Question

Solve the system of equations.
6x + 2y = -32
4x + 3y =-9 (1 point)

Answer 1

To solve the system of equations

1. \( 6x + 2y = -32 \)
2. \( 4x + 3y = -9 \)

we can use the method of substitution or elimination. Here, we will use the elimination method.

First, let's multiply the first equation by 3 and the second equation by 2 to make the coefficients of \(y\) the same:

1. \( 3(6x + 2y) = 3(-32) \) → \( 18x + 6y = -96 \)
2. \( 2(4x + 3y) = 2(-9) \) → \( 8x + 6y = -18 \)

Now we have the new system of equations:

1. \( 18x + 6y = -96 \)
2. \( 8x + 6y = -18 \)

Next, we can subtract the second equation from the first to eliminate \(y\):

\[
(18x + 6y) - (8x + 6y) = -96 - (-18)
\]
\[
18x - 8x = -96 + 18
\]
\[
10x = -78
\]
\[
x = -\frac{78}{10} = -\frac{39}{5} = -7.8
\]

Now that we have \(x\), we substitute it back into one of the original equations to find \(y\). We'll use the first equation:

\[
6(-\frac{39}{5}) + 2y = -32
\]
\[
-\frac{234}{5} + 2y = -32
\]

Next, convert -32 into a fraction with a denominator of 5:

\[
-32 = -\frac{160}{5}
\]

Now set up the equation:

\[
-\frac{234}{5} + 2y = -\frac{160}{5}
\]

To isolate \(2y\), add \(\frac{234}{5}\) to both sides:

\[
2y = -\frac{160}{5} + \frac{234}{5}
\]
\[
2y = \frac{234 - 160}{5} = \frac{74}{5}
\]

Now divide by 2:

\[
y = \frac{74}{10} = \frac{37}{5} = 7.4
\]

Thus, the solution to the system of equations is:

\[
x = -\frac{39}{5}, \quad y = \frac{37}{5}
\]

So the final solution is:

\[
\left(x, y\right) = \left(-\frac{39}{5}, \frac{37}{5}\right)
\] or approximately \( (-7.8, 7.4) \).