Let's solve the system of linear equations step-by-step.
We have two equations:
[tex]\[
x + 2y = 5 \quad \text{(Equation 1)}
\][/tex]
and
[tex]\[
2x + y = 4 \quad \text{(Equation 2)}
\][/tex]
First, let's solve Equation 2 for [tex]\(y\)[/tex]:
[tex]\[
2x + y = 4
\][/tex]
Subtract [tex]\(2x\)[/tex] from both sides:
[tex]\[
y = 4 - 2x \quad \text{(Equation 3)}
\][/tex]
Now, substitute the expression for [tex]\(y\)[/tex] from Equation 3 into Equation 1:
[tex]\[
x + 2(4 - 2x) = 5
\][/tex]
Distribute [tex]\(2\)[/tex] inside the parenthesis:
[tex]\[
x + 8 - 4x = 5
\][/tex]
Combine like terms:
[tex]\[
-3x + 8 = 5
\][/tex]
Subtract 8 from both sides:
[tex]\[
-3x = -3
\][/tex]
Divide both sides by -3:
[tex]\[
x = 1
\][/tex]
Now that we have the value of [tex]\(x\)[/tex], substitute it back into Equation 3 to find [tex]\(y\)[/tex]:
[tex]\[
y = 4 - 2(1)
\][/tex]
Simplify:
[tex]\[
y = 4 - 2
\][/tex]
[tex]\[
y = 2
\][/tex]
So, the solution to the system of equations is:
[tex]\[
x = 1 \quad \text{and} \quad y = 2
\][/tex]
Finally, to find the value of [tex]\(x + y\)[/tex]:
[tex]\[
x + y = 1 + 2
\][/tex]
[tex]\[
x + y = 3
\][/tex]
So, the value of [tex]\(x + y\)[/tex] is:
[tex]\[
\boxed{3}
\][/tex]
