The magic of simultaneous equations comes to life when we see how useful it is in real life applications. Simultaneous equations are often used when we have at least two unknown quantities and at least two pieces of information involving these quantities. Our first step is to define variables to represent these quantities and then translate the pieces of information into equations. Lastly, by using either the substitution method, elimination method or a graphical method, we solve the equations simultaneously.
Worked example
Example 1
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
The sum of two numbers is $56$56 and their difference is $30$30.
Set up two equations by letting $x$x and $y$y be the two numbers.
Use $x$x as the larger of the 2 numbers.
Sum equation: $\editable{}$ Difference equation: $\editable{}$ First solve for $x$x.
Equation 1 $x+y=56$x+y=56 Equation 2 $x-y=30$x−y=30 Now solve for $y$y.
Question 2
The length of a rectangle measures $12$12 units more than the width, and the perimeter of the rectangle is $56$56 units.
Let $y$y be the width and $x$x be the length of the rectangle.
Use the fact that the length of the rectangle is $12$12 units more than the width to set up an equation relating $x$x and $y$y (we'll call this equation (1)).
Use the fact that the perimeter of the rectangle is equal to $56$56 to set up another equation relating $x$x and $y$y (we'll call this equation (2)).
First solve for $y$y to find the width.
$x=y+12$x=y+12 equation (1) $x+y=28$x+y=28 equation (2) Now solve for $x$x to find the length.
$x=y+12$x=y+12 equation (1) $x+y=28$x+y=28 equation (2)