Question

Solve each system using elimination. Show your work.
X + 2y = 3
X - y = 6

Answer 1

To solve the system of equations using the elimination method, we have the following equations:

1) \( x + 2y = 3 \)
2) \( x - y = 6 \)

We can use elimination by manipulating these equations to eliminate \( x \) or \( y \). First, we will align the equations for clarity:

\[
\begin{align*}
1) & \quad x + 2y = 3 \quad \text{(Equation 1)} \\
2) & \quad x - y = 6 \quad \text{(Equation 2)}
\end{align*}
\]

Now, let’s manipulate Equation 2 to align the coefficients of \( x \). We can subtract Equation 2 from Equation 1. First, we rearrange both equations in terms of \( x \):

From Equation 2, we can express \( x \):
\[
x = y + 6
\]

Now substitute this expression for \( x \) into Equation 1:

\[
(y + 6) + 2y = 3
\]

Now simplify this equation:
\[
y + 6 + 2y = 3 \\
3y + 6 = 3
\]

Next, isolate \( y \) by subtracting 6 from both sides:
\[
3y = 3 - 6 \\
3y = -3
\]

Now, divide by 3:
\[
y = -1
\]

Now that we have \( y \), we can substitute it back into either Equation 1 or Equation 2 to solve for \( x \). We will substitute it back into the simplified Equation 2:

\[
x - (-1) = 6 \\
x + 1 = 6
\]

Subtract 1 from both sides:
\[
x = 6 - 1 \\
x = 5
\]

Now we have both solutions:
\[
x = 5, \quad y = -1
\]

Thus, the solution to the system of equations is:
\[
(x, y) = (5, -1)
\]

To verify, we can substitute \( x = 5 \) and \( y = -1 \) back into the original equations to ensure both are satisfied:

1) Substituting into \( x + 2y = 3 \):
\[
5 + 2(-1) = 5 - 2 = 3 \quad \text{(True)}
\]

2) Substituting into \( x - y = 6 \):
\[
5 - (-1) = 5 + 1 = 6 \quad \text{(True)}
\]

Since both equations are satisfied, the solution is correct:
\[
\boxed{(5, -1)}
\]