Question

Solving with Elimination Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
1 of 51 of 5 Items
Question
Solve the system of equations.
6x–5y=27

3x+5y=36
(1 point)
(
,
)

Answer 1

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

1. \(6x - 5y = 27\)
2. \(3x + 5y = 36\)

To eliminate \(y\), we can add the two equations together. However, before doing that, we can manipulate the first equation to make it easier to eliminate \(y\).

Let's multiply the second equation by \(1\):
\[
3x + 5y = 36 \quad (Equation 2)
\]

Now, to eliminate \(y\), we can add both equations directly since the coefficients of \(y\) are negatives in the first equation and positives in the second:

1. \(6x - 5y = 27\)
2. \(3x + 5y = 36\)

Now, adding both equations together:
\[
(6x - 5y) + (3x + 5y) = 27 + 36
\]

This simplifies to:
\[
6x + 3x - 5y + 5y = 63
\]
\[
9x = 63
\]

Now, divide both sides by \(9\):
\[
x = 7
\]

Next, we substitute \(x = 7\) back into one of the original equations to solve for \(y\). We'll use the second equation:
\[
3x + 5y = 36
\]
Substituting \(x\):
\[
3(7) + 5y = 36
\]
\[
21 + 5y = 36
\]
Now, isolate \(y\):
\[
5y = 36 - 21
\]
\[
5y = 15
\]
\[
y = 3
\]

Thus, the solution to the system of equations is:
\[
\boxed{(7, 3)}
\]