−6x+2y=−32

4x+3y=−9
(1 point)
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1 answer

To solve the system of equations given by:

  1. \(-6x + 2y = -32\)
  2. \(4x + 3y = -9\)

we can use either substitution or elimination. Let's solve it using the elimination method.

First, we can manipulate the first equation to eliminate \(y\):

  1. Divide the entire first equation by 2 to simplify it:

\[ -3x + y = -16 \quad \text{(equation 1)} \]

So now we have:

  1. \(-3x + y = -16\)
  2. \(4x + 3y = -9\)

Now, let's express \(y\) in terms of \(x\) from equation 1:

\[ y = 3x - 16 \]

Next, substitute this expression for \(y\) into equation 2:

\[ 4x + 3(3x - 16) = -9 \]

Distributing the 3:

\[ 4x + 9x - 48 = -9 \]

Combine like terms:

\[ 13x - 48 = -9 \]

Add 48 to both sides:

\[ 13x = 39 \]

Now, divide by 13:

\[ x = 3 \]

Now substitute \(x = 3\) back into the expression for \(y\):

\[ y = 3(3) - 16 \] \[ y = 9 - 16 \] \[ y = -7 \]

So the solution to the system of equations is:

\[ (x, y) = (3, -7) \]

Thus, the final answer is:

\[ (3, -7) \]