Solving with Elimination Quick Check

Question

3 of 53 of 5 Items

Question
Solve the system of two equations using elimination with multiplication.
4x−6y=−48

−2x+4y=30
(1 point)
Responses

(−33,−9)
left parenthesis negative 33 comma negative 9 right parenthesis

(−3, 6)
left parenthesis negative 3 comma 6 right parenthesis

(−21,−6)
left parenthesis negative 21 comma negative 6 right parenthesis

(3, 9)

Answer 1

To solve the system of equations using elimination, we'll manipulate the equations to eliminate one of the variables. The given equations are:

1. \(4x - 6y = -48\)

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

First, we can multiply the second equation by 2 to make the coefficients of \(x\) in both equations match:

Multiplying the second equation by 2 gives:

\[ -4x + 8y = 60 \]

Now we rewrite the system of equations:

1. \(4x - 6y = -48\)

2. \(-4x + 8y = 60\)

Next, we add the two equations together to eliminate \(x\):

\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \]

This simplifies to:

\[ 0x + 2y = 12 \]

This means:

\[ 2y = 12 \]

Dividing both sides by 2 gives us:

\[ y = 6 \]

Now that we have \(y\), we can substitute this value back into one of the original equations to find \(x\). We'll use the first equation:

\[ 4x - 6(6) = -48 \]

This simplifies to:

\[ 4x - 36 = -48 \]

Adding 36 to both sides:

\[ 4x = -48 + 36 \] \[ 4x = -12 \]

Dividing both sides by 4 gives:

\[ x = -3 \]

Thus, the solution to the system of equations is:

\((-3, 6)\)

So the correct response is:

(−3, 6)