Sure, let's solve the system of equations step by step.
We have the following system of equations:
[tex]\[
\begin{cases}
x + 3y = 10 & \quad \text{(1)} \\
-x + y = 2 & \quad \text{(2)}
\end{cases}
\][/tex]
### Step 1: Solve one of the equations for one variable
Let's solve equation (2) for [tex]\(y\)[/tex]:
[tex]\[
-x + y = 2 \implies y = x + 2
\][/tex]
### Step 2: Substitute into the other equation
Now, substitute [tex]\(y = x + 2\)[/tex] into equation (1):
[tex]\[
x + 3(x + 2) = 10
\][/tex]
### Step 3: Simplify and solve for [tex]\(x\)[/tex]
Simplify the equation:
[tex]\[
x + 3x + 6 = 10 \implies 4x + 6 = 10
\][/tex]
Subtract 6 from both sides:
[tex]\[
4x = 4
\][/tex]
Divide both sides by 4:
[tex]\[
x = 1
\][/tex]
### Step 4: Substitute [tex]\(x\)[/tex] back to find [tex]\(y\)[/tex]
Substitute [tex]\(x = 1\)[/tex] back into [tex]\(y = x + 2\)[/tex]:
[tex]\[
y = 1 + 2 = 3
\][/tex]
### Step 5: Verify the solution
It's a good practice to verify the solution by substituting [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex] back into the original equations.
Substitute into equation (1):
[tex]\[
1 + 3(3) = 1 + 9 = 10 \quad \text{(True)}
\][/tex]
Substitute into equation (2):
[tex]\[
-1 + 3 = 2 \quad \text{(True)}
\][/tex]
Both equations are satisfied, so the solution to the system is:
[tex]\[
\boxed{(1, 3)}
\][/tex]
