Solve the system of equations.

8x−4y=−28

2x−6y=8
(1 point)
Responses

(−5,−3)
left parenthesis negative 5 comma negative 3 right parenthesis

(−2, 3)
left parenthesis negative 2 comma 3 right parenthesis

(−4,−1)
left parenthesis negative 4 comma negative 1 right parenthesis

(7, 1)

1 answer

To solve the system of equations:

  1. \( 8x - 4y = -28 \)
  2. \( 2x - 6y = 8 \)

we can use either the substitution method or the elimination method. Here, I'll use the elimination method.

First, we can simplify the first equation:

\[ 8x - 4y = -28 \implies 2x - y = -7 \quad \text{(dividing the entire equation by 4)} \]

Now our system of equations looks like this:

  1. \( 2x - y = -7 \)
  2. \( 2x - 6y = 8 \)

Next, we'll express \( y \) in terms of \( x \) from the first equation:

\[ y = 2x + 7 \]

Now we will substitute \( y \) in the second equation:

\[ 2x - 6(2x + 7) = 8 \]

Expanding this gives:

\[ 2x - 12x - 42 = 8 \]

Combining like terms results in:

\[ -10x - 42 = 8 \]

Next, we will isolate \( x \):

\[ -10x = 8 + 42 \implies -10x = 50 \implies x = -5 \]

Now that we have \( x \), we can find \( y \) by substituting \( x = -5 \) back into \( y = 2x + 7 \):

\[ y = 2(-5) + 7 = -10 + 7 = -3 \]

Therefore, the solution to the system of equations is:

\[ (x, y) = (-5, -3) \]

So the correct response is:

\[ \text{(−5,−3)} \]