Question

question 9 of 15 (1 point) | question attempt: 1 of 3
solve the following system of equations.
$5x + 4y = -2$
$3x + 4y = 10$
$x = \square$
$y = \square$

Explanation:

Step1: Subtract the two equations

We have the system:

$$
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Subtract the second equation from the first to eliminate \(y\):
\((5x + 4y)-(3x + 4y)=-2 - 10\)
Simplify: \(5x + 4y-3x - 4y=-12\)
Which gives: \(2x=-12\)

Step2: Solve for \(x\)

Divide both sides of \(2x = - 12\) by 2:
\(x=\frac{-12}{2}=-6\)

Step3: Substitute \(x\) to find \(y\)

Substitute \(x = - 6\) into the second equation \(3x+4y = 10\):
\(3\times(-6)+4y = 10\)
Simplify: \(-18 + 4y=10\)
Add 18 to both sides: \(4y=10 + 18=28\)
Divide by 4: \(y=\frac{28}{4}=7\)

Answer:

\(x=-6\), \(y = 7\)