Elimination Method – Steps, Techniques, and Examples

The elimination method is an important technique widely used when we’re working with systems of linear equations. It is essential to add this to your toolkit of Algebra techniques to help you work with different word problems involving systems of linear equations.

The elimination method allows us to solve a system of linear equations by “eliminating” variables. We eliminate variables by manipulating the given system of equations.

Knowing the elimination method by heart allows you to work on different problems such as mixture, work and number problems with ease. In this article, we’ll break down the process of solving a system of equations using the elimination method. We’ll also show you applications of this method when solving word problems.

What Is the Elimination Method?

The elimination method is a process that uses elimination to reduce the simultaneous equations into one equation with a single variable. This leads to the system of linear equations being reduced to a single-variable equation, making it easier for us.

This is one of the most helpful tools when solving systems of linear equations.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}&{\color{red} \cancel{-40x}} &+ 12 y&=-400\phantom{x}\\+&{\color{red} \cancel{40x}}&+ 2y&=-300\phantom{1}\end{array}}\\ &\begin{array}{cccc}\phantom{+xx} &\phantom{7xxx}&14y&=-700\\&&y&=\phantom{}-50\end{array}\end{matrix}\end{aligned}

Take a look at the equations shown above. By adding the equations, we’ve managed to eliminate $x$ and leave a simpler linear equation, $14y = -700$. From this, it will be easier for us to find the value of $y$ and eventually find the value of $x$. This example shows how easy it is for us to solve a system of equations by manipulating the equations.

The elimination method is possible all thanks to the following algebraic properties:

• Multiplication Properties

In the next section, we’ll show you how these properties are applied. We’ll also break down the process of solving a system of equations by using the elimination method.

How To Solve System of Equations by Elimination?

To solve a system of equations, rewrite the equations so that when these two equations are added or subtracted, one or two variables can be eliminated. The goal is to rewrite the equation so that it will be easier for us to eliminate the terms.

These steps will help you rewrite the equations and apply the elimination method:

1. Multiply one or both of the equations by a strategic factor.
• Focus on making one of the terms be the negative equivalent or be identical to the term found in the remaining equation.
• Our goal is to eliminate the terms sharing the same variable.
1. Add or subtract the two equations depending on the result from the previous step.
• If the terms we want to eliminate are negative equivalents of each other, add the two equations.
• If the terms we want to eliminate are identical, subtract the two equations.
2. Now that we’re working with a linear equation, solve for the remaining variable’s value.
3. Use the known value and substitute it into either of the original equations.
• This results in another equation with one unknown.
• Use this equation to solve for the remaining unknown variable.

Why don’t we apply these steps to solve the system of linear equation $\begin{array}{ccc}x&+\phantom{x}y&=5\\-4x&+3y&= -13 \end{array}$?

1. Multiply both sides of the first equation by $4$ so that we end with $4x$.

\begin{aligned}\begin{array}{ccc}{\color{Teal}4}x&+{\color{Teal}4}y&={\color{Teal}4}(5)\\-4x&+3y&= -13 \\&\downarrow\phantom{x}\\4x&+ 4y&= 20\\ -4x&+3y&= -13\end{array} \end{aligned}

We want $4x$ on the first equation so that we can eliminate $x$ in this equation. We can also eliminate $y$ first by multiplying the first equation’s sides by $3$. That’s for you to work on your own, but for now, let’s continue by eliminating $x$.

1. Since we’re working with $4x$ and $-4x$, add the equations to eliminate $x$ and have one equation in terms of $y$.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{Teal}4x}&+4y &=\phantom{+}20\\+\phantom{xx}\bcancel{\color{Teal}-4x} &+ 3y&= -13\end{array}}\\ &\begin{array}{cccc}\phantom{+} & \phantom{xxxx}&7y&=\phantom{+}7\end{array}\end{matrix} \end{aligned}

1. Solve for $y$ from the resulting equation.

\begin{aligned}7y &= 7\\y &= 1\end{aligned}

1. Substitute $y =1$ into either of the equations from $\begin{array}{ccc}x&+\phantom{x}y&=5\\-4x&+3y&= -13 \end{array}$. Use the resulting equation to solve for $x$.

\begin{aligned}x + y&= 5\\ x+ {\color{Teal} 1} &= 5\\x& =4\end{aligned}

This means that the given system of linear equations is true when $x = 4$ and $y = 1$. We can also write its solution as $(4, 5)$. To double-check the solution, you can substitute these values into the remaining equation.

\begin{aligned}-4x + 3y&= -13\\-4(4) + 3(1)&= -13\\-13&= -13 \checkmark\end{aligned}

Since the equation holds true when $x = 4$ and $y =1$, this further confirms that the solution to the system of equation is indeed $(4, 5)$. When working a system of linear equations, apply a similar process as we have done in this example. The level of difficulty may change but the fundamental concepts needed to use the elimination method remains constant.

In the next section, we’ll cover more examples to help you master the elimination method. We’ll also include word problems involving systems of linear equations to make you appreciate this technique more.

Example 1

Use the elimination method to solve the system of equations, $\begin{array}{ccc}4x- 6y&= \phantom{x}26 \,\,(1)\\12x+8y&= -12 \,\,(2)\end{array}$.

Solution

Inspect the two equations to see which equation would be easier for us to manipulate.

\begin{aligned} \begin{array}{ccc}4x- 6y&= \phantom{x}26\,\,(1)\\12x+8y&= -12\,\,(1)\end{array} \end{aligned}

Since $12x$ is a multiple of $4x$, we can multiply $3$ on both sides of Equation (1) so we’ll have $12x$ in the resulting equation. This leads to us having a $12x$ on both equations, making it possible for us to eliminate later.

\begin{aligned} \begin{array}{ccc}{\color{DarkOrange}3}(4x)& -{\color{DarkOrange}3}(6)y&={\color{DarkOrange}3}(26)\\12x&+8y&= -12\,\, \\&\downarrow\phantom{x}\\12x&- 18y&= 78\,\,\,\, \\ 12x&+8y&= -12\end{array}\end{aligned}

Since the two resulting equations have $12x$, subtract the two equations to eliminate $12x$. This leads to a single equation with one variable.

\begin{aligned}\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{DarkOrange}12x}& -18y &=\phantom{+}78\\-\phantom{xx}\bcancel{\color{DarkOrange}12x} &+ 8y&= -12\end{array}}\\ &\begin{array}{cccc}\phantom{+} & \phantom{xxxx}&-26y&=\phantom{+}90\end{array}\end{matrix}\end{aligned}

Find the value of $y$ using the resulting equation by dividing both sides by $-26$.

\begin{aligned}-26y&= 90\\y&= -\dfrac{90}{26}\\&= -\dfrac{45}{13}\end{aligned}

Now, substitute $y = -\dfrac{45}{13}$ into one of the equations from $\begin{array}{ccc}4x- 6y&= \phantom{x}26 \,\,(1)\\12x+8y&= -12 \,\,(2)\end{array}$.

\begin{aligned}4x – 6y&= 26\\4x -6\left(-\dfrac{45}{13}\right)&= 26\\4x + \dfrac{270}{13}&= 26\end{aligned}

Use the resulting equation to solve $x$ then write down the solution to our system of linear equations.

\begin{aligned}4x + \dfrac{270}{13}&= 26\\52x + 270&= 338\\52x&=68\\x&= \dfrac{17}{13}\end{aligned}

Hence, we have $x = \dfrac{17}{13}$ and $y = -\dfrac{45}{13}$. We can double-check our solution by substituting these values into the remaining equation and see if the equation still holds true.

\begin{aligned}12x+8y&= -12\\ 12\left({\color{DarkOrange}\dfrac{17}{13}}\right)+ 8\left({\color{DarkOrange}-\dfrac{45}{13}}\right)&= -12\\-12 &= -12 \checkmark\end{aligned}

This confirms that the solution to our system of equations is $\left(\dfrac{17}{13}, -\dfrac{45}{13}\right)$.

We’ve shown you examples where we only manipulate one equation to eliminate one term. Let’s now try out an example where we’re required to multiply different factors on both equations.

Example 2

Use the elimination method to solve the system of equations $\begin{array}{ccc}3x- 4y&= \phantom{x}12\,\,(1)\\4x+3y&= \phantom{x}16\,\,(2)\end{array}$.

Solution

This example shows that we sometimes need to work on both linear equations before we can eliminate either $x$ or $y$. Since our first two examples show you how to eliminate the terms with $x$, let’s make it our goal to eliminate $y$ first this time.

Rewrite the terms with $y$ in both equations by multiplying $3$ on both sides of Equation (1) and $4$ on both sides of Equation (2).

\begin{aligned} \begin{array}{ccc}{\color{Orchid}3}(3x)& -{\color{Orchid}3}(4y)&={\color{Orchid}3}(12)\\{\color{Orchid}4}(4x)& -{\color{Orchid}4}(3y)&={\color{Orchid}4}(16)\,\, \\&\downarrow\phantom{x}\\9x&- 12y&= 36\,\, \\ 16x&+ 12y&= 64\,\,\end{array}\end{aligned}

Now that we have $-12y$ and $12y$ on both resulting equations, add the two equations to eliminate $y$.

\begin{aligned} \begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}9x& -\bcancel{\color{Orchid}12y} &=\phantom{+}36\\+\phantom{xx}16x &+ \bcancel{\color{Orchid}12y} &= \phantom{x}64\end{array}}\\ &\begin{array}{cccc}\phantom{+} &25x&\phantom{xxxxx}&=100\end{array}\end{matrix}\end{aligned}

The system of equations has now been reduced to a linear equation with $x$ as the only unknown. Divide both sides of the equation by $25$ to solve for $x$.

\begin{aligned}25x &= 100\\x&= \dfrac{100}{25}\\&= 4\end{aligned}

Substitute $x =4$ into either of the system of linear equations to solve for $y$. In our case, let’s use Equation (1).

\begin{aligned}3x-4y&= 12\\3(4) -4y&= 12\\-4y&= 0\\y &=0\end{aligned}

Hence, the solution to our system of linear equations is $(4, 0)$.

Feel free to substitute these values into either Equation (1) or Equation (2) to double-check the solution. For now, let’s try out a word problem involving systems of linear equations to help you appreciate this topic even more!

Example 3

Amy has a favorite pastry shop where she often buys donuts and coffee. On Tuesday, she paid $\$12$for two boxes of donuts and one cup of coffee. On Thursday, she purchased one box of donuts and two cups of coffee. She paid$\$9$ this time. How much does each box of donuts cost? How about one cup of coffee?

Solution

First, let’s set up the system of linear equations that represent the situation.

• Let $d$ represent the cost of one box of donuts.
• Let $c$ represent the cost of one cup of coffee.

Each equation’s right-hand side represents the total cost in terms of $d$ and $c$. Hence, we have $\begin{array}{ccc}2d+ c&= \phantom{x}12\,\,(1)\\d+2c&= \phantom{xc}9\,\,(2)\end{array}$. Now that we have a system of linear equations, apply the elimination method to solve for $c$ and $d$.

\begin{aligned} \begin{array}{ccc}2d& + c\phantom{xxx}&= 12\phantom{xx}\\{\color{Green}2}(d)& +{\color{Green}2}(2c)&={\color{Green}2}(9)\,\, \\&\downarrow\phantom{x}\\2d&+ c\,\,&= 12\,\, \\ 2d&+ 4c&= 18\,\,\end{array}\end{aligned}

Once we’ve eliminated one of the variables (for our case, it’s $d$), solve the resulting equation to find $c$.

\begin{matrix}&\underline{\begin{array}{cccc}\phantom{+xxx}\bcancel{\color{Green}2d} & + c&=\phantom{+}12\\-\phantom{xx}\bcancel{\color{Green}2d} &+ 4c&= \phantom{x}18\end{array}}\\ &\begin{array}{cccc}\phantom{+} &\phantom{xxxx}&-3c&=-6\\&\phantom{xx}&c&= 2\end{array}\end{matrix}

Substitute $c = 2$ into either of the system of linear equations to solve for $d$.

\begin{aligned}2d + c &= 12\\2d + 2&= 12\\2d&= 10\\d&= 5\end{aligned}

This means that one box of donuts costs $\$5$while a cup of coffee costs$\$2$ at Amy’s favorite pastry shop.

Practice Question

1. Which of the following shows the solution to the system of equations $\begin{array}{ccc}3a – 4b&= \phantom{x}18\\3a – 8b&= \phantom{x}26\end{array}$?
A.$a=-2,b=\dfrac{10}{3}$
B. $a=\dfrac{10}{3},b=-2$
C. $a=-2,b=-\dfrac{10}{3}$
D. $a=\dfrac{10}{3},b=2$

2. Which of the following shows the solution to the system of equations $\begin{array}{ccc}4x + 5y&= \phantom{x}4\\5x- 4y&= -2\end{array}$?
A. $\left(-\dfrac{28}{41},-\dfrac{6}{41}\right)$
B. $\left(-\dfrac{6}{41},-\dfrac{28}{41}\right)$
C. $\left(\dfrac{28}{41},\dfrac{6}{41}\right)$
D. $\left(\dfrac{6}{41},\dfrac{28}{41}\right)$