How to Solve Simultaneous Equations by Substitution or Elimination
Key takeaways
- Use substitution when one equation is already written as y = or x = an expression. Substitute that expression into the other equation and solve.
- Use elimination when both equations are in the form ax + by = c. Make one variable’s coefficient match in both equations, then add or subtract the equations to remove it.
- Check the signs in front of the matching coefficients. If the signs are the same, subtract the second equation from the first. If they are opposite, add the two equations.
- When no coefficients match, multiply one or both equations so that they do. If there is no easy multiple, multiply each equation by the other equation’s coefficient, much like finding a common denominator.
- A worded problem needs a worded answer. Writing d = 4 on its own doesn’t answer the question, so finish with a sentence that says what each value means.
In this video
- 0:00 Choosing substitution or elimination
- 1:15 Substitution with two y = equations
- 2:37 Substituting a whole equation
- 4:10 Elimination by subtracting
- 6:10 Elimination by adding
- 7:53 Multiplying one equation to match coefficients
- 10:23 A quick subtraction example
- 11:15 Multiplying both equations
- 13:45 Setting out the equations in order
- 15:30 Word problem: pies and drinks
- 18:37 Word problem: adult and child tickets
- 23:00 Recap of the elimination steps
This maths lesson shows you how to choose between substitution and elimination when you solve simultaneous equations. It then works through eight examples and two word problems. You will keep using this skill over the next couple of years of school.
Should you use substitution or elimination?
Watch this part from 0:00
There are two methods for solving simultaneous equations: substitution and elimination. To choose, look at how the two equations are written.
If either equation has y or x on its own on one side, equal to an expression on the other side, use substitution. For example, y = 2x + 1 is written this way.
If neither equation is written like that, you shouldn’t use substitution. Use elimination when both equations are in the form ax + by = c.
A pair of simultaneous equations is really two linear equations. Their graphs are two straight lines, and the solution is the point of intersection where they cross. That is why the answer is written as a point, such as (3, 7).
How do you solve simultaneous equations by substitution?
Watch this part from 1:15
- Find the equation written as y = or x = an expression.
- Substitute that expression into the other equation.
- Solve for the remaining variable by grouping the variables on one side and the constants on the other.
- Substitute that value into the simpler equation to find the other variable.
- Write the solution as a point and check it in both equations.
Example: y = 2x + 1 and y = x + 4
Both equations give y, so substitute x + 4 for y in the first equation.
x + 4 = 2x + 1
4 = x + 1
x = 3
Substitute x = 3 into the simpler equation:
y = 3 + 4
y = 7
The solution is (3, 7), the point where the two lines cross.
Substituting a whole equation
In the next example, y = 3x − 2 and 2x + y = 8. Substitute the whole expression 3x − 2 for y in the second equation to get 2x + 3x − 2 = 8. That gives 5x = 10, so x = 2. Then y = 3 × 2 − 2 = 4, and the solution is (2, 4). Check it in the other equation: 2 × 2 + 4 = 8.
How do you solve simultaneous equations by elimination?
Watch this part from 4:10
- Check that both equations are in the form ax + by = c.
- Look for a variable with the same coefficient in both equations.
- If there isn’t one, multiply one or both equations so that a coefficient matches.
- Look at the signs in front of the matching coefficients. If they are the same, subtract the second equation from the first. If they are opposite, add the second equation to the first.
- Solve for the remaining variable.
- Substitute that value into either equation to find the other variable.
- Check the solution in both equations.
When you subtract, put brackets around the whole second equation with a minus sign in front. The brackets remind you that subtracting a negative term gives a positive.
Example: x + y = 10 and x − y = 4
Both x terms have a coefficient of 1 and the same sign, so subtract the second equation from the first.
(x + y) − (x − y) = 10 − 4
x − x = 0, so x is eliminated.
y − (−y) = 2y
2y = 6
y = 3
Substitute y = 3 into the second equation:
x − 3 = 4
x = 7
The solution is (7, 3). Check: 7 + 3 = 10 and 7 − 3 = 4.
When do you add the equations in elimination?
Watch this part from 6:10
Add when the signs in front of the matching coefficients are opposite. In 2x + 3y = 12 and 4x − 3y = 6, the x coefficients don’t match, but the y coefficients do. One is +3y and the other is −3y, so add the equations. The y terms cancel and you get 6x = 18, so x = 3. Substituting into the first equation gives 6 + 3y = 12, so y = 2 and the solution is (3, 2).
When the coefficients already match with the same sign, you can subtract straight away. In 4x + 3y = 26 and x + 3y = 11, subtracting gives 3x = 15, so x = 5. Then 5 + 3y = 11 gives y = 2, so the solution is (5, 2).
What do you do when no coefficients match?
Watch this part from 7:53
Multiply one or both equations so that a coefficient matches. This doesn’t change the solution. You are only changing how the equation is written, so the new equations have the same answer as the original ones.
In 3x + 2y = 16 and x + 4y = 12, you could multiply the second equation by 3 to match the x terms. You could also multiply the first equation by 2 to match the y terms. Either works, so pick the option that keeps the numbers smaller. Multiplying the first equation by 2 gives 6x + 4y = 32. Subtracting x + 4y = 12 gives 5x = 20, so x = 4 and y = 2.
Example: 2x + 3y = 13 and 3x − 2y = 0
There is no easy way to turn a 2 into a 3, so multiply both equations. It works much like making a common denominator: multiply each equation by the coefficient from the other one. Match the x terms, because they already have the same sign.
Equation 1 × 3: 6x + 9y = 39
Equation 2 × 2: 6x − 4y = 0
Subtract the second equation from the first. The 6x terms are eliminated.
9y − (−4y) = 13y and 39 − 0 = 39
13y = 39
y = 3
Some students don’t realise you can substitute into the original equations here. You can, because they have the same solution as the multiplied ones.
2x + 3 × 3 = 13
2x + 9 = 13
2x = 4
x = 2
The solution is (2, 3). Check: 4 + 9 = 13 and 6 − 6 = 0.
Does the order of the equations matter?
Watch this part from 13:45
No. Either equation can go first. Set the equations out in whatever way makes the subtracting or rearranging easier. It also helps to write the x term first in each equation.
In 3x + y = 7 and y − 2x = −3, rewrite the second equation as −2x + y = −3. The y terms now match with the same sign, so subtract. That gives 5x = 10, so x = 2. Substituting into 3x + y = 7 gives y = 1, and the solution is (2, 1).
How do you solve a simultaneous equations word problem?
Watch this part from 15:30
Start by finding the variables. Ask what the question actually wants you to find. Then write one equation for each piece of information. In many questions, each sentence gives you one equation.
Here is the first word problem. Three pies and two drinks cost $23. One pie and two drinks cost $13. Find the price of a pie and the price of a drink.
A lot of students overcomplicate this, but the variables are simply those two prices.
Let p = the price of a pie and d = the price of a drink.
3p + 2d = 23
p + 2d = 13
The 2d terms match with the same sign, so subtract. That gives 2p = 10, so p = 5. Then 5 + 2d = 13, so d = 4.
A big mistake is to stop at p = 5 and d = 4. A worded question needs a worded answer that says what the values mean: the price of a pie is $5 and the price of a drink is $4.
Check the values in both equations: 3 × 5 + 2 × 4 = 15 + 8 = 23, and 5 + 2 × 4 = 5 + 8 = 13.
How do you solve a word problem when the equations aren’t obvious?
Watch this part from 18:37
Some word problems are more mixed up, so the two equations don’t sit in two clear sentences. This kind is more likely to come up in an exam.
Word problem: adult and child tickets
A school buys 12 tickets. Adult tickets are $15 and child tickets are $9. The total is $132. How many of each did the school buy?
The pies problem asked for prices. This one gives the prices and asks how many, so the variables are the numbers of tickets.
Let a = the number of adult tickets and c = the number of child tickets.
The 12 tickets are some mix of adult and child tickets:
a + c = 12
Each price times the number of those tickets adds up to the $132 total:
15a + 9c = 132
Look for a way to make the numbers easier first. 15 and 9 are both divisible by 3, and a calculator shows 132 is too. Divide the second equation by 3:
5a + 3c = 44
Multiply the first equation by 3 so the c terms match:
3a + 3c = 36
Subtract the first equation from the second, because that order is easier here.
5a − 3a = 2a and 44 − 36 = 8
2a = 8
a = 4
Substitute into the original first equation, which is the easiest to work with:
4 + c = 12
c = 8
Give the worded answer: the school bought 4 adult tickets and 8 child tickets.
Check: 4 + 8 = 12, and 5 × 4 + 3 × 8 = 20 + 24 = 44.
Frequently asked questions
How do I know whether to use substitution or elimination?
Look at how the equations are written. If either equation is in the form y = or x = an expression, use substitution. If both are in the form ax + by = c, use elimination.
When do you add and when do you subtract in elimination?
Look at the signs in front of the matching coefficients. If the signs are the same, subtract the second equation from the first. If they are opposite, add the two equations.
Does multiplying an equation change the answer?
No. Multiplying a whole equation by a number gives an equation with the same solution. You can substitute your values into the original equations to find the other variable or to check your answer.
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