A. {
−𝒙 + 𝟑𝒚 = 𝟓; 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟏
𝒙 + 𝒚 = 𝟒; 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟐

Let's solve this system of linear equations:

\[
\text{Ecuación 1: } -x + 3y = 5
\]
\[
\text{Ecuación 2: } x + y = 4
\]

### Step 1: Solve for \( x \) from Ecuación 2.
From the second equation:

\[
x + y = 4 \quad \Rightarrow \quad x = 4 - y
\]

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

Now substitute \( x = 4 - y \) into the first equation:

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

Simplifying:

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

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

Now substitute the value of \( y \) into the expression for \( x \):

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

### Solution:

\[
x = \frac{7}{4}, \quad y = \frac{9}{4}
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this:
1. How do you verify the solution of a system of equations?
2. Can systems of linear equations have more than one solution?
3. What methods can be used to solve systems of equations, besides substitution?
4. How can these equations be solved graphically?
5. What happens if the system has no solution or infinite solutions?

**Tip:** When solving a system of equations, substitution is often useful when one variable is easy to isolate in one of the equations.

A. {−𝒙 + 𝟑𝒚 = 𝟓; 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟏, 𝒙 + 𝒚 = 𝟒; 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟐}

Explore how to solve the system of linear equations −x + 3y = 5 and x + y = 4 using the substitution method. This example walks through the process step-by-step, highlighting key algebraic concepts. Suitable for students in Grades 8-10.

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