When solving problems that involve missing or unknown values, we will often replace that value with a variable in order to make an equation.
For example: Jill's age is equal to $5$5 more than twice the age of Jack. If Jill's age is $23$23, how old is Jack?
If we replace Jill's age with $23$23 and Jack's age with the variable $x$x, we can solve this question using the equation:
$23=5+2x$23=5+2x
To find Jack's age, we can solve this equation to find the value of $x$x.
Reverse operations
To solve equations using algebra, the most important rule to remember is that if we apply operations to one side of the equation, we must also apply it to the other.
When applying operations to equations, we always apply the same step to both sides of the equation. This way, both sides of the equation will be equal once we solve the equation.
Making sure to follow this rule, we can isolate the variable in an equation by applying operations to both sides of the equation which reverse the operations applied to the variable.
To do this, we need to know which operations are reversed by which.
| Operation | Reverse operation | Example |
|---|---|---|
|
Addition Subtraction |
Subtraction Addition |
$x+4-4=x$x+4−4=x |
|
Multiplication Division |
Division Multiplication |
$y\times4\div4=y$y×4÷4=y |
Worked Examples
Example 1
Solve the equation: $x+\frac{2}{5}=\frac{17}{20}$x+25=1720
Think: To isolate $x$x, we want to reverse the addition of $\frac{2}{5}$25.
Do: Since subtraction reverses addition, we can isolate $x$x by subtracting $\frac{2}{5}$25 from both sides of the equation. This gives us:
$x=\frac{17}{20}-\frac{2}{5}$x=1720−25
Noting that $\frac{2}{5}$25 is equivalent to $\frac{8}{20}$820, we can perform the subtraction to find that:
$x=\frac{9}{20}$x=920
Example 2
Solve the equation: $\frac{x}{3.5}=4$x3.5=4
Think: To isolate $x$x, we want to reverse the division by $3.5$3.5.
Do: Since multiplication reverses division, we can isolate $x$x by multiplying both sides of the equation by $3.5$3.5. This gives us:
$x=4\times3.5$x=4×3.5
Performing the multiplication gives us:
$x=14$x=14
Reflect: In both cases, the original operation applied to the variable was cancelled out by the reverse operation which we applied to both sides of the equation, leaving just $x$x on the left hand side.
The order of reverse operations
If we return to the case of finding Jack's age, we can see that there are two operations that have been applied to the variable in the equation:
$23=5+2x$23=5+2x
As we can see, to get $23$23 we multiplied $x$x by $2$2 and then added $5$5. In order to isolate $x$x we need to reverse these operations in the correct order. Which operation should we try to reverse first?
If we try to reverse the multiplication first, we would divide both sides by $2$2 to get:
$\frac{23}{2}=\frac{5+2x}{2}$232=5+2x2
This only makes the variable less isolated and is not what we want.
When reversing operations, we apply them in the reverse of the usual order of operations.
When applying reverse operations to isolate a variable, we apply them according to the order of reverse operations:
- Addition and subtraction
- Multiplication and division
- Expressions inside brackets
Knowing this, we can now solve the equation to find Jack's age.
| $23$23 | $=$= | $5+2x$5+2x |
|
| $18$18 | $=$= | $2x$2x |
Reverse the addition |
| $9$9 | $=$= | $x$x |
Reverse the multiplication |
As such, we find that Jack's age is $9$9.
Practice questions
Question 1
Consider the equation $2\left(p+6\right)=-8$2(p+6)=−8.
Which pair of operations will make $p$p the subject of the equation?
Step $1$1 Step $2$2 $2\left(p+6\right)$2(p+6) 
$p+6$p+6 $p$p Divide by $2$2, then subtract $6$6
ADivide by $6$6, then subtract $2$2
BMultiply by $2$2, then add $6$6
CDivide by $2$2, then add $6$6
DApply these operations to the right-hand side of the equation as well.
Divide by $2$2 Subtract $6$6 $-8$−8 $\editable{}$ $\editable{}$ Using your answer from part (b), what value of $p$p will make the equation $2\left(p+6\right)=-8$2(p+6)=−8 true?
Question 2
Solve the equation $\frac{x+15}{2}=5$x+152=5.
Moving the variable first
With the skills we now have, we can reverse as many operations as we need to isolate our variable. But what if we want to solve equations like $4-x=17$4−x=17 or $\frac{5}{x}=20$5x=20?
In both of these cases we can see that the variable is actually part of the operation being applied, so we can't isolate it so easily.
However, if we reverse the operation containing the variable, we can move it so that the equation can be solved using the skills we just learned.
Worked Example
Solve the equation: $4-x=17$4−x=17
Think: We want to first reverse the subtraction of $x$x first and then solve the equation as per usual.
Do: We can reverse the subtraction by adding $x$x to both sides of the equation. This gives us:
$4=17+x$4=17+x
Since the order of numbers in addition doesn't matter, we can use the equivalent equation:
$4=x+17$4=x+17
Which we can solve by reversing the addition to get:
$-13=x$−13=x
Reflect: Another way to solve this equation would be to reverse the addition of $4$4, then reverse the negative sign on the variable by multiplying both sides of the equation by $-1$−1. Doing this would follow the steps:
| $4-x$4−x | $=$= | $17$17 |
|
| $-x$−x | $=$= | $13$13 |
Reverse the addition |
| $x$x | $=$= | $-13$−13 |
Reverse the negative |
We can solve equations where the variable is the denominator using a similar method. To solve these equations we can multiply both sides of the equation by the variable and then perform the usual reverse operations to isolate it.
Practice question
Question 3
Solve the equation $\frac{-44}{p}=11$−44p=11.