Sure! Let's solve the system of equations using the elimination method. We have the following system of equations:
[tex]\[
\begin{cases}
x + 2y = 3 & \text{(Equation 1)} \\
-x + y = 3 & \text{(Equation 2)}
\end{cases}
\][/tex]
Step 1: Add the two equations together to eliminate [tex]\( x \)[/tex]
When we add Equation 1 and Equation 2 directly, we get:
[tex]\[
(x + 2y) + (-x + y) = 3 + 3
\][/tex]
Simplify the left side:
[tex]\[
x - x + 2y + y = 6
\][/tex]
This reduces to:
[tex]\[
3y = 6
\][/tex]
Step 2: Solve for [tex]\( y \)[/tex]
Divide both sides of the equation by 3:
[tex]\[
y = \frac{6}{3}
\][/tex]
Thus:
[tex]\[
y = 2
\][/tex]
Step 3: Substitute [tex]\( y = 2 \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex]
We will use Equation 2 to find [tex]\( x \)[/tex]:
[tex]\[
-x + y = 3
\][/tex]
Substitute [tex]\( y = 2 \)[/tex] into the equation:
[tex]\[
-x + 2 = 3
\][/tex]
Subtract 2 from both sides:
[tex]\[
-x = 3 - 2
\][/tex]
This simplifies to:
[tex]\[
-x = 1
\][/tex]
Step 4: Solve for [tex]\( x \)[/tex]
Multiply both sides by -1 to isolate [tex]\( x \)[/tex]:
[tex]\[
x = -1
\][/tex]
Final Solution:
The solution to the system of equations is:
[tex]\[
(x, y) = (-1, 2)
\][/tex]
So, [tex]\( x = -1 \)[/tex] and [tex]\( y = 2 \)[/tex].
