We can solve the system of linear equations:

\[
-x + 3y = 5 \quad (1)
\]
\[
x + y = 4 \quad (2)
\]

### Step 1: Solve one equation for one variable.
From equation (2), solve for \( x \):

\[
x = 4 - y
\]

### Step 2: Substitute \( x = 4 - y \) into equation (1).

\[
-(4 - y) + 3y = 5
\]

Simplifying:

\[
-4 + y + 3y = 5
\]
\[
-4 + 4y = 5
\]
\[
4y = 9
\]
\[
y = \frac{9}{4}
\]

### Step 3: Substitute \( y = \frac{9}{4} \) into \( x = 4 - y \).

\[
x = 4 - \frac{9}{4} = \frac{16}{4} - \frac{9}{4} = \frac{7}{4}
\]

### Solution:
The solution to the system is:
\[
x = \frac{7}{4}, \quad y = \frac{9}{4}
\]

Would you like more details on any part of this solution?

Here are 5 related questions to explore further:
1. How do you solve systems of equations by substitution?
2. How do you verify solutions to systems of equations?
3. What are alternative methods for solving systems, like elimination?
4. How do you solve systems of equations with fractions or decimals?
5. Can graphing help visualize solutions to linear systems?

**Tip**: To check if your solution is correct, substitute the values back into the original equations.

Learn how to solve the system of linear equations –x + 3y = 5 and x + y = 4 using the substitution method. This step-by-step guide walks through the solution process, covering essential algebra concepts. Ideal for students in Grades 8-10.

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