Question

Solve using elimination.

6x + 8y = 18
–7x − 6y = –1

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

1. \( 6x + 8y = 18 \) (Equation 1)
2. \( -7x - 6y = -1 \) (Equation 2)

The goal of elimination is to eliminate one of the variables so we can solve for the other. We can do this by manipulating both equations so that when we add or subtract them, one of the variables cancels out.

### Step 1: Align the equations

Currently, we have:

\[
6x + 8y = 18 \quad \text{(1)}
\]

\[
-7x - 6y = -1 \quad \text{(2)}
\]

### Step 2: Make the coefficients of \(y\) the same

To eliminate \(y\), we can multiply Equation 1 by \(3\) and Equation 2 by \(4\) so that the coefficients of \(y\) become the same.

\[
3(6x + 8y) = 3(18) \quad \Rightarrow \quad 18x + 24y = 54 \quad \text{(3)}
\]

\[
4(-7x - 6y) = 4(-1) \quad \Rightarrow \quad -28x - 24y = -4 \quad \text{(4)}
\]

### Step 3: Add the equations to eliminate \(y\)

Now we add Equation 3 and Equation 4 together:

\[
(18x + 24y) + (-28x - 24y) = 54 + (-4)
\]

This simplifies to:

\[
18x - 28x = 54 - 4
\]

\[
-10x = 50
\]

### Step 4: Solve for \(x\)

Divide both sides by \(-10\):

\[
x = -5
\]

### Step 5: Substitute \(x\) back into one of the original equations

Let's substitute \(x = -5\) back into Equation 1:

\[
6(-5) + 8y = 18
\]

\[
-30 + 8y = 18
\]

### Step 6: Solve for \(y\)

Add \(30\) to both sides:

\[
8y = 48
\]

Now divide by \(8\):

\[
y = 6
\]

### Conclusion

The solution to the system of equations is:

\[
\boxed{(-5, 6)}
\]