Solving a System of Linear Equations – Substitution Method

We are given the following system of linear equations:
\[
\left\{
\begin{array}{l}
5x + 3y = -4 \\
x - 3y = 10
\end{array}
\right.
\]
We can solve this system using substitution or elimination. Here, we will use the substitution method.

Step 1: Solve one equation for one variable

Let's solve the second equation for \( x \):

\[
x - 3y = 10
\]
Add \( 3y \) to both sides:

\[
x = 10 + 3y
\]
Now we have \( x \) in terms of \( y \).

Step 2: Substitute this expression for \( x \) into the first equation

Substitute \( x = 10 + 3y \) into the first equation:

\[
5(10 + 3y) + 3y = -4
\]

Step 3: Simplify and solve for \( y \)

Expand the equation:

\[
50 + 15y + 3y = -4
\]
Combine like terms:

\[
50 + 18y = -4
\]
Subtract 50 from both sides:

\[
18y = -4 - 50 \quad \Rightarrow \quad 18y = -54
\]

Divide both sides by 18:

\[
y = \frac{-54}{18} \quad \Rightarrow \quad y = -3
\]

Step 4: Substitute \( y = -3 \) back into the equation for \( x \)

Now substitute \( y = -3 \) into the equation

\[
x = 10 + 3y \quad \Rightarrow \quad x = 10 + 3(-3) \quad \Rightarrow \quad x = 1
\]

Step 5: Final solution

The solution to the system of equations is:

\[
x = 1, \, y = -3
\]
This is the point where the two lines intersect: \( (1, -3) \).