Question

Solving with Elimination Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
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Question
Solve the system of equations.
3x+6y=21

–5x+2y=13
(1 point)
(
,
)
Remaining Attempts : 3

Answer 1

To solve the system of equations using the elimination method, we can start by rewriting the equations:

1. \(3x + 6y = 21\) (Equation 1)
2. \(-5x + 2y = 13\) (Equation 2)

First, we can simplify Equation 1. Dividing all terms by 3:

\[
x + 2y = 7 \quad \text{(Simplified Equation 1)}
\]

Now we have:

1. \(x + 2y = 7\)
2. \(-5x + 2y = 13\)

Next, let's eliminate \(y\) by subtracting Equation 1 from Equation 2. We can rewrite Equation 1 and then subtract:

Subtract Equation 1 from Equation 2:

\[
(-5x + 2y) - (x + 2y) = 13 - 7
\]

This simplifies to:

\[
-5x - x + 2y - 2y = 6
\]

Combining like terms gives:

\[
-6x = 6
\]

Dividing by -6:

\[
x = -1
\]

Now that we have \(x\), we can substitute it back into one of the original equations to find \(y\). We can use the simplified Equation 1:

\[
x + 2y = 7
\]

Substituting \(x = -1\):

\[
-1 + 2y = 7
\]

Adding 1 to both sides:

\[
2y = 8
\]

Dividing by 2:

\[
y = 4
\]

So, the solution to the system of equations is:

\[
(x, y) = (-1, 4)
\]

Final answer:

\[
\boxed{(-1, 4)}
\]