Let's solve the given system of linear equations step-by-step in order to find the solution:
1. The system of equations is:
[tex]\[
\begin{cases}
2x + y = 3 & \text{(Equation 1)} \\
x - 2y = -1 & \text{(Equation 2)}
\end{cases}
\][/tex]
2. First, we will multiply Equation 1 by 2 to facilitate the elimination method. By multiplying Equation 1 by 2, we get:
[tex]\[
2 \cdot (2x + y) = 2 \cdot 3
\][/tex]
This simplifies to:
[tex]\[
4x + 2y = 6 \quad \text{(Equation 3)}
\][/tex]
3. Now, we will add Equation 3 to Equation 2:
[tex]\[
(4x + 2y) + (x - 2y) = 6 + (-1)
\][/tex]
4. Combining like terms on the left-hand side, we get:
[tex]\[
4x + x + 2y - 2y = 6 - 1
\][/tex]
This further simplifies to:
[tex]\[
5x = 5
\][/tex]
5. Solving for [tex]\(x\)[/tex], we divide both sides by 5:
[tex]\[
x = 1
\][/tex]
6. Now that we have [tex]\(x = 1\)[/tex], we substitute this value back into Equation 1 to find [tex]\(y\)[/tex]:
[tex]\[
2(1) + y = 3
\][/tex]
This simplifies to:
[tex]\[
2 + y = 3
\][/tex]
Solving for [tex]\(y\)[/tex], we subtract 2 from both sides:
[tex]\[
y = 1
\][/tex]
7. Therefore, the solution to the system of equations is:
[tex]\[
x = 1, \quad y = 1
\][/tex]
