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 transpose (or 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 also sometimes 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
Transpose the equation $-5\left(y-1\right)=25$−5(y−1)=25 to find the value of $y$y.
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 the equation $5x+9=x+21$5x+9=x+21 for $x$x.
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 value of $x$x if:
$x+4=9$x+4=9
Question 2
Solve the following equation: $8m+9=65$8m+9=65
Transposing equations of more than one variable
Quite often we will have an equation or formula that includes more than one variable. We may want to transpose such an equation to make a different variable the subject. As an example, consider the formula for the circumference $C$C of a circle with radius $r$r:
$C=2\pi r$C=2πr
In this formula, $C$C is currently the subject. By rearranging the formula, however, we can make $r$r the subject. This can be done with the same methods we used for solving equations. In this case, after dividing by $2\pi$2π the transposed formula becomes:
$r=\frac{C}{2\pi}$r=C2π
The subject of an equation or formula is the single variable by itself on one side of the equals sign, usually the left-hand side.
Worked examples
Example 3
Transpose the equation $y=mx+c$y=mx+c to make $c$c the subject.
Solution:
| $y$y | $=$= | $mx+c$mx+c |
|
| $y-mx$y−mx | $=$= | $mx+c-mx$mx+c−mx |
Subtracting $mx$mx from both sides |
| $y-mx$y−mx | $=$= | $c$c |
Simplifying |
| $c$c | $=$= | $y-mx$y−mx |
Rewriting with the subject on the LHS |
example 4
Transpose $ax=b-cx$ax=b−cx to make $x$x the subject.
Think: In this equation, the variable x appears on both sides. So we will start by rearranging so that all terms containing $x$x are on one side of the equation (we'll use the left-hand side), and all other terms are on the other side. We can then factorise $x$x out of the left-hand side, so that we only have one x in the equation, and divide by its coefficient.
Do:
| $ax$ax | $=$= | $b-cx$b−cx |
|
| $ax+cx$ax+cx | $=$= | $b-cx+cx$b−cx+cx |
Adding $cx$cx to both sides |
| $ax+cx$ax+cx | $=$= | $b$b |
Simplifying |
| $x\left(a+c\right)$x(a+c) | $=$= | $b$b |
Taking out the common factor of $x$x |
| $\frac{x\left(a+c\right)}{a+c}$x(a+c)a+c | $=$= | $\frac{b}{a+c}$ba+c |
Dividing both sides by $\left(a+c\right)$(a+c) |
| $x$x | $=$= | $\frac{b}{a+c}$ba+c |
|
example 5
Rearrange the equation $\frac{1}{p+2}=\frac{q}{r}$1p+2=qr to make $p$p the subject.
Think: In this equation, the variable $p$p appears in the denominator. To deal with this, we can first take the reciprocal of both sides (that is, flip the fractions on either side upside down), so that $p$p is in the numerator. We can then finish transposing as normal.
Do:
| $\frac{1}{p+2}$1p+2 | $=$= | $\frac{q}{r}$qr |
|
| $p+2$p+2 | $=$= | $\frac{r}{q}$rq |
Taking the reciprocal of both sides |
| $p+2-2$p+2−2 | $=$= | $\frac{r}{q}-2$rq−2 |
Subtracting $2$2 from both sides |
| $p$p | $=$= | $\frac{r}{q}-2$rq−2 |
Simplifying |
Practice questions
Question 3
Solve $D=RT$D=RT for $R$R.
Question 4
Solve $15x^2+5y=20$15x2+5y=20 for $y$y.