So far we've only been solving equations with one variable. We know that rearranging $x+2=5$x+2=5 will give us a unique answer of $3$3 for $x$x - but what happens when we have more than one variable in our equation?
Exploration
Consider the equation $x+y=6$x+y=6. This could have the solution $x=2,y=4$x=2,y=4, or it could have the solution of $x=40,y=-34$x=40,y=−34. In fact, there are infinitely many solutions to this equation! Just choose any value for $x$x, and then use the equation to find the corresponding value of $y$y required to make the equation hold true.
If we have two equations with the same two variables in them (often $x$x and $y$y), we call the set of equations system of equations. They are also commonly referred to as simultaneous equations.
We might be interested in finding a common pair of $x$x- and $y$y-values that satisfy both equations simultaneously. If we can find a pair of values for $x$x and $y$y that successfully do this, then we will have found a unique solution to our system.
The graphical method
On a coordinate plane, the solution is represented by the point of intersection of the two equations' graphs (where the two graphs cross over). So the $x$x- and $y$y-values of our solution will take the form of the coordinates of the intersection point $\left(x,y\right)$(x,y).
If the straight lines representing the two equations are not parallel, then there will be exactly one point of intersection between them (as pictured above).
If the lines are parallel and distinct, then there will be no points of intersection between them. That is, there are no corresponding $x$x- and $y$y-values that satisfy both equations simultaneously.
The final case to consider is when two different equations have the same graphical representation. For example, if the graphs of $x+y=5$x+y=5 and $2x+2y=10$2x+2y=10 were placed on the same set of axes, we would end up with two lines lying perfectly on top of one another. Every point on that line will satisfy both equations and is a point of intersection, meaning there are infinitely many solutions to this type of system of equations.
Practice question
Question 1
The substitution method
One way to solve simultaneous equations is called the substitution method. The simultaneous equations we are going to work with usually involve $2$2 variables and $2$2 equations, and this method works by solving for one variable first through 'substituting' one equation into the other.
Consider the equations $3y+4=x$3y+4=x and $y=2+x$y=2+x, for example. We can combine the two equations by substituting $y=2+x$y=2+x into $3y+4=x$3y+4=x. This gives us:
| $3\left(2+x\right)+4$3(2+x)+4 | $=$= | $x$x |
| $6+3x+4$6+3x+4 | $=$= | $x$x |
| $3x-x$3x−x | $=$= | $-4-6$−4−6 |
| $2x$2x | $=$= | $-10$−10 |
| $x$x | $=$= | $-5$−5 |
Now that we know the value of $x$x, we can solve for $y$y by using the equation we already have that has $y$y as the subject: $y=2+x$y=2+x. So we have $y=2+\left(-5\right)$y=2+(−5) which is equal to $-3$−3. So our answer is $\left(-5,-3\right)$(−5,−3).
- Name the equations by writing (1) and (2) next to them, and whenever we create new equations out of one or both of them, we can name them (3), (4), etc. When adding two equations we can write it in shorthand as (3) + (1), and similarly for all the other operations. This helps to keep our ideas in order and not get confused.
- When dealing with equations with big numbers, see if they can be simplified before beginning to solve them. For example, $2x-4=6y$2x−4=6y can be simplified to $x-2=3y$x−2=3y without changing the values of $x$x and $y$y.
Remember to solve for the values of both variables! Check that you have both parts of the solution at the end of every simultaneous equation problem, unless you are specifically solving for just one of the variables.
Practice questions
Question 2
We want to solve the following system of equations using the substitution method.
| Equation 1 | $y=3x-18$y=3x−18 |
| Equation 2 | $x+y=-2$x+y=−2 |
First solve for $x$x.
Now solve for $y$y.
Question 3
We want to solve the following system of equations using the substitution method.
| Equation 1 | $y=5x+34$y=5x+34 |
| Equation 2 | $y=3x+18$y=3x+18 |
First solve for $x$x.
Hence, solve for $y$y.
The elimination method
There's another equally useful method to solve simultaneous equations - 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.
We can eliminate variables from a system of simultaneous equations by adding or subtracting:
For example, we sould add $2x-y=1$2x−y=1 and $5x+y=2$5x+y=2 to get $7x=3$7x=3, which doesn't have any $y$y terms.
As another example, we could subtract $3y-x=2$3y−x=2 from $3y-2x=0$3y−2x=0 to get $-x=-2$−x=−2, which also doesn't have any $y$y terms.
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 and 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, consider $x+3y=5$x+3y=5 (equation (1)) and $2y+2x=1$2y+2x=1 (equation (2)). Suppose we want to eliminate the $x$x terms first. Notice that the coefficient of $x$x is $1$1 in equation (1) and $2$2 in equation (2). The idea then is to multiply equation (1) by $2$2 so that the coefficient of $x$x is $2$2 in both equations. Then we can apply the same logic as before and add or subtract the equations. Let's see this in action:
Multiplying equation (1) by $2$2:
| $2\left(x+3y\right)$2(x+3y) | $=$= | $2\times5$2×5 |
Multiply both sides of the equation by $2$2 |
| $2x+6y$2x+6y | $=$= | $10$10 |
Simplify and we will call this new equation, equation (3) |
Now let's find the difference between equation (3) 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. Let's use 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 | $=$= | $5-\frac{27}{4}$5−274 |
| $x$x | $=$= | $\frac{-7}{4}$−74 |
Practice questions
Question 4
Use the elimination method by adding both equations to solve for $x$x first and then $y$y .
| Equation 1 | $2x+5y=44$2x+5y=44 |
| Equation 2 | $6x-5y=-28$6x−5y=−28 |
Solve for $x$x.
Now find the value of $y$y
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
Solving simultaneous equations using a CAS calculator
A CAS calculator can be utilised in two different ways to solve a linear system of equations. We can use a graphical approach, where we graph both lines on an $xy$xy-plane and use the CAS calculator to find the point of intersection. Or we can use the solve function of the CAS calculator to find the point of intersection immediately with no graphing involved.
Select your brand of calculator below to view instructions for these two approaches.
Casio Classpad
How to use the CASIO Classpad to complete the following tasks regarding simultaneous equations
Consider the following system of equations:
| Equation 1 | $y=x-3$y=x−3 |
| Equation 2 | $2x+5y=20$2x+5y=20 |
Solve the system of linear equations using the graphing functionality of your CAS calculator, leaving your answer as a pair of coordinates.
Solve the system of linear equations using the solving functionality of your CAS calculator, leaving your answer as a pair of coordinates.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding simultaneous equations
Consider the following system of equations:
| Equation 1 | $y=x-3$y=x−3 |
| Equation 2 | $2x+5y=20$2x+5y=20 |
Solve the system of linear equations using the graphing functionality of your CAS calculator, leaving your answer as a pair of coordinates.
Solve the system of linear equations using the solving functionality of your CAS calculator, leaving your answer as a pair of coordinates.
Practice questions
Question 7
We want to find any solutions to the following system of equations.
| Equation 1 | $-9x+8y=-5$−9x+8y=−5 |
| Equation 2 | $5x+7y=-9$5x+7y=−9 |
Graph the two linear functions using the graph mode of your CAS calculator.
How many solutions are there to the system of linear equation?
None
AInfinitely many
BOne
CWhat can you conclude from your answer in part (a)?
The equations have the same gradient and y-intercept.
AThe equations have the same gradient but different y-intercepts.
BThe equations have different gradients.
C
Question 8
Consider the following system of equations.
| Equation 1 | $y=4x+35$y=4x+35 |
| Equation 2 | $y=2x+21$y=2x+21 |
Solve the system of linear equations using the solving functionality of your CAS calculator, leaving your answer as a pair of coordinates.
$\left(\editable{},\editable{}\right)$(,)
Question 9
Consider the system of equations below.
| Equation 1 | $3x-1.6y=5.9$3x−1.6y=5.9 |
| Equation 2 | $0.45x+0.7y=-0.02$0.45x+0.7y=−0.02 |
Solve the system of linear equations using the solving functionality of your CAS calculator, leaving your answer as a pair of coordinates, correct to two decimal places.
$\left(\editable{},\editable{}\right)$(,)