Solving System of Linear Equations: (lesson 1 of 5)

## Substitution Method

The **substitution method** is most useful for systems of 2 equations in 2 unknowns.
The main idea here is that we solve one of the equations for one of the
unknowns, and then substitute the result into the other equation.

**Substitution method can be applied in four steps**

**Step 1:**

Solve one of the equations for either **x = ** or **y =** .

**Step 2:**

Substitute the solution from step 1 into the other equation.

**Step 3:**

Solve this new equation.

**Step 4:**

Solve for the second variable.

Example 1: Solve the following system by substitution

$$
\begin{aligned}
2x + 3y &= 5 \\
x + y &= 5
\end{aligned}
$$

Solution:

**Step 1:** Solve one of the equations for either **x = ** or **y =** . We will solve
second equation for y.

$$
\begin{aligned}
x + y &= 5 \\
\color{blue}{y} &\color{blue}{=} \color{blue}{5 - x}
\end{aligned}
$$

**Step 2:** Substitute the solution from step 1 into the second equation.

$$
\begin{aligned}
2x + 3\color{blue}{y} &= 5 \\
2x + 3\color{blue}{( 5 - x )} &\color{blue}{=} \color{blue}{5}
\end{aligned}
$$

**Step 3:** Solve this new equation.

$$
\begin{aligned}
2x + 3( 5 - x) &= 5 \\
2x + 15 - 3x &= 5 \\
- x + 15 &= 5 \\
- x &= 5 - 15 \\
\color{red}{x} &\color{red}{=} \color{red}{10}
\end{aligned}
$$

**Step 4:** Solve for the second variable

$$
\begin{aligned}
y &= 5 - \color{red}{x} \\
y &= 5 - \color{red}{10} \\
y &= - 5
\end{aligned}
$$

**The solution is: (x, y) = (10, -5)**

Note: It does not matter which equation we choose
first and which second. Just choose the most convenient one first!

Example 2: Solve by substitution

$$
\begin{aligned}
2x + 5y &= 12 \\
4x - y &= 2
\end{aligned}
$$

Solution:

**Step 1: Solve one of the equations for either x = or y =.**
Since the coefficient of y in equation 2 is -1, it is easiest to solve for y in equation 2.

$$
\begin{aligned}
4x - y &= 2 \\
- y &= 2 - 4x \\
\color{blue}{y} &\color{blue}{=} \color{blue}{4x - 2}
\end{aligned}
$$

**Step 2:** Substitute the solution from step 1 into the second equation.

$$
\begin{aligned}
2x + 5\color{blue}{y} &= 12 \\
2x + 5 \color{blue}{(4x - 2)} &= 12 \\
\end{aligned}
$$

**Step 3:** Solve this new equation ( for x ).

$$
\begin{aligned}
2x + 5\color{blue}{(4x - 2)} &= 12 \\
2x + 2x + 20x - 10 &= 12 \\
22x &= 22\\
\color{red}{x} &\color{red}{=} \color{red}{1}
\end{aligned}
$$

**Step 4:** Solve for the second variable

$$
\begin{aligned}
y &= 4 \color{red}{x} - 2 \\
y &= 4 \cdot \color{red}{x} - 2 \\
y &= 2
\end{aligned}
$$

**The solution is: $(x, y) = (1, 2)$**

Exercise: Solve the following systems by substitution