), we get \( x = 4 - y \).

Step2: Substitute x into second equation

Substitute \( x = 4 - y \) into \( 2x + 3y = 0 \):
\( 2(4 - y) + 3y = 0 \)
\( 8 - 2y + 3y = 0 \)
\( 8 + y = 0 \)
\( y = -8 \)? Wait, no, wait—wait, let's recalculate. Wait, no, the options have D as \( x=12, y=-8 \). Wait, let's check substitution again. Wait, first equation: \( x + y = 4 \). If \( y = -8 \), then \( x = 4 - (-8) = 12 \). Then check second equation: \( 2x + 3y = 2(12) + 3(-8) = 24 - 24 = 0 \). Yes, that works. Wait, but let's check other options. Option A: \( x=-6, y=2 \). \( x + y = -4
eq 4 \). Option B: \( x=-1, y=2 \). \( x + y = 1
eq 4 \). Option C: \( x=11/5, y=9/5 \). Sum is \( 20/5 = 4 \), but check second equation: \( 2(11/5) + 3(9/5) = 22/5 + 27/5 = 49/5
eq 0 \). So D works.

Answer:

D. \( x = 12, y = -8 \)