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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

Level 1

 $$\color{blue}{x + y = 4}\\\color{blue}{2x - 3y = 18}$$ $( x , y ) = ( 6 , 2 )$ $( x , y ) = ( -6 , 2 )$ $( x , y ) = ( 6 , -2 )$ $( x , y ) = ( -6 , -2 )$

Level 2

 $$\color{blue}{3x + 5y = -2}\\\color{blue}{2x - y = 3}$$ $( x , y ) = ( 1 , -1 )$ $( x , y ) = ( -1 , -1 )$ $( x , y ) = ( 1 , 1 )$ $( x , y ) = ( -1 , 1 )$