To solve the system of equations using the elimination method, follow these steps:

Given system:
\[
\begin{aligned}
-2x + 2y &= -12 \quad \text{(Equation 1)} \\
2x + 20y &= 12 \quad \text{(Equation 2)}
\end{aligned}
\]

### Step 1: Add the equations
To eliminate \( x \), add Equation 1 and Equation 2:

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

Simplifying this, we get:

\[
2y + 20y = 0
\]

\[
22y = 0
\]

### Step 2: Solve for \( y \)
\[
y = \frac{0}{22} = 0
\]

### Step 3: Substitute \( y = 0 \) back into one of the original equations
Let's use Equation 1:

\[
-2x + 2(0) = -12
\]

\[
-2x = -12
\]

### Step 4: Solve for \( x \)
\[
x = \frac{-12}{-2} = 6
\]

### Final Answer:
The solution to the system is \( x = 6 \) and \( y = 0 \).

Would you like more details, or do you have any questions?

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Here are 5 related questions:
1. What happens if you multiply one of the equations by a constant before elimination?
2. How can you solve the system using substitution instead of elimination?
3. What does it mean if, after elimination, you get a false statement (e.g., \( 0 = 5 \))?
4. Can you solve the system by graphing the two equations?
5. How do you handle systems with more than two variables?

**Tip:** When using the elimination method, ensure the coefficients of one variable are opposites to eliminate it easily.

-2x + 2y = -12, 2x + 20y = 12 elimination

Discover how to solve the system of linear equations -2x + 2y = -12 and 2x + 20y = 12 using the elimination method. This step-by-step guide covers key algebraic concepts, the elimination method, and addition of equations. Suitable for students in Grades 8-10.

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