Solving System of Linear Equations: (lesson 2 of 5)

## Elimination Method

The **elimination method** of solving systems of equations is also called the addition method.
To solve a system of equations by elimination we transform the system such that one variable "cancels out".

Example 1: Solve the system of equations by elimination

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

Solution:

In this example we will "cancel out" the y term. To do so, we can add the equations together.

$$
\begin{aligned}
&\underline{\left. \begin{aligned}
3x \color{red}{- y} = 5\\
x \color{red}{+ y} = 3
\end{aligned}
\right\}}
\text{Add equations}\\
&4x = 8
\end{aligned}
$$

Now we can find: **$x = 2$**

In order to solve for y, take the value for **x** and substitute it back into either one of the original
equations.

$$
\begin{aligned}
\color{red}{x} + y &= 3 \\
\color{red}{2} + y &= 3 \\
y &= 1
\end{aligned}
$$

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

Example 2: Solve the system using elimination

$$
\begin{aligned}
x + 3y &= -5 \\
4x - y &= 6
\end{aligned}
$$

Solution:

Look at the x - coefficients. Multiply the first equation by -4, to set up the x-coefficients to cancel.

$$
\begin{aligned}
&x + 3y = -5
\color{red}{\,\,\,\,\text {multiply by -4}}\\
&\underline{4x - y = 6}
\end{aligned}\\
\begin{aligned}
&\underline{\left. \begin{aligned}
\color{red}{- 4x} - 12y= 20\\
\color{red}{4}x - y = 6
\end{aligned}
\right\}}
\text{Add equations}\\
&-13x = 26
\end{aligned}
$$

Now we can find: **$y = -2$**

Take the value for **y** and substitute it back into either one of the original equations.

$$
\begin{aligned}
x + 3y &= -5 \\
x + 3\cdot(\color{red}{-2}) &= -5\\
x - 6 &= -5\\
x &= 1
\end{aligned}
$$

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

Example 3: Solve the system using elimination method

$$
\begin{aligned}
2x - 5y &= 11 \\
3x + 2y &= 7
\end{aligned}
$$

Solution:

In this example, we will multiply the first row by **-3** and the second row by **2**; then we will add down as before.

$$
\begin{aligned}
&2x - 5y = 11
\color{red}{\,\,\,\, \text {multiply by -3}}\\
&\underline{3x + 2y = 7 \color{red}{\,\,\,\, \text {multiply by 2}}}
\end{aligned}\\
\begin{aligned}
&\underline{\left. \begin{aligned}
\color{red}{- 6} + 15y = -33\\
\color{red}{6}x + 4y = 14
\end{aligned}
\right\}}
\text{Add equations}\\
&19y = -19
\end{aligned}
$$

Now we can find: y = -1

Substitute y = -1 back into first equation:

$$
\begin{aligned}
2x - 5\color{blue}{y} &= 11 \\
2x - 5\cdot\color{blue}{(-1)} &= 11\\
2x + 5 &= 11\\
\color{blue}{x} &\color{blue}{=} \color{blue}{3}
\end{aligned}
$$

**The solution is $(x, y) = (3, -1)$.**

Exercise: Solve the following systems using elimination method