If we have two or more equations containing the same variables, we call them a system of equations. Sometimes we may also refer to them as simultaneous equations.
Given two variables, usually (but not always) $x$x and $y$y, and two equations involving those variables, we can try to find a common pair of $x$x and $y$y values that satisfy both of the equations simultaneously. If we can find such a pair of values, then we will have found a solution to the system of equations.
There are a number of different methods for solving systems of equations.
Solving graphically
Simultaneous equations with two variables can be represented as graphs on a number plane. The most common systems of equations involve linear equations, which we can represent with straight line graphs. Our aim is to find a solution that solves both equations, and graphically this corresponds to finding a point of intersection of the two straight lines.
Worked example
Example 1
Find the solution to the simultaneous equations $y=5x$y=5x and $y=x+2$y=x+2.
Think: We can solve this system of equations by sketching the two lines on the same number plane, then reading off the coordinates of the point of intersection.
Do: Here is a sketch of the two lines:
Reading off the graph, we can see that the point of intersection is $\left(0.5,2.5\right)$(0.5,2.5). Therefore the solution to this system of equations is the pair $x=0.5$x=0.5 and $y=2.5$y=2.5.
Practice questions
Question 1
Consider the following linear equations:
$y=2x-4$y=2x−4 and $y=-2x-4$y=−2x−4
What are the gradient and $y$y-intercept of the line $y=2x-4$y=2x−4?
gradient $\editable{}$ $y$y-value of $y$y-intercept $\editable{}$ What are the intercepts of the line $y=-2x-4$y=−2x−4?
$x$x-value of $x$x-intercept $\editable{}$ $y$y-value of $y$y-intercept $\editable{}$ Plot the lines of the 2 equations on the same graph.
Loading Graph...State the values of $x$x and $y$y which satisfy both equations.
$x$x = $\editable{}$
$y$y = $\editable{}$
question 2
Consider the following linear equations:
$y=5x-7$y=5x−7 and $y=-x+5$y=−x+5
Plot the lines of the two equations on the same graph.
Loading Graph...State the values of $x$x and $y$y which satisfy both equations.
$x$x = $\editable{}$
$y$y = $\editable{}$
Parallel lines
If two (or more) lines are parallel and distinct then they will never cross, and as such there will be no solution to a system of equations that contains parallel lines. For example, the simultaneous equations $y=3x-1$y=3x−1 and $y=3x+6$y=3x+6 both have a gradient of $3$3, and so we know that they are parallel. If we were to try and solve this system, we would find that any pair of $x$x- and $y$y-values that is a solution to one of the equations is not a solution to the other.
Worked example
Example 2
Consider the two equations $2x-6+y=0$2x−6+y=0 and $15-2y=4x$15−2y=4x. Is there a solution to this system of equations?
Think: If the two lines are distinct and have the same gradient then they are parallel and the system does not have a solution. If we rearrange both equations into gradient-intercept form, it will be easier to see their gradients.
Do:
Rearranging the equation $2x-6+y=0$2x−6+y=0, we get:
| $2x-6+y$2x−6+y | $=$= | $0$0 |
| $-6+y$−6+y | $=$= | $-2x$−2x |
| $y$y | $=$= | $-2x+6$−2x+6 |
Rearranging the equation $15-2y=4x$15−2y=4x, we get:
| $15-2y$15−2y | $=$= | $4x$4x |
| $-2y$−2y | $=$= | $4x-15$4x−15 |
| $y$y | $=$= | $-2x+\frac{15}{2}$−2x+152 |
So we can see that these two equations have the same gradient of $-2$−2 (and have different $y$y-intercepts). Therefore they are parallel, and the system has no solutions.
Solving algebraically
Sometimes it is not so easy to sketch a graph from an equation. Other times, the graphs may be easy enough to sketch but it can be tricky to read the exact point of intersection off of the graph (especially if it doesn't line up at nice integer coordinates). In these cases, we want to know a different method of solution that doesn't depend on the graphs.
There are a few ways to solve simultaneous equations algebraically. The two main methods are the method of substitution, and the method of elimination. We will take a look at both of these methods.
Solving by substitution
One efficient way to solve simultaneous equations is the method of substitution. This method works by solving for one variable first by 'substituting' one equation into the other, then using the value for that variable to find the other.
Exploration
Consider the pair of simultaneous equations $3y+4=x$3y+4=x and $2+x-y=0$2+x−y=0. Let's start by labelling these equations to make them easier to refer to:
| $3y+4=x$3y+4=x | equation (1) | |
| $2+x-y=0$2+x−y=0 | equation (2) |
We now want to solve this pair of equations algebraically. Notice that in equation (1) we have an expression for $x$x in terms of $y$y. Any $x$x-value that is a solution to this system of equations must solve both equations, by definition. So we can use the fact that $x$x is equal to $3y+4$3y+4 from the first equation, and use this to rewrite the value of $x$x in the second equation - that is, we are going to substitute equation (1) into equation (2):
| $2+x-y$2+x−y | $=$= | $0$0 |
|
| $2+\left(3y+4\right)-y$2+(3y+4)−y | $=$= | $0$0 |
substituting (1) into (2) |
Now that we have replaced $x$x with an expression in terms of $y$y, the resulting equation has only the variable $y$y and constants - so we can solve this equation using our usual methods:
| $2+\left(3y+4\right)-y$2+(3y+4)−y | $=$= | $0$0 |
| $2y+6$2y+6 | $=$= | $0$0 |
| $2y$2y | $=$= | $-6$−6 |
| $y$y | $=$= | $-3$−3 |
Now that we have a value of $y$y, we can substitute it back into either of the original equations to find the corresponding value of $x$x. Let's use equation (1):
| $x$x | $=$= | $3y+4$3y+4 |
reverse of equation (1) |
| $=$= | $3\times\left(-3\right)+4$3×(−3)+4 |
substituting $y=-3$y=−3 |
|
| $=$= | $-9+4$−9+4 |
|
|
| $=$= | $-5$−5 |
|
- Does this solution also work for equation (2)?
Let's check by substituting in the values:
| $2+x-y$2+x−y | $=$= | $2+\left(-5\right)-\left(-3\right)$2+(−5)−(−3) |
substituting |
| $=$= | $2-5+3$2−5+3 |
|
|
| $=$= | $0$0 |
|
Therefore the pair $x=-5$x=−5, $y=-3$y=−3 is the solution to this system of equations.
- Is it possible that there are other solutions that we didn't find?
Since the two equations that we started with were distinct linear equations, their graphs are straight lines and so they can only have one possible point of intersection. If we had more complicated equations that would result in more solutions, the algebra would also show this - so we can be confident that $\left(-5,-3\right)$(−5,−3) is the only solution to this system of equations.
- If we had solved the equations in a different way, would we get a different solution?
In this case it was convenient to substitute $x$x in terms of $y$y because of the form of equation (1). We didn't have to do this, however - we could have (for instance) rearranged equation (2) to make $y$y the subject, and then substitute it into equation (1) to get an equation involving only $x$x. Let's take a quick look at this alternate method:
| $2+x-y=0$2+x−y=0 | equation (2) | |
| $2+x=y$2+x=y | equation (3) |
By rearranging we find that $y=2+x$y=2+x, and we labelled this new equation as equation (3). We can now substitute into equation (1) and solve:
| $3y+4$3y+4 | $=$= | $x$x |
|
| $3\left(2+x\right)+4$3(2+x)+4 | $=$= | $x$x |
substituting (3) into (2) |
| $6+3x+4$6+3x+4 | $=$= | $x$x | |
| $3x+10$3x+10 | $=$= | $x$x | |
| $2x+10$2x+10 | $=$= | $0$0 | |
| $2x$2x | $=$= | $-10$−10 | |
| $x$x | $=$= | $-5$−5 |
That took a bit more effort, but we have gotten to the same value of $x$x. Substituting this back into any of our original equations will result in $y=-3$y=−3 again. So it doesn't matter which equation we substitute into the other, we will end up with the same solution!
- Label the equations by writing "equation (1)" and "equation (2)" next to them. Whenever you create new equations out of one or both of them, you can label them "equation (3)" and so on. In this way, when substituting one equation into another you can write it in shorthand as "substitute (3) into (1)". This helps to keep your ideas in order and clearly demonstrate your solution.
- When dealing with equations with big numbers, see if any simplification can be done before trying to solve them simultaneously. For example, suppose we have $24x+36y=60$24x+36y=60 as equation (2). We could divide through by $12$12 to obtain $2x+3y=5$2x+3y=5 (which we might label as equation (3)), and these numbers are much more manageable for substitution and manipulation.
- Using brackets when performing substitutions can help to avoid making errors.
Remember that a solution to a system of equations involves finding a set of values for all of the variables in the equations. Make sure that you fully answer each question before moving on to the next.
Concurrent lines
We know that the solution of a system of two simultaneous equations is represented graphically as the two lines' point of intersection, i.e. where they cross each other. Simultaneous equations can also tell us whether a set of three or more lines are concurrent. When a set of lines are concurrent it means that they all cross over at the same point.
An example of three concurrent lines
Having the same intersection point between them means that if you take any two of these three lines, then their intersection point will be same as that of any other two lines.
Practice questions
Question 3
We want to solve the following system of equations using the substitution method.
| Equation 1 | $y=-2x-1$y=−2x−1 |
| Equation 2 | $x+2y=13$x+2y=13 |
First solve for $x$x.
Hence, solve for $y$y.
Question 4
We want to solve the following system of equations using the substitution method.
| Equation 1 | $-7p+2q=-\frac{13}{10}$−7p+2q=−1310 |
| Equation 2 | $-21p+10q=-\frac{9}{10}$−21p+10q=−910 |
First solve for $q$q.
Now solve for $p$p.
Solving by elimination
Another way to solve simultaneous equations is by using the method of elimination. This method works by adding or subtracting equations from one another in order to eliminate one variable, leaving the other variable to solve on its own. This method is most useful when there are only two linear equations in the system.
As a quick example, consider the two equations
| $2x+5y=13$2x+5y=13 | equation (1) | |
| $2x-5y=3$2x−5y=3 | equation (2) |
Notice that the coefficient of $x$x in both equations is the same. So we could solve this system by subtracting one equation from the other (that is, subtract the left hand sides to form the new left hand side, and subtract the right hand sides to form the new right hand side) to eliminate the variable $x$x. Doing so, we get:
| $2x+5y-\left(2x-5y\right)$2x+5y−(2x−5y) | $=$= | $13-3$13−3 |
(1)$-$−(2) |
| $2x+5y-2x+5y$2x+5y−2x+5y | $=$= | $10$10 |
|
| $10y$10y | $=$= | $10$10 |
|
| $y$y | $=$= | $1$1 |
|
At this point, we could substitute back into either equation to find the corresponding value of $x$x.
Alternatively, notice that the coefficients of $y$y are also the same, but with opposite signs. So we could have instead solved this system by adding the equations together, to eliminate the variable $y$y. Doing so, we get:
| $2x+5y+\left(2x-5y\right)$2x+5y+(2x−5y) | $=$= | $13+3$13+3 |
(1)$+$+(2) |
| $4x$4x | $=$= | $16$16 |
|
| $x$x | $=$= | $4$4 |
|
We can see that we should use addition when the coefficients of the variable we want to eliminate are equal and have opposite signs, and subtraction when they're equal and have the same sign.
Simultaneous equations don't always come with nicely paired coefficients like this, however. For example, consider the equations
| $x+3y=5$x+3y=5 | equation (1) | |
| $2x+2y=1$2x+2y=1 | equation (2) |
How could we use the method of elimination to solve this system?
Using multiplication and division
When we we are dealing with equations that don't already have the same coefficients for one or both of the variables, we can first multiply or divide each equation by a constant until one of the variables shares a coefficient across both equations.
Using the above example, notice that the coefficient of $x$x in equation (2) is double the coefficient of $x$x in equation (1). So let's multiply equation (1) by $2$2:
| $x+3y=5$x+3y=5 | equation (1) | |
| $2x+6y=10$2x+6y=10 | equation (3) |
Here we have labelled the new equation as equation (3). Comparing equations (2) and (3), we can now solve them with the method of elimination, by subtracting equation (2) from equation (3):
| $2x+6y-\left(2x+2y\right)$2x+6y−(2x+2y) | $=$= | $10-1$10−1 |
(3)$-$−(2) |
| $2x+6y-2x-2y$2x+6y−2x−2y | $=$= | $9$9 |
|
| $4y$4y | $=$= | $9$9 |
|
| $y$y | $=$= | $\frac{9}{4}$94 |
|
We can now substitute this value back into any of our original equations to find the corresponding value of $x$x. Let's use equation (1):
| $x+3y$x+3y | $=$= | $5$5 |
equation (1) |
| $x+3\times\left(\frac{9}{4}\right)$x+3×(94) | $=$= | $5$5 |
substituting $y=\frac{9}{4}$y=94 |
| $x+\frac{27}{4}$x+274 | $=$= | $5$5 |
|
| $x$x | $=$= | $5-\frac{27}{4}$5−274 |
|
| $x$x | $=$= | $-\frac{7}{4}$−74 |
|
So we find that the solution to this system of equations is the pair $x=-\frac{7}{4}$x=−74, $y=\frac{9}{4}$y=94.
- As with the method of substitution, labelling the equations as "equation (1)", "equation (2)", and so on will help you to keep track of your work and clearly demonstrate your solution method.
- When adding or subtracting two equations, we can write it in shorthand using the labels, as "(3)$+$+(1)" for instance.
- Again, when dealing with equations with large numbers, see if any simplification can be done before beginning to solve them.
-
Check your answers by substituting the solutions back into the original two equations.
Practice questions
Question 5
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.
Question 6
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 |
First solve for $y$y
Now solve for $x$x