An equation is a mathematical statement of equality. It uses an equals sign to show that the expression on the left-hand side of the statement is equal to the expression on the right-hand side. The equals sign means that the equation is balanced, like a set of scales.
The diagram above represents the equation $3x=x+4$3x=x+4. Only one value of $x$x will make this equation true and keep the scales balanced.
If we were to subtract an $x$x from only the right-hand side of the equation, the scales would become unbalanced. To keep the scales balanced we would need to subtract an $x$x from the left-hand side as well.
Whenever we add, subtract, multiply, divide, or apply any other operation to one side of an equation, we must do exactly the same to the other side.
To solve an equation, our aim is to find the possible value(s) of the variable (most commonly $x$x). To do so, we rearrange the equation to get the variable by itself on one side of the equation, equal to its value on the other side. This is called isolating the variable. In the example above, the solution to the equation $3x=x+4$3x=x+4 would be written as $x=2$x=2.
For any equation, we can check the solution by substituting it back into the original equation. If the solution is correct, the left-hand side (LHS) of the equation will be equal to the right-hand side (RHS).
With equations that are solved using more than one step, we need to think carefully about the order in which we perform the operations to both sides.
Worked examples
example 1
Solve $-5\left(y-1\right)=25$−5(y−1)=25
Think: In this equation, the brackets tell us that the quantity $y-1$y−1 has been multiplied by $-5$−5, so our first step will be to divide both sides by $-5$−5.
Do:
| $-5\left(y-1\right)$−5(y−1) | $=$= | $25$25 |
|
| $\frac{-5\left(y-1\right)}{-5}$−5(y−1)−5 | $=$= | $\frac{25}{-5}$25−5 |
Divide both sides by $-5$−5 |
| $y-1$y−1 | $=$= | $-5$−5 |
|
| $y-1+1$y−1+1 | $=$= | $-5+1$−5+1 |
Add $1$1 to both sides |
| $y$y | $=$= | $-4$−4 |
|
Reflect: An alternative approach to solving this question would be to expand the brackets first, taking care to multiply the negative numbers correctly.
example 2
Solve $5x+9=x+21$5x+9=x+21
Think: In this equation there are variables on both sides. Our first step is to use an inverse operation to remove the variable from one side of the equation. We do this by subtracting $x$x from both sides. We can then subtract $9$9 from both sides to get the variables on one side and the numbers on the other side.
Do:
| $5x+9$5x+9 | $=$= | $x+21$x+21 |
|
| $5x+9-x$5x+9−x | $=$= | $x+21-x$x+21−x |
Subtract $x$x from both sides |
| $4x+9$4x+9 | $=$= | $21$21 |
|
| $4x+9-9$4x+9−9 | $=$= | $21-9$21−9 |
Subtract $9$9 from both sides |
| $4x$4x | $=$= | $12$12 |
|
| $\frac{4x}{4}$4x4 | $=$= | $\frac{12}{4}$124 |
Divide both sides by $4$4 |
| $x$x | $=$= | $3$3 |
Simplify |
Practice questions
Question 1
Find the solution for the following equation: $\frac{x+9}{7}=4$x+97=4
Question 2
Solve the following equation: $3\left(x+6\right)+3\left(x+24\right)=12$3(x+6)+3(x+24)=12
Question 3
Solve the following equation: $\frac{8x-2}{3}=\frac{6x-3}{4}$8x−23=6x−34
Question 4
Solve the following equation: $x+\frac{5x-1}{4}=1$x+5x−14=1
Solving word problems
Now that we know how to solve equations, the next step is to create our own equations to solve a particular situation or problem.
Many problems that we encounter will come from a description of the situation, without any equations present. Our first task will be to translate the description into an equation that models the problem. It reduces the complexity of the written problem and allows us to find a solution using our equation-solving techniques.
Noticing keywords
Certain keywords in a problem description indicate the mathematical operations or concepts we might need to use in our equations. Here is a list of the more common ones:
There are many different ways the four basic operations of addition, subtraction, multiplication and division are described:
| Addition ($+$+) | add, plus, sum, total, increase, more, combined |
| Subtraction ($-$−) | subtract, minus, difference, take-away, decrease, less, deduct |
| Multiplication ($\times$×) | multiply, times, product, multiple, double, triple, twice, of, by, groups of |
| Division ($\div$÷) | divide, quotient, share, goes into, how many times, per, split, halve |
The following terms require a bit more explanation:
- The word product indicates multiplication. For example, the product of $3$3 and $x$x means $3\times x$3×x and is written as $3x$3x.
- A quotient is the name we use for the fraction we get from the division of two numbers. For example, the quotient of $x$x and $5$5 is written as $\frac{x}{5}$x5. The quotient of $5$5 and $x$x is $\frac{5}{x}$5x. Notice that the order is important.
- Consecutive numbers are numbers that follow each other in order. For example $5,6,7,8$5,6,7,8 are consecutive numbers, and $45,47,49$45,47,49 are consecutive odd numbers, but $1,3,16,25$1,3,16,25 are not consecutive.
There may also be specific words related to mathematics that indicate which concept to use. For example:
- The word average indicates that we add a series of values together, then divide by the number of values.
- For a rectangle the word perimeter indicates that we add side lengths together, whereas area means we multiply the length by the width.
In many cases, the variable used to represent the unknown quantity will not be specified. In this case we choose a letter of the alphabet and state which variable it represents - this is called defining the variable. Note that it doesn't matter which letter we use, but it is important that we define what the letter represents.
Worked examples
Example 3
Translate the following statement into an equation, then solve the equation to find the number.
Twenty-one less than the product of a number and nine, is double the number.
Think: Let $x$x be the unknown number and highlight the keywords:
Twenty-one less than the product of a number and nine, is double the number.
Do:
| $9x-21$9x−21 | $=$= | $2x$2x |
Translate the problem description into an equation |
| $9x-21-2x$9x−21−2x | $=$= | $2x-2x$2x−2x |
Now solve by subtracting $2x$2x from both sides |
| $7x-21$7x−21 | $=$= | $0$0 |
|
| $7x-21+21$7x−21+21 | $=$= | $0+21$0+21 |
Add $21$21 to both sides |
| $7x$7x | $=$= | $21$21 |
|
| $\frac{7x}{7}$7x7 | $=$= | $\frac{21}{7}$217 |
Divide both sides by 7 |
| $x$x | $=$= | $3$3 |
Reflect: We can check that $x=3$x=3 is correct by using $3$3 as the number in the original problem description. The product of $3$3 and $9$9 is $27$27. Subtracting $21$21 gives us $6$6, which is double the value of $3$3.
example 4
Tim is building a rectangular yard for his chickens. The yard is to have a perimeter of $72$72 metres, and its length is to be $6$6 metres more than double its width.
Let $w$w be the width of the yard.
(a) Write an expression for the length of the yard.
Solution:
The length of the yard is $6$6 metres more than double the width. Therefore the length can be written as $2w+6$2w+6.
(b) Write an equation in terms of $w$w and solve it to find the width of the yard.
Solution:
The yard is a rectangle with a perimeter equal to $72$72 metres. The perimeter is equal to two times the length added to two times the width.
| Perimeter | $=$= | $2\times\text{length }+2\times\text{width }$2×length +2×width |
Write formula for perimeter |
| $72$72 | $=$= | $2\times\left(2w+6\right)+2\times w$2×(2w+6)+2×w |
Substitute known values and expressions |
| $72$72 | $=$= | $2\left(2w+6\right)+2w$2(2w+6)+2w |
|
| $72$72 | $=$= | $4w+12+2w$4w+12+2w |
Expand brackets |
| $72$72 | $=$= | $6w+12$6w+12 |
Simplify |
Now solving for the unknown:
| $72-12$72−12 | $=$= | $6w+12-12$6w+12−12 |
Subtract $12$12 from both sides |
| $60$60 | $=$= | $6w$6w |
|
| $\frac{60}{6}$606 | $=$= | $\frac{6w}{6}$6w6 |
Divide both sides by $6$6 |
| $10$10 | $=$= | $w$w |
Simplify |
| $\therefore w$∴w | $=$= | $10$10 |
|
So the width of the yard is $10$10 metres.
(c) Hence, state the length of the yard.
Think: Here we substitute our value for $w$w into our expression for the length of the yard, $2w+6$2w+6.
Do:
| length | $=$= | $2w+6$2w+6 |
| $=$= | $2\times10+6$2×10+6 | |
| $=$= | $26$26 |
So the length of the yard is $26$26 metres.
Reflect: As a final check, we can work out the perimeter of the yard using our values for length and width:
Perimeter$=$=$2\times26+2+10$2×26+2+10$=$=$72$72 m
Practice questions
Question 5
The sum of three consecutive even numbers is $36$36.
Let $x$x be the smallest of the numbers. Form an equation and solve it for $x$x.
Hence the three numbers, in ascending order, are:
$\editable{}$ $\editable{}$ $\editable{}$
Question 6
Kate and Isabelle do some fundraising for their sporting team. Together they raised $\$600$$600.
If Isabelle raised $\$p$$p, and Kate raised $\$272$$272 more than Isabelle:
Solve for the value of $p$p.
Write each line of working as an equation.
Now calculate how much Kate raised.
Question 7
A commercial airplane has a total mass at take off of $51000$51000 kg. The luggage and fuel are $\frac{1}{3}$13 the mass of the unloaded plane, and the crew and passengers are $\frac{1}{4}$14 the mass of the fuel and luggage. Solve for $p$p, the mass of the unloaded plane.