Certainly! Let's solve the system of equations using the elimination method.
We start with our system of linear equations:
[tex]\[
\left\{\begin{array}{l}
2x + y = 5 \quad \text{(Equation 1)}\\
3x - 2y = 4 \quad \text{(Equation 2)}
\end{array}\right.
\][/tex]
### Step 1: Eliminate one variable
To eliminate the variable [tex]\( y \)[/tex], we need the coefficients of [tex]\( y \)[/tex] in both equations to be opposites. We can do this by multiplying Equation 1 by 2:
[tex]\[
\begin{array}{c}
2(2x + y = 5) \\
\Rightarrow 4x + 2y = 10 \quad \text{(Equation 3)}
\end{array}
\][/tex]
Now, we have the system:
[tex]\[
\left\{\begin{array}{l}
4x + 2y = 10 \quad \text{(Equation 3)}\\
3x - 2y = 4 \quad \text{(Equation 2)}
\end{array}\right.
\][/tex]
### Step 2: Add the equations to eliminate [tex]\( y \)[/tex]
Now, add Equation 3 and Equation 2:
[tex]\[
\begin{array}{c}
(4x + 2y) + (3x - 2y) = 10 + 4 \\
4x + 2y + 3x - 2y = 14 \\
7x = 14
\end{array}
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides of the equation by 7:
[tex]\[
\begin{array}{c}
x = \frac{14}{7} \\
x = 2
\end{array}
\][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]
Substitute [tex]\( x = 2 \)[/tex] into Equation 1:
[tex]\[
\begin{array}{c}
2(2) + y = 5 \\
4 + y = 5 \\
y = 5 - 4 \\
y = 1
\end{array}
\][/tex]
### Step 5: Write the solution
The solution to the system of equations is:
[tex]\[
(x, y) = (2, 1)
\][/tex]
So, [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex].
