Answer :

Answer:

THE ANSWER IS GIVEN BELOW

Step-by-step explanation:

To solve the system of equations:

\[ \begin{cases}

x + 3y = 6 \\

2x - 2y = -20

\end{cases} \]

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

1. Multiply the first equation by 2 to make the coefficients of \(x\) in both equations equal:

\[ \begin{cases}

2x + 6y = 12 \\

2x - 2y = -20

\end{cases} \]

2. Now, subtract the second equation from the first equation:

\[ (2x + 6y) - (2x - 2y) = 12 - (-20) \]

\[ 2x + 6y - 2x + 2y = 12 + 20 \]

\[ 8y = 32 \]

3. Divide both sides by 8:

\[ y = \frac{32}{8} \]

\[ y = 4 \]

4. Substitute \(y = 4\) into the first equation to find \(x\):

\[ x + 3(4) = 6 \]

\[ x + 12 = 6 \]

\[ x = 6 - 12 \]

\[ x = -6 \]

So, the solution to the system of equations is \(x = -6\) and \(y = 4\).