Question

solve the following system of equations algebraically.
$x - y = 6$
$x + 3y = -6$
which of the following describes the solution to the system?

• x-coordinate is -3
• x-coordinate is 6
• x-coordinate is 3
• x-coordinate is -2
• y-coordinate is 3
• y-coordinate is -3
• y-coordinate is -6
• y-coordinate is 2

Explanation:

Step1: Solve for x from first equation

From \( x - y = 6 \), we get \( x = y + 6 \).

Step2: Substitute x into second equation

Substitute \( x = y + 6 \) into \( x + 3y = -6 \):
\( (y + 6) + 3y = -6 \)
Simplify: \( 4y + 6 = -6 \)
Subtract 6: \( 4y = -12 \)
Divide by 4: \( y = -3 \).

Step3: Find x using y

Substitute \( y = -3 \) into \( x = y + 6 \):
\( x = -3 + 6 = 3 \). Wait, no—wait, let's re - check. Wait, original equations:
First equation: \( x - y = 6 \)
Second: \( x + 3y = -6 \)
Subtract first equation from second: \( (x + 3y)-(x - y)=-6 - 6 \)
Simplify: \( 4y=-12\), so \( y = - 3 \). Then \( x=y + 6=-3 + 6 = 3 \). So x - coordinate is 3, y - coordinate is - 3. Wait, but the yellow - highlighted option was x - coordinate is - 2, which was wrong. Let's re - solve:

Wait, maybe I made a mistake. Let's use elimination:

Equation 1: \( x - y = 6 \)

Equation 2: \( x + 3y = -6 \)

Subtract Equation 1 from Equation 2:

\( (x + 3y)-(x - y)=-6 - 6 \)

\( x+3y - x + y=-12 \)

\( 4y=-12\)

\( y = - 3 \)

Then plug y = - 3 into Equation 1: \( x-(-3)=6\Rightarrow x + 3 = 6\Rightarrow x = 3 \). So x - coordinate is 3, y - coordinate is - 3. So the correct options are "x - coordinate is 3" and "y - coordinate is - 3".

Answer:

x - coordinate is 3, y - coordinate is - 3