Question

Solving systems by elimination -x+3y=9 5x+4y=-26

Answer 1

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

1. \(-x + 3y = 9\) (Equation 1)
2. \(5x + 4y = -26\) (Equation 2)

First, let's manipulate the first equation to eliminate \(x\):

1. From Equation 1, we can express \(x\) in terms of \(y\):
\[
-x + 3y = 9 \implies x = 3y - 9
\]

Next, we'll substitute this expression for \(x\) into Equation 2:

2. Substitute \(x\) in Equation 2:
\[
5(3y - 9) + 4y = -26
\]
Simplifying this:
\[
15y - 45 + 4y = -26
\]
Combine like terms:
\[
19y - 45 = -26
\]
Add 45 to both sides:
\[
19y = 19
\]
Divide by 19:
\[
y = 1
\]

Now that we have \(y\), we can find \(x\) by substituting \(y\) back into the expression for \(x\):
\[
x = 3(1) - 9
\]
\[
x = 3 - 9 = -6
\]

Thus, the solution to the system of equations is:
\[
(x, y) = (-6, 1)
\]

You can verify the solution by substituting \(x = -6\) and \(y = 1\) back into the original equations:

For Equation 1:
\[
-x + 3y = 9 \implies -(-6) + 3(1) = 6 + 3 = 9 \quad \text{(True)}
\]

For Equation 2:
\[
5x + 4y = -26 \implies 5(-6) + 4(1) = -30 + 4 = -26 \quad \text{(True)}
\]

Both equations are satisfied, confirming the solution is correct:
\[
\boxed{(-6, 1)}
\]