Question

Solve the following system of equations using the elimination method.

3x−4y=−4, x+3y=−10

(1 point)
Responses

(−4,−2)
left parenthesis negative 4 comma negative 2 right parenthesis

(−2,−4)
left parenthesis negative 2 comma negative 4 right parenthesis

(−16,2)
left parenthesis negative 16 comma 2 right parenthesis

(−16,−2)

Answer 1

To solve the system of equations using the elimination method, we have the following equations:

1. \(3x - 4y = -4\) (Equation 1)
2. \(x + 3y = -10\) (Equation 2)

First, we can manipulate one of the equations to align the terms for elimination. Let's solve Equation 2 for \(x\):

\[
x = -10 - 3y
\]

Now we substitute \(x\) in Equation 1:

\[
3(-10 - 3y) - 4y = -4
\]

Distributing the \(3\):

\[
-30 - 9y - 4y = -4
\]

Combine like terms:

\[
-30 - 13y = -4
\]

Now, add \(30\) to both sides:

\[
-13y = 26
\]

Now, divide by \(-13\):

\[
y = -2
\]

Now that we have \(y\), substitute it back into the expression for \(x\):

\[
x = -10 - 3(-2)
\]

Calculate:

\[
x = -10 + 6 = -4
\]

So the solution to the system of equations is:

\[
(x, y) = (-4, -2)
\]

Therefore, the answer is:

\((−4,−2)\)