Solve the following system of equations using the elimination method.

Question

3x−4y=−4 , x+3y=−10

(1 point)
Responses

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

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

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

(−4,−2)

Answer 1

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

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

We aim to eliminate one of the variables. To do this, we can manipulate the equations so they have the same coefficient for either \(x\) or \(y\).

Let's eliminate \(x\). We can multiply Equation 2 by 3 to make the coefficients of \(x\) in both equations the same:

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

This gives us:

\[ 3x + 9y = -30 \quad \text{(Equation 3)} \]

Now, we have:

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

Next, we can subtract Equation 1 from Equation 3 to eliminate \(x\):

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

This simplifies to:

\[ 9y + 4y = -30 + 4 \] \[ 13y = -26 \]

Now, divide both sides by 13:

\[ y = -2 \]

Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\). We'll use Equation 2:

\[ x + 3(-2) = -10 \] \[ x - 6 = -10 \] \[ x = -10 + 6 \] \[ x = -4 \]

Thus, the solution to the system of equations is:

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

Therefore, the final answer is:

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