Question

Solve using elimination.

–7x + 8y = –10
–x + 6y = –16

Answer 1

To solve the system of equations using the elimination method, we have:

1. \(-7x + 8y = -10\) (Equation 1)
2. \(-x + 6y = -16\) (Equation 2)

We can first manipulate Equation 2 to make the coefficients of \(x\) match so we can eliminate \(x\). We can multiply Equation 2 by 7 to align the \(x\) terms with those in Equation 1:

\[
7(-x + 6y) = 7(-16)
\]
This simplifies to:

\[
-7x + 42y = -112 \quad \text{(Equation 3)}
\]

Now, we can set up our system with Equation 1 and the modified Equation 3:

1. \(-7x + 8y = -10\) (Equation 1)
2. \(-7x + 42y = -112\) (Equation 3)

Next, we will subtract Equation 1 from Equation 3 to eliminate \(x\):

\[
(-7x + 42y) - (-7x + 8y) = -112 - (-10)
\]

This simplifies to:

\[
(42y - 8y) = -112 + 10
\]
\[
34y = -102
\]

Now, solving for \(y\):

\[
y = \frac{-102}{34} = -3
\]

Now that we have \(y\), we can substitute this value back into one of the original equations to find \(x\). We'll use Equation 2:

\[
-x + 6(-3) = -16
\]
\[
-x - 18 = -16
\]

Adding 18 to both sides:

\[
-x = 2
\]

Thus,

\[
x = -2
\]

The solution to the system of equations is:

\[
(x, y) = (-2, -3)
\]

In conclusion, the solution is:
\[
\boxed{(-2, -3)}
\]