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Wize High School Algebra I Textbook (Common Core) > Systems of Linear Equations

Solving by Elimination (Part 1)

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Solving by Simple Elimination


To solve a system of linear equations using elimination, we want to add or subtract the entire equations in order to eliminate (remove or cancel out) one variable.
  1. Decide on which variable to eliminate and how to eliminate it (add or subtract the equations)
  2. Add or subtract the equations and simplify
  3. Solve for the remaining variable
  4. Use this answer to go back and solve for the other variable

It sounds more complicated than it is! Let's take a look at an example.
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Elimination by Adding

Let's go through the steps for elimination by adding using this system of linear equations as an example:

3x+y=91◯9x−y=152◯\begin{array}{cc} 3x+y=9&\text{\textcircled 1}\\ 9x-y=15&\text{\textcircled 2} \end{array}

Step 1: Decide on a variable to eliminate
Since the xx variables have different numbers in front of them, it might be harder to eliminate.

But looking at the yy variable, since we have +y+y in equation 1◯\text{\textcircled 1} and −y-y in equation 2◯\text{\textcircled 2}, we can eliminate yy by adding the two equations!

Step 2: Add or subtract the equations
Let's add the two equations:

3x+y=9+(9x−y=15)3x+9x+y−y=9+1512x∅=24\begin{array}{crrcl} &\colorFour{3x}&\colorThree{+y}&=&\colorTwo{9}\\ +&(\colorFour{9x}&\colorThree{-y}&=&\colorTwo{15})\\ \hline\\ &\colorFour{3x+9x}&\colorThree{\cancel{+y-y}}&=&\colorTwo{9+15}\\\\ &\colorFour{12x}&\colorThree{\empty}&=&\colorTwo{24} \end{array}
We get 12x=2412x=24

Watch Out!
Be careful when you add! You need to add each item in the equation, not just the first ones!


Step 3: Solve for the remaining variable

12x=24x=2412x=2\begin{array}{lcr} 12x&=&24\\ x&=&\dfrac{24}{12}\\ x&=&2 \end{array}

Step 4: Go back and solve for the other variable
Use x=2\colorbox{yellow}{$x$}=\colorbox{yellow}{$2$} to solve for yy in equation 1◯\text{\textcircled 1}:

3x+y=93(2)+y=96+y=9y=9−6y=3\begin{array}{rcl} 3\colorbox{yellow}{$x$}+y&=&9\\ 3(\colorbox{yellow}{$2$})+y&=&9\\ 6+y&=&9\\ y&=&9-6\\ y&=&3 \end{array}

Therefore, the answer is (2,3)\boxed{(2,3)}.

Optional
You can always check your answer by putting the value for xx and yy back into both equations to see if they "fit" both equations.

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Elimination by Subtracting

Let's go through the steps for elimination by subtracting using this system of linear equations as an example:

3x+4y=311◯3x−5y=−52◯\begin{array}{cc} 3x+4y=31&\text{\textcircled 1}\\ 3x-5y=-5&\text{\textcircled 2} \end{array}

Step 1: Decide on a variable to eliminate
Since the xx variables in both equation are 3x3x, we can subtract the two equations to eliminate xx.


Step 2: Add or subtract the equations
Let's subtract the two equations:

3x+4y=31−(3x−5y=−5)3x−3x+4y−(−5y)=31−(−5)∅9y=36\begin{array}{crcl} &\colorFour{3x}&+\colorThree{4y}&=&\colorTwo{31}\\ -&(\colorFour{3x}&\colorThree{-5y}&=&\colorTwo{-5})\\ \hline\\ &\colorFour{\cancel{3x-3x}}&\colorThree{+4y-(-5y)}&=&\colorTwo{31-(-5)}\\\\ &\colorFour{\empty}&\colorThree{9y}&=&\colorTwo{36} \end{array}
We get 9y=369y=36

Watch Out!
Be very careful when you subtract! You need to subtract each item in the equation, not just the first ones!

Step 3: Solve for the remaining variable

9y=36y=369y=4\begin{array}{rcl} 9y&=&36\\ y&=&\dfrac{36}{9}\\ y&=&4 \end{array}

Step 4: Go back and solve for the other variable
Use y=4\colorbox{yellow}{$y$}=\colorbox{yellow}{$4$} to solve for xx in equation 1◯\text{\textcircled 1}:

3x+4y=313x+4(4)=313x+16=313x=31−163x=15x=153x=5\begin{array}{rcl} 3x+4\colorbox{yellow}{$y$}&=&31\\ 3x+4(\colorbox{yellow}{$4$})& =&31\\ 3x+16&=&31\\ 3x&=&31-16\\ 3x&=&15\\ x&=&\dfrac{15}{3}\\ x&=&5 \end{array}

Therefore, the answer is (5,4)\boxed{(5,4)}.

Optional
You can always check your answer by putting the value for xx and yy back into both equations to see if they "fit" both equations.

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Example: Solving by Simple Elimination

Solve the following system of linear equations using elimination.
3x−y=91◯3x+y=−32◯\begin{array}{cc} 3x-y=9&\text{\textcircled 1}\\ 3x+y=-3&\text{\textcircled 2} \end{array}

Step 1: Decide which variable to eliminate

For the xx variable, we have 3x3x in both equations -- so we can subtract the equations to eliminate xx

For the yy variable, we have −y-y in equation 1◯\text{\textcircled 1} and +y+y in equation 2◯\text{\textcircled 2} -- so we can add the equations to eliminate yy.

So for this question, we can either add or subtract the equations and it will both work!


If we subtract


Step 2: Add or subtract the equations

3x−y=9−(3x+y=−3)3x−3x−y−y=9−(−3)∅−2y=12\begin{array}{crrcl} &\colorFour{3x}&\colorThree{-y}&=&\colorTwo{9}\\ -&(\colorFour{3x}&\colorThree{+y}&=&\colorTwo{-3})\\ \hline\\ &\colorFour{\cancel{3x-3x}}&\colorThree{-y-y}&=&\colorTwo{9-(-3)}\\\\ &\colorFour{\empty}&\colorThree{-2y}&=&\colorTwo{12} \end{array}
So we have −2y=12-2y=12.

Step 3: Solve for the remaining variable

−2y=12y=12−2y=−6\begin{array}{rcl} -2y&=&12\\ y&=&\dfrac{12}{-2}\\ y&=&-6 \end{array}

Step 4: Go back and solve for the other variable

Use y=−6\colorbox{yellow}{$y$}=\colorbox{yellow}{$-6$} to solve for xx in equation 1◯\text{\textcircled 1}
3x−y=93x−(−6)=93x+6=93x=9−63x=3x=33x=1\begin{array}{rcl} 3x-\colorbox{yellow}{$y$}&=&9\\ 3x-(\colorbox{yellow}{$-6$})&=&9\\ 3x+6&=&9\\ 3x&=&9-6\\ 3x&=&3\\ x&=&\dfrac{3}{3}\\ x&=&1 \end{array}

Therefore, the solution to this system of linear equations is (1,−6)\boxed{(1,-6)}.

If we add

Step 2: Add or subtract the equations

3x−y=9+(3x+y=−3)3x+3x−y+y=9+(−3)6x∅=6\begin{array}{crrcl} &\colorFour{3x}&\colorThree{-y}&=&\colorTwo{9}\\ +&(\colorFour{3x}&\colorThree{+y}&=&\colorTwo{-3})\\ \hline\\ &\colorFour{3x+3x}&\colorThree{\cancel{-y+y}}&=&\colorTwo{9+(-3)}\\\\ &\colorFour{6x}&\colorThree{\empty}&=&\colorTwo{6} \end{array}
So we have 6x=66x=6.

Step 3: Solve for the remaining variable


6x=6x=66x=1\begin{array}{rcl} 6x&=&6\\ x&=&\dfrac{6}{6}\\ x&=&1 \end{array}

Step 4: Go back and solve for the other variable

Use x=1\colorbox{yellow}{$x$}=\colorbox{yellow}{$1$} to solve for yy in equation 2◯\text{\textcircled 2} (you get the same answer if you use equation 1◯\text{\textcircled 1}):

3x+y=−33(1)+y=−33+y=−3y=−3−3y=−6\begin{array}{rcl} 3\colorbox{yellow}{$x$}+y&=&-3\\ 3(\colorbox{yellow}{$1$})+y&=&-3\\ 3+y&=&-3\\ y&=&-3-3\\ y&=&-6 \end{array}

Therefore, the solution to this system of linear equations is (1,−6)\boxed{(1,-6)}.

We can confirm the solution by putting the solution into the original formulas.

Practice: Simple Elimination

For each of the following systems of linear equations, decide which variable is the easiest to eliminate, and if adding or subtracting the equations will eliminate that variable.
2x+3y=71◯−2x+5y=02◯\begin{array}{cc} 2x+3y=7&\text{\textcircled 1}\\ -2x+5y=0&\text{\textcircled 2} \end{array}

Practice: Solving by Simple Elimination

Solve the following system of linear equations using elimination.

x+2y=4x+2y=4
3x−2y=43x-2y=4


Practice: Solving by Simple Elimination

Solve the following system of linear equations using elimination.

3x+y=373x+y=37
3x−3y=93x-3y=9