Question

solve the system of linear equations below.
$2x + 3y = 2$
$x + 6y = 4$
a. $x = 0, y = \frac{2}{3}$
b. $x = 2, y = -\frac{2}{3}$
c. $x = \frac{4}{3}, y = \frac{2}{9}$
d. $x = 4, y = -2$

Explanation:

Step1: Simplify the second equation

From \( x + 6y = 4 \), we can express \( x \) as \( x = 4 - 6y \).

Step2: Substitute into the first equation

Substitute \( x = 4 - 6y \) into \( 2x + 3y = 2 \):
\( 2(4 - 6y) + 3y = 2 \)
Expand: \( 8 - 12y + 3y = 2 \)
Simplify: \( 8 - 9y = 2 \)
Subtract 8: \( -9y = 2 - 8 = -6 \)
Divide: \( y = \frac{-6}{-9} = \frac{2}{3} \)? Wait, no, wait, let's check again. Wait, maybe I made a mistake. Wait, the second equation is \( x + 6y = 4 \), so \( x = 4 - 6y \). Substitute into first equation: \( 2(4 - 6y) + 3y = 2 \) → \( 8 - 12y + 3y = 2 \) → \( 8 - 9y = 2 \) → \( -9y = -6 \) → \( y = \frac{2}{3} \)? But that's not matching options. Wait, maybe the second equation is \( x + 6y = 4 \) or is it \( x + 6y = 4 \) or maybe a typo? Wait, no, let's check the options. Wait, option A is \( x=0, y=2/3 \). Let's test option A: plug into first equation: \( 2(0) + 3(2/3) = 0 + 2 = 2 \), which matches. Second equation: \( 0 + 6(2/3) = 0 + 4 = 4 \), which matches. Oh! I see, I misread the second equation. The second equation is \( x + 6y = 4 \), so when \( x=0 \), \( 6y=4 \) → \( y=4/6=2/3 \). Yes, so option A. Wait, but let's redo the substitution correctly. Let's use elimination. Multiply the second equation by 2: \( 2x + 12y = 8 \). Subtract the first equation \( 2x + 3y = 2 \) from this: \( (2x + 12y) - (2x + 3y) = 8 - 2 \) → \( 9y = 6 \) → \( y = 6/9 = 2/3 \). Then substitute \( y=2/3 \) into second equation: \( x + 6*(2/3) = 4 \) → \( x + 4 = 4 \) → \( x=0 \). So that's option A.

Answer:

A. \( x = 0, y = \frac{2}{3} \)