Question

Solve the equation with elimination (algebra 1)
-6x - 8y = -28
9x + 5y = -14

Answer 1

To solve the system of equations using the elimination method, we can follow these steps:

1. Write down the equations:
\[
-6x - 8y = -28 \quad \text{(Equation 1)}
\]
\[
9x + 5y = -14 \quad \text{(Equation 2)}
\]

2. We want to eliminate one of the variables. Let's eliminate \(y\). To do this, we can manipulate each equation so that the coefficients of \(y\) are opposites.

We can multiply Equation 1 by 5 and Equation 2 by 8 to make the coefficients of \(y\) match:

Multiply Equation 1 by 5:
\[
5(-6x - 8y) = 5(-28)
\]
This results in:
\[
-30x - 40y = -140 \quad \text{(Equation 3)}
\]

Multiply Equation 2 by 8:
\[
8(9x + 5y) = 8(-14)
\]
This results in:
\[
72x + 40y = -112 \quad \text{(Equation 4)}
\]

3. Now we can add Equation 3 and Equation 4:
\[
(-30x - 40y) + (72x + 40y) = -140 + (-112)
\]

Simplifying gives:
\[
(72x - 30x) + (-40y + 40y) = -252
\]
\[
42x = -252
\]

4. Solve for \(x\):
\[
x = \frac{-252}{42} = -6
\]

5. Now that we have \(x\), we substitute \(x = -6\) into one of the original equations to find \(y\). We'll use Equation 2:
\[
9(-6) + 5y = -14
\]
\[
-54 + 5y = -14
\]

Add 54 to both sides:
\[
5y = -14 + 54
\]
\[
5y = 40
\]

6. Solve for \(y\):
\[
y = \frac{40}{5} = 8
\]

7. The solution to the system of equations is:
\[
x = -6, \quad y = 8
\]

Thus, the solution is:
\[
\boxed{(-6, 8)}
\]