Solve the following system of equations using the elimination method.%0D%0A%0D%0A3x−4y=−4, x+3y=−10%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0A(−16,−2)%0D%0Aleft parenthesis negative 16 comma negative 2 right parenthesis%0D%0A%0D%0A(−16,2)%0D%0Aleft parenthesis negative 16 comma 2 right parenthesis%0D%0A%0D%0A(−4,−2)%0D%0Aleft parenthesis negative 4 comma negative 2 right parenthesis%0D%0A%0D%0A(−2,−4)

Answer 1

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

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

Our goal is to eliminate one variable. To do that, we can manipulate the second equation to match the coefficients of \(x\) or \(y\) in the first equation.

Firstly, let's multiply the second equation by 3, so that the coefficients of \(x\) become the same:

\[ 3(x + 3y) = 3(-10) \]

This gives us:

\[ 3x + 9y = -30 \]

Now we have an updated system of equations:

1. \( 3x - 4y = -4 \)
2. \( 3x + 9y = -30 \)

Next, we can eliminate \(3x\) by subtracting the first equation from the second:

\[ (3x + 9y) - (3x - 4y) = -30 - (-4) \]

Simplifying this gives:

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

\[ 13y = -26 \]

Now, we can solve for \(y\):

\[ y = \frac{-26}{13} = -2 \]

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

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

This simplifies to:

\[ x - 6 = -10 \]

Now, solve for \(x\):

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

Thus, the solution to the system of equations is:

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

So, the answer is:

\(\boxed{(-4, -2)}\)