7.03 Solving systems of equations with the elimination method
We've already had a look at two ways of solving a system of equations: graphically and by the substitution method. However, there's another equally useful method - the elimination method. It works by adding or subtracting equations from one another to eliminate one variable, leaving us with the other variable to solve on its own.
Worked examples
Question 1
Solve the following system of equations using the elimination method:
$2x-y=1$2x−y=1 and $5x+y=2$5x+y=2
Think: Look at the $x$x and $y$y terms in each equation: $2x$2x, $-y$−y, $5x$5x, $y$y. Are there any like terms that have the opposite coefficients? Because the $y$y terms have opposite coefficients, the equations can be added together to eliminate the $y$y term.
Do: Add $2x-y=1$2x−y=1 and $5x+y=2$5x+y=2 .
| $2x-y+5x+y$2x−y+5x+y | $=$= | $1+2$1+2 |
| $2x+5x-y+y$2x+5x−y+y | $=$= | $3$3 |
| $7x$7x | $=$= | $3$3 |
| $x$x | $=$= | $\frac{3}{7}$37 |
Now, substitute $x$x-value into one of the equations from the system:
| $5x+y$5x+y | $=$= | $2$2 |
| $5\times\frac{3}{7}+y$5×37+y | $=$= | $2$2 |
| $\frac{15}{7}+y$157+y | $=$= | $2$2 |
| $y$y | $=$= | $\frac{14}{7}-\frac{15}{7}$147−157 |
| $y$y | $=$= | $\frac{-1}{7}$−17 |
The solution: $\frac{3}{7},\frac{-1}{7}$37,−17
Reflect: Check your answer.
| equation 1 | equation 2 |
| $2x-y=1$2x−y=1 | $5x+y=2$5x+y=2 |
| $2\times\frac{3}{7}-\frac{-1}{7}=1$2×37−−17=1 | $5\times\frac{3}{7}+\frac{-1}{7}=2$5×37+−17=2 |
| $\frac{6}{7}+\frac{1}{7}=1$67+17=1 | $\frac{15}{7}-\frac{1}{7}=2$157−17=2 |
| $1=1$1=1 | $2=2$2=2 |
Question 2
Solve the following system of equations using the elimination method:
$3y-x=2$3y−x=2 and $3y-2x=0$3y−2x=0
Think: Look at the $x$x and $y$y terms in each equation: $3y$3y, $-x$−x, $3y$3y, $-2x$−2x. There are two like terms with the same coefficients: $3y$3y. Should we add the equations or subtract them? Because the $y$y terms have the same coefficients, the equations can be subtracted to eliminate the $y$y term.
Do: Subtract $3y-x=2$3y−x=2 and $3y-2x=0$3y−2x=0 .
| $3y-x-\left(3y-2x\right)$3y−x−(3y−2x) | $=$= | $2-0$2−0 | remember to distribute the (-1) |
| $3y-x-3y+2x$3y−x−3y+2x | $=$= | $2$2 | combine like terms |
| $x$x | $=$= | $2$2 |
Now, substitute $x$x-value into one of the equations from the system:
| $3y-x$3y−x | $=$= | $2$2 |
| $3y-2$3y−2 | $=$= | $2$2 |
| $3y$3y | $=$= | $4$4 |
| $y$y | $=$= | $\frac{4}{3}$43 |
The solution: $2,\frac{4}{3}$2,43
Reflect: Check your answer.
| equation 1 | equation 2 |
| $3y-x=2$3y−x=2 | $3y-2x=0$3y−2x=0 |
| $3\times\frac{4}{3}-2=2$3×43−2=2 | $3\times\frac{4}{3}-2\times2=0$3×43−2×2=0 |
| $4-2=2$4−2=2 | $4-4=0$4−4=0 |
| $2=2$2=2 | $0=0$0=0 |
We can see that we should use addition when the coefficients of the variable we want to eliminate are equal and opposite in sign, and subtraction when they're equal and the same sign. But what happens when they don't have the same coefficients at all?
Using multiplication & division
When we don't have the same value coefficients for like terms we want to eliminate, we can multiply or divide the whole equation by a constant until it gets to the coefficient that we want.
Worked example
Question 3
Solve the following system of equations using the elimination method.
$x+3y=5$x+3y=5 and $2y+2x=1$2y+2x=1
Think: Look at the $x$xand $y$y terms in each equation: $x$x , $3y$3y , $2y$2y , $2x$2x . Since there are no like terms, we need to find the LCM for each pair of terms' coefficients ($1$1and $2$2: LCM =$2$2; $3$3 and $2$2: LCM = $6$6). To eliminate a term, a pair of like terms' coefficients must be opposites.
Do:
Option 1 - To eliminate the $x$x term, use the LCM for $1$1 and $2$2 (LCM = 2). This requires the equation 1 to be multiplied by $-2$−2.
Option 2 - To eliminate the y term, use the LCM for $3$3 and $2$2 (LCM 6). This requires the equation 1 to be multiplied by $-2$−2 and equation 2 by$3$3.
Option 1 is the most efficient because it involves the manipulation of one equation.
| Equation 1 | ||
|---|---|---|
| $x+3y$x+3y | $=$= | $5$5 |
| $-2\left(x+3y\right)$−2(x+3y) | $=$= | $-2\times5$−2×5 |
| $-2x-6y$−2x−6y | $=$= | $-10$−10 |
Now, add the modified equation 1 and equation 2:
| $-2x-6y+2y+2x$−2x−6y+2y+2x | $=$= | $-10+1$−10+1 |
| $-2x+2x-6y+2y$−2x+2x−6y+2y | $=$= | $-9$−9 |
| $-4y$−4y | $=$= | $-9$−9 |
| $y$y | $=$= | $\frac{9}{4}$94 |
Now, let's find the x value by using equation 1:
| $x+3y$x+3y | $=$= | $5$5 |
| $x+\frac{3\times9}{4}$x+3×94 | $=$= | $5$5 |
| $x+\frac{27}{4}$x+274 | $=$= | $5$5 |
| $x$x | $=$= | $\frac{20}{4}-\frac{27}{4}$204−274 |
| $x$x | $=$= | $\frac{-7}{4}$−74 |
The solution is : $\frac{-7}{4},\frac{9}{4}$−74,94
Reflect: Another way to approach this problem is to divide Equation 2 by $2$2, make the coefficient $1$1 and we'll get the same answer, but we might be dealing with more fractions and it won't be as easy.
- Being organized is key. Name your equations by writing $(1)$(1) & $(2)$(2) next to them, and whenever you create new equations out of one or both of them, you can name them $(3)$(3), $(4)$(4), etc. When adding two equations you can write it in shorthand as $\text{(3) + (1)}$(3) + (1), and similarly for all the other operations. This helps to keep your ideas in order and not get confused.
- When dealing with equations with big numbers, see if you can simplify them before beginning to solve them. Eg. $2x-4=6y$2x−4=6y can be simplified to $x-2=3y$x−2=3y without changing the values of $x$x & $y$y.
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If you have time, remember to check your answers by substituting your $x$x and $y$y values back into the original two equations.
Remember to solve for the values of both $x$x and $y$y! Check if you have both at the end of every system of equations problem, unless told otherwise.
Practice questions
Questions 4
How would we eliminate a variable in the following system of equations?
| $2x$2x | $+$+ | $3y$3y | $=$= | $4$4 |
| $5x$5x | $-$− | $y$y | $=$= | $3$3 |
Eliminate $y$y by multiplying the second equation by $-3$−3 and then adding the equations.
AEliminate $y$y by multiplying the second equation by $3$3 and then adding the equations.
BEliminate $x$x by multiplying the second equation by $2$2 and then subtracting the equations.
CEliminate $x$x by multiplying the first equation by $-5$−5 and then adding the equations.
D
Question 5
Use the elimination method by subtraction to solve for $x$x and $y$y.
| Equation 1 | $3x-2y=-3$3x−2y=−3 |
| Equation 2 | $3x+5y=39$3x+5y=39 |
First solve for $y$y.
Now solve for $x$x.
Question 6
Use the elimination method to solve for $x$x and $y$y.
| Equation 1 | $-6x-2y=46$−6x−2y=46 |
| Equation 2 | $-30x-6y=246$−30x−6y=246 |
First solve for $x$x.
Now solve for $y$y.