Question

find the solution of the system of equations.\

$$\begin{cases}-2x + 3y = 8\\\\-x - 3y = 4\\end{cases}$$

\
show your work here\
x =\
y =\
check

Explanation:

Step1: Add the two equations

We have the system:

$$\begin{cases} -2x + 3y = 8 \\ -x - 3y = 4 \end{cases}$$

Adding the two equations together to eliminate \(y\):
\((-2x + 3y)+(-x - 3y)=8 + 4\)
Simplify the left - hand side: \(-2x - x+3y - 3y=-3x\)
Simplify the right - hand side: \(12\)
So we get the equation \(-3x = 12\)

Step2: Solve for \(x\)

Divide both sides of the equation \(-3x = 12\) by \(-3\):
\(x=\frac{12}{-3}=- 4\)

Step3: Substitute \(x = - 4\) into one of the original equations to solve for \(y\)

Let's substitute \(x=-4\) into the second equation \(-x - 3y = 4\)
We have \(-(-4)-3y = 4\)
Simplify \(-(-4)\) to get \(4\), so the equation becomes \(4-3y = 4\)
Subtract \(4\) from both sides: \(4-4-3y=4 - 4\)
Which simplifies to \(-3y = 0\)
Divide both sides by \(-3\): \(y = 0\)

Answer:

\(x=-4\), \(y = 0\)