 Elimination method

Lesson

We've already had a look at one way of solving simultaneous equations: 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.

So now we know how to eliminate variables from a system of simultaneous equations by adding or subtracting.

For example I can add $2x-y=1$2xy=1 and $5x+y=2$5x+y=2 to get $7x=3$7x=3, which doesn't have any $y$y's.

OR

I can subtract $3y-x=2$3yx=2 from $3y-2x=0$3y2x=0 to get $-x=-2$x=2, which also doesn't have any $y$y's.

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 the variables we want to eliminate, we can multiply or divide the whole equation by a constant until it gets to the coefficient that we want.

For example, we have $x+3y=5$x+3y=5 & $2y+2x=1$2y+2x=1. Say we want to eliminate the $x$x's first. However, the coefficient of $x$x is $1$1 in one equation and $2$2 in the other. The idea then is to multiply the first equation by $2$2 so that we have an $x$x coefficient of $2$2 for both equations. Then we can apply the same logic as before and add or subtract the equations. Let's see this in action:

Multiplying the first equation by $2$2:

 $2\left(x+3y\right)$2(x+3y) $=$= $2\times5$2×5 $2x+6y$2x+6y $=$= $10$10

Now let's find the difference of this new equation and Equation 2:

 $2x+6y-\left(2y+2x\right)$2x+6y−(2y+2x) $=$= $10-1$10−1 $2x+6y-2y-2x$2x+6y−2y−2x $=$= $9$9 $4y$4y $=$= $9$9 $y$y $=$= $\frac{9}{4}$94​

Now that we have our value for $y$y it's a simple case of substituting it back in any of the equations to get our value for $x$x:

 $x+3y$x+3y $=$= $5$5 $x+\frac{3\times9}{4}$x+3×94​ $=$= $5$5 $x+\frac{27}{4}$x+274​ $=$= $5$5 $x$x $=$= $5-\frac{27}{4}$5−274​ $x$x $=$= $\frac{-7}{4}$−74​

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.

Tips
• 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$2x4=6y can be simplified to $x-2=3y$x2=3y without changing the values of $x$x & $y$y.
• If you have time, remember to check your answers by substituting your $x$x and $y$y values back into the original two equations.

Careful!

Remember to solve for the values of both $x$x AND $y$y! Check if you have both at the end of every simultaneous question problem, unless told otherwise.

Examples

question 1

Use the elimination method to solve for $x$x & $y$y:

 Equation 1 $4x-5y=6$4x−5y=6 - $(1)$(1) Equation 2 $2x-y=3$2x−y=3 - $(2)$(2)

a) Solve for $x$x

Think: What's the easiest way to to make the coefficients of the $y$y terms be the same value, of equal or opposite sign?

Do:

The $y$y coefficients of the two equations are $-5$5 and $-1$1, so let's multiply Equation 2 by $5$5 to make it $-5$5 as well.

 $(2)\times5$(2)×5: $5\left(2x-y\right)$5(2x−y) $=$= $5\times3$5×3 $10x-5y$10x−5y $=$= $15$15 - $(3)$(3)

Let's name this Equation 3. Now that the coefficients of $y$y in Equations 1 & 3 are equal in value and sign, let's subtract one from the other. Let's try Equation 3 take away Equation 1:

 $(3)-(1)$(3)−(1): $10x-5y-\left(4x-5y\right)$10x−5y−(4x−5y) $=$= $15-6$15−6 $10x-5y-4x+5y$10x−5y−4x+5y $=$= $9$9 $6x$6x $=$= $9$9 $x$x $=$= $\frac{3}{2}$32​

b) Solve for $y$y

Think: Which of the $3$3 equations we have would be easiest to derive $y$y from, if we have $x$x?

Do:

Equation 2 seems the most simple so let's substitute our $x$x value in there:

 $x$x→$(2)$(2): $2\times\frac{3}{2}-y$2×32​−y $=$= $3$3 $3-y$3−y $=$= $3$3 $3$3 $=$= $y+3$y+3 $y$y $=$= $0$0

Check:

Subbing in our $x$x and $y$y values into our original equations:

 $4x-5y$4x−5y $=$= $6$6 $\frac{4\times3}{2}-5\times0$4×32​−5×0 $=$= $6$6 $2\times3-0$2×3−0 $=$= $6$6 $6$6 $=$= $6$6
 $2x-y$2x−y $=$= $3$3 $\frac{2\times3}{2}-0$2×32​−0 $=$= $3$3 $3$3 $=$= $3$3
question 2

Use the elimination method to solve for $x$x & $y$y

 Equation 1 $2x-y=1$2x−y=1 - $(1)$(1) Equation 2 $3x+2y=5$3x+2y=5 - $(2)$(2)

a) Solve for $y$y

Think: To solve for $y$y we would need to eliminate $x$x terms, so we need the same coefficient for the $x$x terms for both equations

Do:

The coefficients of $x$x terms in the two equations are $2$2 & $3$3, which are not equal. To make them equal we can look to common multiples of $2$2 & $3$3. The LCM is $6$6, so let's try making the coefficients equal to that.

First we multiply Equation 1 by $3$3:

 $(1)\times3$(1)×3: $3\left(2x-y\right)$3(2x−y) $=$= $3\times1$3×1 $6x-3y$6x−3y $=$= $3$3 - $(3)$(3)

Let's name this Equation 3.

Then we multiply Equation 2 by $2$2:

 $(2)\times2$(2)×2: $2\left(3x+2y\right)$2(3x+2y) $=$= $2\times5$2×5 $6x+4y$6x+4y $=$= $10$10 - $(4)$(4)

Let's name this Equation 4.

Now that the coefficients of $x$x terms are the same with same sign we can eliminate them in Equations 3 & 4 by finding the difference between them (here Equation 4 is subtracting 3):

 $(4)-(3)$(4)−(3): $6x+4y-\left(6x-3y\right)$6x+4y−(6x−3y) $=$= $10-3$10−3 $6x+4y-6x+3y$6x+4y−6x+3y $=$= $7$7 $7y$7y $=$= $7$7 $y$y $=$= $1$1

b) Solve for $x$x

Think: which of the $4$4 equations we have so far would be the easiest to derive $x$x from, knowing $y$y?

Do:

I'm going to use Equation 1 $2x-y=1$2xy=1 to find $x$x as it has the smallest $x$x coefficient and thus will be easier to isolate the $x$x term.

 $y$y→$(1)$(1): $2x-y$2x−y $=$= $1$1 $2x-1$2x−1 $=$= $1$1 $2x$2x $=$= $2$2 $x$x $=$= $1$1

Check:

Subbing in our $x$x and $y$y values into our original equations:

 $2x-y$2x−y $=$= $1$1 $2\times1-1$2×1−1 $=$= $1$1 $2-1$2−1 $=$= $1$1 $1$1 $=$= $1$1
 $3x+2y$3x+2y $=$= $5$5 $3\times1+2\times1$3×1+2×1 $=$= $5$5 $3+2$3+2 $=$= $5$5 $5$5 $=$= $5$5

Question 3

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
1. First solve for $x$x.

2. Now solve for $y$y.

Question 4

Use the elimination method to solve for $x$x and $y$y.

 Equation 1 $-\frac{x}{4}+\frac{y}{5}=8$−x4​+y5​=8 Equation 2 $\frac{x}{5}+\frac{y}{3}=1$x5​+y3​=1
1. First solve for $y$y

2. Now solve for $x$x

Outcomes

10P.LR3.02

Solve systems of two linear equations involving two variables with integer coefficients, using the algebraic method of substitution or elimination