We can see how useful it is to be able to solve simultaneous equations once we take a look at some real life applications. Whenever we have at least two unknown quantities, and the same number of pieces of information involving those quantities, we can represent the quantities and information using a system of equations. To do so, we first represent the quantities using variables, then use the information to construct a set of equations involving those variables (one equation for each piece of information). At this point, we can use our knowledge of simultaneous equations to solve for the possible values of those variables, then translate this information back into the context of the quantities we started with.
Worked examples
Example 1
Leah gets a quote from two photographers for an event. Adam charges $\$48$$48 for a booking fee plus $\$17$$17 per hour, while Brianna charges $\$28$$28 for a booking fee plus $\$21$$21 per hour.
Determine an equation for each photographer in the form $\text{cost of hiring}=\text{booking fee}+\text{hourly rate}\times n$cost of hiring=booking fee+hourly rate×n, where $n$n is the number of hours, and use this to create a table comparing the total cost of hiring each photographer for up to $6$6 hours.
Solution:
For Adam, we know that his booking fee is $\$48$$48 and his hourly rate is $\$17$$17. So the equation representing his charges is
$\text{cost of hiring Adam}=48+17n$cost of hiring Adam=48+17n
For Brianna, we know that her booking fee is $\$28$$28 and her hourly rate is $\$21$$21. So the equation representing her charges is
$\text{cost of hiring Brianna}=28+21n$cost of hiring Brianna=28+21n
We can now use these equations to create a table comparing their costs:
| Hours | Hiring cost for Adam ($\$$$) | Hiring cost for Brianna ($\$$$) |
|---|---|---|
| $1$1 | $65$65 | $49$49 |
| $2$2 | $82$82 | $70$70 |
| $3$3 | $99$99 | $91$91 |
| $4$4 | $116$116 | $112$112 |
| $5$5 | $133$133 | $133$133 |
| $6$6 | $150$150 | $154$154 |
Comparing the columns of the table, we can see that the two photographers both charge $\$133$$133 for $5$5 hours. For durations shorter than $5$5 hours Adam is more expensive, and for durations longer than $5$5 hours Brianna is more expensive.
Example 2
Rachel is at a cafe buying drinks for her friends. Some of them want coffee, while others want tea. At this particular cafe, a regular cup of coffee costs $\$4$$4 and a regular cup of black tea costs $\$3$$3. If she buys $7$7 drinks for a total cost of $\$26$$26, how many of each type of drink did she purchase?
Think:
We want to determine the number of cups of coffee and the number of cups of tea that Rachel bought. These are two unknowns, so we can choose variables to represent them. We then also know the total number of drinks that Rachel bought, and the total cost of the drinks - so we have two pieces of information, and can use this to construct two simultaneous equations.
Do:
Let $x$x be the number cups of tea, and let $y$y be the number of cups of coffee.
Using the first piece of information, we know that the total number of drinks purchased is $7$7 - that is, the sum of $x$x and $y$y is $7$7. So we have our first equation:
| $x+y=7$x+y=7 | equation (1) |
Using the second piece of information, we know that the total cost of the drinks is $\$26$$26. Now the price of a cup of tea is $\$3$$3, so the total cost of the cups of tea will be $3x$3x dollars. Similarly, the total cost of the cups of coffee will be $4y$4y. Putting this together, we have the second equation:
| $3x+4y=26$3x+4y=26 | equation (2) |
Now that we have our two equations, we can either use substitution or elimination to solve the system. Let's use the method of substitution - to do so, however, we will first rearrange equation (1) to make $y$y the subject:
| $x+y=7$x+y=7 | equation (1) | |
| $y=7-x$y=7−x | equation (3) |
We can now substitute this into equation (2):
| $3x+4y$3x+4y | $=$= | $26$26 |
|
| $3x+4\left(7-x\right)$3x+4(7−x) | $=$= | $26$26 |
substituting (3) into (2) |
| $3x+28-4x$3x+28−4x | $=$= | $26$26 |
|
| $28-x$28−x | $=$= | $26$26 |
|
| $-x$−x | $=$= | $-2$−2 |
|
| $x$x | $=$= | $2$2 |
|
So we have that Rachel purchased $2$2 cups of tea (remember, this is what we chose $x$x to represent). We can now use this to find $y$y (the number of cups of coffee) by substituting back into any of our original equations. In this case, it will be simplest to use equation (3):
| $y$y | $=$= | $7-x$7−x |
|
| $=$= | $7-2$7−2 |
substituting $x=2$x=2 |
|
| $=$= | $5$5 |
|
So we have found that Rachel bought $2$2 cups of tea and $5$5 cups of coffee.
Practice questions
question 1
When comparing some test results Christa noticed that the sum of her Geography test score and Science test score was $172$172, and that their difference was $18$18.
Given that her Geography score is $x$x and her Science score is $y$y and she scored higher for the Geography test:
Use the sum of the test scores to form an equation. We will refer to this as equation (1).
Use the difference of the test scores to form an equation. We will refer to this as equation (2).
Use these two equations to find her Geography score.
Now find her Science score.
question 2
The perimeter of the triangle is $65$65 cm, and the same values for $x$x and $y$y are used to construct the rectangle shown, where the length is $7$7 cm longer than the width. Find the values of $x$x and $y$y.
Complete an equation for the perimeter of the triangle. Call this equation $1$1.
Equation 1: $\editable{}x+\editable{}y+3=\editable{}$x+y+3=
Use the rectangle to find $x$x in terms of $y$y. Call this equation $2$2.
Using two equations relating $x$x and $y$y, solve for the value of $y$y.
Finally, use the above value for $y$y to find $x$x.
Break-even
It is important for businesses to make a profit! If a business is running at a loss, they won't be able to exist for very long. An important figure for businesses in this regard is the break-even point - the amount of money they need to take in to cover all their expenses. At this point there is no profit or loss, and their income is equal to their expenses.
Break-even analysis looks at the graphs of cost and revenue together to determine where they meet. The point of intersection is the break-even point where income equals expenses.
Here is an example of a break-even analysis for a single day of operation for the company "Lovely Lemonade".
The horizontal axis is the number of units (drinks, in this case) that they sell. The expenses line (red) starts off higher than the income line (green). So for a low number of units sold, Lovely Lemonade will run at a loss. For example, if they only sell $1000$1000 units, the green line tells us they earn $\$200$$200, the red line tells us they spend $\$600$$600, and so overall they make $\$200-\$600=-\$400$$200−$600=−$400. The negative sign means that overall they lose $\$400$$400.
But eventually the income line becomes higher than the expenses line, so for a high number of units sold Lovely Lemonade will make a profit. For example, if they sell $5000$5000 units, the graph tells us that they will earn $\$1100$$1100 but only spend $\$1000$$1000. Overall they make $\$1100-\$1000=\$100$$1100−$1000=$100, and the positive sign means that they make a profit of $\$100$$100.
The important point is when the two lines meet - this is the break-even point. In this example we can see that selling $4000$4000 units means they don't make any profit, but they don't lose any money either - both their income and their expenses are $\$900$$900. Reaching this amount should be an important first goal for Lovely Lemonade every single day!
Practice question
Question 3
Consider the following phone plans:
GO SMALL plan: This plan has a $\$20$$20 monthly base charge and charges $90$90 cents per minute for all calls.
GO MEDIUM plan: This plan has a $\$26$$26 monthly base charge and then charges $70$70 cents per minute for all calls.
Complete the following table of values for various total monthly call times for the two plans:
Call time (in minutes) Total cost for GO SMALL plan Total cost for GO MEDIUM plan $20$20 $\editable{}$ $\editable{}$ $30$30 $\editable{}$ $\editable{}$ $40$40 $\editable{}$ $\editable{}$ $50$50 $\editable{}$ $\editable{}$ Sketch the graph of the two plans.
Loading Graph...Using the graphs, determine how many minutes of calls would need to be made so that the monthly bill costs the same on both plans.
Number of solutions
Depending on the particular system of equations, it is possible for there to be different numbers of solutions.
One solution
If two graphs intersect at one point, then there is a unique simultaneous solution to the pair of equations. For example, consider the following two linear equations:
| $2x+\frac{1}{2}y$2x+12y | $=$= | $14$14 |
| $y$y | $=$= | $-3x+7$−3x+7 |
The two graphs are shown below. Their intersection is at the point $\left(21,-56\right)$(21,−56).
This happens whenever the two lines have different gradients. Try and draw pairs of lines with different gradients. Can you find any pairs which don't intersect at exactly one point if you extend them from both ends?
Infinitely many solutions
It can happen that the graphs of two equations coincide. That is to say, every point that satisfies one equation also satisfies the other, and so there are infinitely many solutions. This is the case for the following pair of equations:
| $y$y | $=$= | $-3x+7$−3x+7 |
| $2y+6x-14$2y+6x−14 | $=$= | $0$0 |
Using this example, observe what happens if we rearrange the second equation to put it into gradient-intercept form as well:
| $2y+6x-14$2y+6x−14 | $=$= | $0$0 |
| $2y$2y | $=$= | $-6x+14$−6x+14 |
| $y$y | $=$= | $\frac{-6x+14}{2}$−6x+142 |
| $y$y | $=$= | $-3x+7$−3x+7 |
This is the same as the first equation! In fact, this is always the case - if two equations represent the same line and have the same set of solutions, then one can be rearranged into the same form as the other.
No solutions
It can also happen that a pair of lines do not intersect. We are familiar with the concept of parallel lines, and this is exactly the case here - two distinct lines which are parallel (and have the same gradient) will never intersect each other. Such equations are said to be inconsistent.
For example, the lines represented by the following two equations are parallel and do not intersect:
| $y$y | $=$= | $-3x+7$−3x+7 |
| $y$y | $=$= | $-3x+28$−3x+28 |
The graphs of these lines are shown below. We can observe that one is a vertical translation of the other:
Practice question
Question 4
Consider the following system of equations:
| $32x+12y$32x+12y | $=$= | $20$20 |
| $3y-5$3y−5 | $=$= | $-8x$−8x |
Which of the following best describes the graphs of the two lines?
The two lines are identical.
AThe two lines intersect at a single point.
BThe two lines are parallel.
CHow many solutions will this system of equations have?
No solutions.
AInfinitely many solutions.
BOne unique solution.
C
Question 5
Solve the following system of equations.
| $4x$4x | $-$− | $9y$9y | $=$= | $5$5 |
| $-12x$−12x | $+$+ | $27y$27y | $=$= | $5$5 |
Is it an inconsistent system or does it have infinitely many solutions?
It is an inconsistent system.
AIt has infinitely many solutions.
B
Solving using technology
Consider the following system of linear equations:
| $x+y$x+y | $=$= | $7$7 |
| $2x-3y$2x−3y | $=$= | $14$14 |
It is possible to quickly solve linear equations using technology, such as a CAS calculator. To do so, we can either use a graphical approach or the solve function on the CAS calculator.
Graphical approach
Go to the graphing section of your CAS calculator (the following screenshots are from TI-spire) and enter both linear equations. Note that the equation here is entered as a function of $x$x - that is, the equation is in gradient-intercept form $y=mx+c$y=mx+c.
If the equations are in standard form $ax+by=c$ax+by=c or otherwise, we can first rearrange them to make $y$y the subject so that they can be entered in as above. This isn't compulsory, however, as tour CAS calculator will have the ability to input your equation in standard form.
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Once both equations have been entered, you might need to adjust the visible domain range so that you can see both lines as well as their point of intersection (if it exists). This can be achieved by zooming in and out of the graph, or you can adjust the window settings manually.
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Once the point of intersection is visible, you can then analyse the graph to determine the solutions for $x$x and $y$y. To do this, open the Menu, select Analyse Graph and then select Intersection. Because it is possible for more complicated functions to have several points of intersection, select $x$x-values to the left and right of your point of intersection. The $x$x and $y$y values for the point of intersection should appear. It may be an exact solution or it could be rounded to a certain number of decimal places, depending on the settings of your calculator.
As we can see, the point of intersection is $\left(7,0\right)$(7,0). So the pair $x=7$x=7 and $y=0$y=0 simultaneously solve both equations in the above system.
Using the solve function
The solve function of CAS can be used to find the solution of a system of equations. This function can solve systems of equations that have many variables and equations, not just two equations of two variables.
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From the calculator section of your CAS calculator use the Menu, Algebra or Equations and you should find Solve System of Equations and/or Solve Linear System or Equations. Either one of these options will allow you to solve the system of equations.
Select how many variables the system has and how many equations are being solved. For the example above, there are only $2$2 variables and $2$2 equations.
Enter both of your equations into your CAS calculator, press ENTER.
So $x=7$x=7 and $y=0$y=0 simultaneously solve both equations in the above system of linear equations.
If using technology gives an answer that has other parameters such as $n$n or $t$t, this means that the system has an infinite amount of solutions rather than just one unique solution.
Systems of linear equations in three variables
We have seen so far that pairs of equations in two variables can have $0,1$0,1 or infinitely many solutions. The same can be said for systems of equations in three variables - though in order for there to be a unique solution, we will need to have (at least) three equations in those variables.
As the number of equations and variables increases above two, it becomes more difficult to visualise and solve by graphical methods. It is still possible to solve such systems algebraically, however, and by using the solve functionality of a CAS calculator this can be done quite quickly.
Below is an example of a system of three linear equations in three variables, which can be solved using technology.
Worked example
Example 3
I have a collection of $124$124 coins. Their denominations are $5$5c, $10$10c and $20$20c. The value of the $5$5c and $10$10c coins together is $\$5.75$$5.75 and the value of the $10$10c and $20$20c coins together is $\$6.90$$6.90. How many of each kind of coin do I have?
The solution involves numbers of coins. So, we choose variables to represent the numbers of the three types of coin. Let $x$x be the number of $5$5c coins, let $y$y be the number of $10$10c coins and let $z$z be the number of $20$20c coins.
From the given information we can write the following equations
$x+y+z=124$x+y+z=124
$0.05x+0.10y+0z=5.75$0.05x+0.10y+0z=5.75
$0x+0.10y+0.20z=6.9$0x+0.10y+0.20z=6.9
This is a system of three equations in the three unknown quantities. Each equation is represented by a plane and, in this case, the three planes intersect at a single point. You should confirm, using your device, that the intersection point has coordinates $(x,y,z)=(81,17,26)$(x,y,z)=(81,17,26).
It is possible to confirm by hand that these numbers satisfy all three of the equations.
Practice questions
Question 6
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 7
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.
$\left(\editable{},\editable{}\right)$(,)