Question

-x + 2y = 18\
7x + 8y = 6\
(\square, \square)

Explanation:

Step1: Multiply the first equation by 4

We have the system of equations:
\[

$$\begin{cases} -x + 2y = 18 \\ 7x + 8y = 6 \end{cases}$$

\]
Multiply the first equation \(-x + 2y = 18\) by 4 to make the coefficients of \(y\) in both equations have a relationship for elimination.
\(4\times(-x + 2y)=4\times18\)
\(-4x + 8y = 72\)

Step2: Subtract the new first equation from the second equation

Now we have two equations:
\(-4x + 8y = 72\) (new first equation)
\(7x + 8y = 6\) (second equation)
Subtract the new first equation from the second equation:
\((7x + 8y)-(-4x + 8y)=6 - 72\)
\(7x + 8y + 4x - 8y=- 66\)
\(11x=-66\)

Step3: Solve for \(x\)

Divide both sides of the equation \(11x=-66\) by 11:
\(x=\frac{-66}{11}=-6\)

Step4: Substitute \(x = - 6\) into the first original equation to solve for \(y\)

Substitute \(x=-6\) into \(-x + 2y = 18\):
\(-(-6)+2y = 18\)
\(6 + 2y = 18\)
Subtract 6 from both sides:
\(2y=18 - 6=12\)
Divide both sides by 2:
\(y = \frac{12}{2}=6\)

Answer:

\((-6,6)\)