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Study Guides > Intermediate Algebra

Solve Systems by Elimination

Learning Outcomes

  • Solve systems of equations by addition
  • Express the solution of a system of dependent equations containing two variables
A third method of solving systems of linear equations is the elimination method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by elimination.

Example

Solve the given system of equations by elimination.
[latex]\begin{array}{l}x+2y=-1\hfill \\ -x+y=3\hfill \end{array}[/latex]

Answer: Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[/latex] in the second equation, [latex]–1[/latex], is the opposite of the coefficient of [latex]x[/latex] in the first equation, [latex]1[/latex]. We can add the two equations to eliminate [latex]x[/latex] without needing to multiply by a constant.

[latex]\dfrac{\begin{array}{l}\hfill \\ \:\:x+2y=-1\hfill \\ -x+y=3\hfill \end{array}}{\text{}\text{}\text{}\text{}\text{}\:\:\:\:\:\:3y=2}[/latex]
  Now that we have eliminated [latex]x[/latex], we can solve the resulting equation for [latex]y[/latex].

[latex]\begin{array}{l}3y=2\hfill \\ \text{ }y=\dfrac{2}{3}\hfill \end{array}[/latex]

Then, we substitute this value for [latex]y[/latex] into one of the original equations and solve for [latex]x[/latex]

[latex]\begin{array}{l}\text{ }-x+y=3\hfill \\ \text{ }-x+\dfrac{2}{3}=3\hfill \\ \text{ }-x=3-\dfrac{2}{3}\hfill \\ \text{ }-x=\dfrac{7}{3}\hfill \\ \text{ }\:\:\:\:\:x=-\dfrac{7}{3}\hfill \end{array}[/latex]

The solution to this system is [latex]\left(-\dfrac{7}{3},\dfrac{2}{3}\right)[/latex]. Check the solution in the first equation.

[latex]\begin{array}{llll}\text{ }x+2y=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }\left(-\dfrac{7}{3}\right)+2\left(\dfrac{2}{3}\right)=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }-\dfrac{7}{3}+\dfrac{4}{3}=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }-\dfrac{3}{3}=-1\hfill & \hfill & \hfill & \hfill \\ \text{ }-1=-1\hfill & \hfill & \hfill & \text{True}\hfill \end{array}[/latex]

We gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, you will see that the equations intersect at the solution. We do not need to ask whether there may be a second solution, because observing the graph confirms that the system has exactly one solution. A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.

In the following video, you will see another example of how to use the method of elimination to solve a system of linear equations. https://youtu.be/M4IEmwcqR3c Sometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system and solve it. In the next example, you will see a technique where we multiply one of the equations in the system by a number that will allow us to eliminate one of the variables.

Example

Solve the given system of equations by the elimination method.

[latex]\begin{array}{l}3x+5y=-11\hfill \\ x - 2y=11\hfill \end{array}[/latex]

Answer: Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[/latex] in it and the second equation has [latex]x[/latex]. So if we multiply the second equation by [latex]-3,\text{}[/latex] the x-terms will add to zero.

[latex]\begin{array}{llll}\text{ }x - 2y=11\hfill & \hfill & \hfill & \hfill \\ -3\left(x - 2y\right)=-3\left(11\right)\hfill & \hfill & \hfill & \text{Multiply both sides by }-3.\hfill \\ -3x+6y=-33\hfill & \hfill & \hfill & \text{Use the distributive property}.\hfill \end{array}[/latex]
Now, let us add them.

[latex]\begin{array}\ \hfill 3x+5y=−11 \\ \hfill −3x+6y=−33 \\ \text{_____________} \\ \hfill 11y=−44 \\ \hfill y=−4 \end{array}[/latex]

For the last step, we substitute [latex]y=-4[/latex] into one of the original equations and solve for [latex]x[/latex].

[latex]\begin{array}{c}3x+5y=-11\\ 3x+5\left(-4\right)=-11\\ 3x - 20=-11\\ 3x=9\\ x=3\end{array}[/latex]

Our solution is the ordered pair [latex]\left(3,-4\right)[/latex]. Check the solution in the original second equation.
[latex]\begin{array}{llll}\text{ }x - 2y=11\hfill & \hfill & \hfill & \hfill \\ \left(3\right)-2\left(-4\right)=11\hfill & \hfill & \hfill & \hfill \\ 11=11\hfill & \hfill & \hfill & \text{True}\hfill \end{array}[/latex]
  A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.

Below is another video example of using the elimination method to solve a system of linear equations. https://youtu.be/_liDhKops2w In the next example, we will see that sometimes both equations need to be multiplied by different numbers in order for one variable to be eliminated.

Example

Solve the given system of equations in two variables by elimination.
[latex]\begin{array}{c}2x+3y=-16\\ 5x - 10y=30\end{array}[/latex]

Answer: One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex], so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex].

[latex]\begin{array}{l} -5\left(2x+3y\right)=-5\left(-16\right)\hfill \\ \text{ }-10x - 15y=80\hfill \\ \text{ }2\left(5x - 10y\right)=2\left(30\right)\hfill \\ \text{ }10x - 20y=60\hfill \end{array}[/latex]
Then, we add the two equations together.

[latex]\begin{array}\ −10x−15y=80 \\ \:\:10x−20y=60 \\ \text{______________} \\ \text{ }\:\:\:\:\:\:\:\:\:\:−35y=140 \\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y=−4 \end{array}[/latex]

Substitute [latex]y=-4[/latex] into the original first equation.

[latex]\begin{array}{c}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{array}[/latex]

The solution is [latex]\left(-2,-4\right)[/latex]. Check it in the second original equation.
[latex]\begin{array}{r}\hfill \text{ }5x - 10y=30\\ \hfill 5\left(-2\right)-10\left(-4\right)=30\\ \hfill \text{ }-10+40=30\\ \hfill \text{ }30=30\end{array}[/latex]
A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.

Below is a summary of the general steps for using the elimination method to solve a system of equations.

How To: Given a system of equations, solve using the elimination method

  1. Write both equations with and y-variables on the left side of the equal sign and constants on the right.
  2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
  3. Solve the resulting equation for the remaining variable.
  4. Substitute that value into one of the original equations and solve for the second variable.
  5. Check the solution by substituting the values into the other equation.
In the next example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.

Example

Solve the given system of equations in two variables by elimination.

[latex]\begin{array}{l}\dfrac{x}{3}+\dfrac{y}{6}=3\hfill \\ \dfrac{x}{2}-\dfrac{y}{4}=\text{ }1\hfill \end{array}[/latex]

Answer: First clear each equation of fractions by multiplying both sides of the equation by the least common denominator

[latex]\begin{array}{l}6\left(\dfrac{x}{3}+\dfrac{y}{6}\right)=6\left(3\right)\hfill \\ \text{ }2x+y=18\hfill \\ 4\left(\dfrac{x}{2}-\dfrac{y}{4}\right)=4\left(1\right)\hfill \\ \text{ }2x-y=4\hfill \end{array}[/latex]

Now multiply the second equation by [latex]-1[/latex] so that we can eliminate the x-variable.

[latex]\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}[/latex]

Add the two equations to eliminate the x-variable and solve the resulting equation.

[latex]\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}[/latex]

Substitute [latex]y=7[/latex] into the first equation.

[latex]\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\dfrac{11}{2}\hfill \end{array}[/latex]

The solution is [latex]\left(\dfrac{11}{2},7\right)[/latex]. Check it in the other equation.

[latex]\begin{array}{c}2x-y=4\\ 2(\dfrac{11}{2})-7=4\\ 11-7=4 \\ 4=4\end{array}[/latex]

In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions. https://youtu.be/s3S64b1DrtQ Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as [latex]0=0[/latex]. The last example includes two equations that represent the same line and are therefore dependent.

Example

Find a solution to the system of equations using the elimination method.

[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]

Answer: With the elimination method, we want to eliminate one of the variables by adding the equations. In this case, focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by [latex]-3[/latex], then we will be able to eliminate the [latex]x[/latex] -variable.

[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}[/latex]

Now add the equations.

[latex]\begin{array} \hfill−3x−9y=−6 \\ \hfill3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}[/latex]

We can see that there will be an infinite number of solutions that satisfy both equations. If we rewrote both equations in slope-intercept form, we might know what the solution would look like before adding. Look at what happens when we convert the system to slope-intercept form.

[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\dfrac{3}{9}x+\dfrac{6}{9}\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \end{array}[/latex]

See the graph below. Notice the results are the same. The general solution to the system is [latex]\left(x, -\dfrac{1}{3}x+\dfrac{2}{3}\right)[/latex]. A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.

In the following video, we show another example of solving a system that is dependent using elimination. https://youtu.be/NRxh9Q16Ulk In our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other. https://youtu.be/z5_ACYtzW98

Licenses & Attributions

CC licensed content, Shared previously

  • Ex 1: Solve a System of Equations Using the Elimination Method. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex 2: Solve a System of Equations Using the Elimination Method. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: Solve a System of Equations Using Eliminations (Fractions). Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: System of Equations Using Elimination (Infinite Solutions). Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: System of Equations Using Elimination (No Solution). Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.

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