Simultaneous equations are incredibly useful for real life applications. They are used for problems that have at least two unknown quantities and at least two pieces of information involving both of these quantities.
- Read the question entirely from start to finish
- Assign variables to each unknown quantity (for example, let $x$x be the number of $\dots$…)
- Use the information given in the question to develop simultaneous linear equations
- Use either substitution, elimination or graphical methods to find solution to simultaneous equations
- Answer the problem given in the question
Worked example
Example 1
Rachel is buying drinks for her friends, and some of them want coffee while others want tea. In the cafe she went to a cup of coffee costs $\$4$$4 and a cup of tea costs $\$3$$3 If she bought $7$7 drinks and spent a total of $\$26$$26, how many teas and coffees did she get of each?
Think: What are the two amounts we need to find and the two bits of information relating the two amounts?
Do:
We need to find the number of teas bought and the number of coffees bought. Let's let these be $x$x and $y$y.
The first bit of information tells us that the total number of drinks is $7$7, therefore:
$x+y=7$x+y=7 (1)
The next bit of information tell us that the total charge was $\$26$$26, where coffees cost $\$4$$4 and teas $\$3$$3. Therefore:
$3x+4y=26$3x+4y=26 (2)
So now that we have our two equations, substitution looks better if we rearrange (1):
$y=7-x$y=7−x (3)
Substituting this equation into the $y$y of (2):
| (3) → (2): | ||
| $3x+4y$3x+4y | $=$= | $26$26 |
| $3x+4\left(7-x\right)$3x+4(7−x) | $=$= | $26$26 |
| $3x+28-4x$3x+28−4x | $=$= | $26$26 |
| $-x+28$−x+28 | $=$= | $26$26 |
| $-x$−x | $=$= | $-2$−2 |
| $x$x | $=$= | $2$2 |
Now substituting in this $x$x-value back into (3):
| $x$x → (3): | ||
| $y$y | $=$= | $7-x$7−x |
| $=$= | $7-2$7−2 | |
| $=$= | $5$5 |
Therefore Rachel bought $2$2 cups of tea and $5$5 cups of coffee.
Practice questions
Question 1
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)