Question

solve the following system of equations. 7x + 6y = -4 4x + 6y = 8 x = \square y = \square

Explanation:

Step1: Subtract the two equations

We have the system of equations:
\[

$$\begin{cases} 7x + 6y = -4 \\ 4x + 6y = 8 \end{cases}$$

\]
Subtract the second equation from the first equation to eliminate \(y\):
\[
(7x + 6y)-(4x + 6y)=-4 - 8
\]
Simplify the left - hand side: \(7x+6y - 4x - 6y=3x\)
Simplify the right - hand side: \(-4-8=-12\)
So we get \(3x=-12\)

Step2: Solve for \(x\)

From \(3x = - 12\), divide both sides by 3:
\[
x=\frac{-12}{3}=-4
\]

Step3: Substitute \(x = - 4\) into one of the original equations

Substitute \(x=-4\) into the second equation \(4x + 6y=8\):
\[
4\times(-4)+6y = 8
\]
Simplify the left - hand side: \(-16 + 6y=8\)

Step4: Solve for \(y\)

Add 16 to both sides of the equation \(-16 + 6y=8\):
\[
6y=8 + 16
\]
\[
6y=24
\]
Divide both sides by 6:
\[
y=\frac{24}{6}=4
\]

Answer:

\(x=-4\), \(y = 4\)