Sure! Let's solve the system of linear equations step-by-step:
Given the system of equations:
1. [tex]\(x + y = 4\)[/tex]
2. [tex]\(x - y = 2\)[/tex]
We'll solve this system using the method of addition (also known as the elimination method).
Step 1: Add the two equations to eliminate [tex]\(y\)[/tex]
[tex]\[
\begin{aligned}
&\quad x + y = 4 &\quad \text{(Equation 1)} \\
+ & \quad x - y = 2 &\quad \text{(Equation 2)} \\
\end{aligned}
\][/tex]
When we add these two equations, the [tex]\(y\)[/tex] terms cancel out:
[tex]\[
(x + y) + (x - y) = 4 + 2
\][/tex]
Simplifying this, we get:
[tex]\[
2x = 6
\][/tex]
Step 2: Solve for [tex]\(x\)[/tex]
[tex]\[
x = \frac{6}{2} = 3
\][/tex]
So, we have determined that [tex]\(x = 3\)[/tex].
Step 3: Substitute [tex]\(x\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]
We can substitute [tex]\(x = 3\)[/tex] into the first equation:
[tex]\[
x + y = 4
\][/tex]
So:
[tex]\[
3 + y = 4
\][/tex]
Step 4: Solve for [tex]\(y\)[/tex]
[tex]\[
y = 4 - 3 = 1
\][/tex]
Thus, the solution to the system of equations is [tex]\(x = 3\)[/tex] and [tex]\(y = 1\)[/tex].
The pair [tex]\((x, y) = (3, 1)\)[/tex] satisfies both of the given equations.
Therefore, the solution to the system of equations is:
[tex]\[
\boxed{x = 3, \, y = 1}
\][/tex]
