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 \(y = 4\).
