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

Solving by Elimination (Part 2)

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Solving by Multiplication before Elimination

Simple elimination works very well when the xx or yy variables have the same number values in both equations or if they are opposites (negative) of one another.

What if the numbers in front of the variables are different?

In this case, we have to first multiply one or both equations by a number to end up with the same or opposite numbers in front of one of our variables, then we can use simple elimination again.


Wize Tip
We're trying to find the lowest common multiple between the two numbers in front of one of the variables -- it's kind of like finding a common denominator!


Here are the basic steps:
  1. Multiply one or both equations to end up with the same or opposite numbers in front of one of the variables. ⭐ Tip that always works⭐ Pick one variable then "cross-multiply" the entire equations by the numbers in front of this variable
  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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Example
Solve these simultaneous equations:
3x+4y=51◯2x+5y=12◯\begin{array}{cc} 3x+4y=5&\text{\textcircled 1}\\ 2x+5y=1&\text{\textcircled 2} \end{array}

Step 1: Pick a variable and "cross-multiply"
Let's pick variable xx (if you picked yy, you would end up with the same answer)

3x+4y=5\colorFour3x+4y=5
2x+5y=1\colorThree2x+5y=1

We can cross multiply these so we end up with 6x6x in each equation once we multiply.

2×(3x+4y=5)\colorThree2\times\left(\colorFour3x+4y=5\right)
3×(2x+5y=1)\colorFour3\times\left(\colorThree2x+5y=1\right)

Simplify:
6x+8y=106x+8y=10
6x+15y=36x+15y=3

Step 2: Add or subtract the equations
Since we have 6x6x in both equations, let's subtract the equations to eliminate xx:

6x+  8y=10−(6x+15y=3)6x−6x+8y−15y=10−3∅−7y=7\begin{array}{crrcl} &6x&+~~8y&=&10\\ -&(6x&+15y&=&3)\\ \hline\\ &\colorFive{\cancel{6x-6x}}&+8y-15y&=&10-3\\\\ &\colorFive{\empty}&-7y&=&7 \end{array}
So we get −7y=7-7y=7.

Step 3: Solve for the remaining variable

−7y=7y=7−7y=−1\begin{array}{rcl} -7y&=&7\\ y&=&\dfrac{7}{-7}\\ y&=&-1 \end{array}

Step 4: Go back and solve for the other variable
Using y=−1\colorbox{yellow}{$y$}=\colorbox{yellow}{$-1$} to solve for xx in 2◯\text{\textcircled 2} (you will get the same answer if you used equation 1◯\text{\textcircled 1})

2x+5y=12x+5(−1)=12x−5=12x=1+52x=6x=62x=3\begin{array}{rcl} 2x+5\colorbox{yellow}{$y$}&=&1\\ 2x+5(\colorbox{yellow}{$-1$})&=&1\\ 2x-5&=&1\\ 2x&=&1+5\\ 2x&=&6\\ x&=&\dfrac{6}{2}\\ x&=&3 \end{array}


Therefore, the answer is (3,−1)\boxed{(3,-1)}.

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 Elimination

Solve using elimination.
3x−2y=121◯5x+3y=12◯\begin{array}{cc} 3x-2y=12&\text{\textcircled 1}\\ 5x+3y=1&\text{\textcircled 2} \end{array}


Step 1: Pick a variable and "cross-multiply"
Let's pick variable xx (if you picked yy, your calculations may look a bit different, but you would end up with the same answer)

3x−2y=12\colorFour3x-2y=12
5x+3y=1\colorThree5x+3y=1

We can cross multiply these so we end up with 15x15x in each equation once we multiply.

5×(3x−2y=12)\colorThree5\times\left(\colorFour3x-2y=12\right)
3×(5x+3y=1)\colorFour3\times\left(\colorThree5x+3y=1\right)

Simplify:
15x−10y=6015x-10y=60
15x+9y=315x+9y=3

Step 2: Add or subtract the equations
Since we have 15x15x in both equations, let's subtract the equations to eliminate xx:

15x−10y=60−(15x+  9y=  3)15x−15x−10y−9y=60−3∅−19y=57\begin{array}{crrcl} &15x&-10y&=&60\\ -&(15x&+~~9y&=&~~3)\\ \hline\\ &\colorFive{\cancel{15x-15x}}&-10y-9y&=&60-3\\\\ &\colorFive{\empty}&-19y&=&57 \end{array}
So we get −19y=57-19y=57.

Step 3: Solve for the remaining variable

−19y=57−19y−19=57−19y=−3\begin{array}{rcl} -19y&=&57\\ \dfrac{-19y}{-19}&=&\dfrac{57}{-19}\\ y&=&-3 \end{array}

Step 4: Go back and solve for the other variable
Using y=−3\colorbox{yellow}{$y$}=\colorbox{yellow}{$-3$} to solve for xx in 1◯\text{\textcircled 1} (you will get the same answer if you used equation 2◯\text{\textcircled 2})

3x−2y=123x−2(−3)=123x+6=123x=12−63x3=63x=2\begin{array}{rcl} 3x-2\colorbox{yellow}{$y$}&=&12\\\\ 3x-2(\colorbox{yellow}{$-3$})&=&12\\\\ 3x+6&=&12\\\\ 3x&=&12-6\\\\ \dfrac{3x}{3}&=&\dfrac{6}{3}\\\\ x&=&2\\ \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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Example: Solving by Elimination

Solve these simultaneous equations using elimination.
5x−6y=21◯7x+3y=372◯\begin{array}{cc} 5x-6y=2&\text{\textcircled 1}\\ 7x+3y=37&\text{\textcircled 2} \end{array}

Step 1: Pick a variable and "cross-multiply"
Our "cross-multiplying" trick still works for this question, but there's actually a slightly faster method for this question.

If we look at the yy variables (ignore the + or - sign for now), notice that we have 6y6y in equation 1◯\text{\textcircled 1} and 3y3y in equation 2◯\text{\textcircled 2}. So if double equation 2◯\text{\textcircled 2}, we'll be able to eliminate the yy variable!

5x−6y=22 ×(7x+3y=37)\begin{array}{rc} &5x-6y=2\\ \colorFour{2~\times}&(7x+3y=37)\\ \end{array}

Simplify:
5x−6y=214x+6y=74\begin{array}{c} 5x-6y=2\\ 14x+6y=74\\ \end{array}

Now we see that we can add the two equations to eliminate the 6y6y's in both equations.

Step 2: Add or subtract the equations
Let's add the equations to eliminate yy:

5x−6y=2+(14x+6y=74)5x+14x−6x+6x=2+7419x∅=76\begin{array}{crrcl} &5x&-6y&=&2\\ +&(14x&+6y&=&74)\\ \hline\\ 5x+14x&&\colorFive{\cancel{-6x+6x}}&=&2+74\\\\ 19x&&\colorFive{\empty}&=&76 \end{array}
So we get 19x=7619x=76.

Step 3: Solve for the remaining variable

19x=7619x19=7619x=4\begin{array}{rcl} 19x&=&76\\[0.5em] \dfrac{19x}{19}&=&\dfrac{76}{19}\\[0.5em] x&=&4 \end{array}

Step 4: Go back and solve for the other variable
Using x=4\colorbox{yellow}{$x$}=\colorbox{yellow}{$4$} to solve for xx in 1◯\text{\textcircled 1} (you will get the same answer if you used equation 2◯\text{\textcircled 2})

5x−6y=25(4)−6y=220−6y=2−6y=2−20−6y−6=18−6y=−3\begin{array}{rcl} 5\colorbox{yellow}{$x$}-6y&=&2\\\\ 5(\colorbox{yellow}{$4$})-6y&=&2\\\\ 20-6y&=&2\\\\ -6y&=&2-20\\\\ \dfrac{-6y}{-6}&=&\dfrac{18}{-6}\\\\ y&=&-3 \end{array}


Therefore, the answer is (4,3)\boxed{(4,3)}.

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.

Practice: Solving by Elimination

Solve this system of linear equations using elimination.

7x+3y=47x+3y=4
5x+2y=45x+2y=4

Practice: Solving by Multiplication before Elimination


Solve these simultaneous equations using elimination.

3x+6y=123x+6y=12
−2x+2y=10-2x+2y=10

Practice: Solving by Elimination

A high school sports committee is selling raffle tickets at a football game to raise money for a local charity. They decide to sell two types of tickets -- adults and students. The adult tickets are priced at $7 each and the student tickets are priced at $5 each. If the sports committee sold 275 tickets in total and raised $1615, how much of each type of raffle ticket did they sell?