Question

1. \\(\frac{1}{2}x + y = -6\\)\\(6x + 2y = 8\\)

Explanation:

Step1: Simplify the first equation

Multiply the first equation \(\frac{1}{2}x + y=-6\) by 2 to eliminate the fraction:
\(2\times(\frac{1}{2}x + y)=2\times(-6)\)
\(x + 2y=-12\)

Step2: Subtract the two equations

We now have the system:
\(

$$\begin{cases}x + 2y=-12\\6x + 2y=8\end{cases}$$

\)

Subtract the first equation from the second equation:
\((6x + 2y)-(x + 2y)=8-(-12)\)
\(6x + 2y - x - 2y=8 + 12\)
\(5x=20\)

Step3: Solve for \(x\)

Divide both sides by 5:
\(x=\frac{20}{5}=4\)

Step4: Solve for \(y\)

Substitute \(x = 4\) into the simplified first equation \(x + 2y=-12\):
\(4+2y=-12\)
Subtract 4 from both sides:
\(2y=-12 - 4=-16\)
Divide by 2:
\(y=\frac{-16}{2}=-8\)

Answer:

The solution to the system is \(x = 4\), \(y=-8\)