Rearranging Formulas

A formula is an equation that shows how two or more quantities are related, like $d = st$ for distance, speed and time. Formulas are usually written to give you one particular quantity, but real problems often ask for a different one. Rearranging a formula means solving it for a different variable, using the same balancing steps you use to solve linear equations.

Key ideas

The subject of a formula

The variable by itself on one side is called the subject. In $A = lw$, the subject is $A$ (area). If you know the area and the width and want the length, you make $l$ the subject:

$$A = lw \quad\Rightarrow\quad l = \frac{A}{w}$$

Rearranging is solving

To isolate a variable, treat every other letter as if it were a number, and use inverse operations, doing the same thing to both sides. Undo the operations in reverse order: addition and subtraction first, then multiplication and division.

It helps to compare with an equation you already know how to solve:

Equation	Formula
$2l + 10 = 50$	$2l + 2w = P$
$2l = 40$ (subtract $10$)	$2l = P - 2w$ (subtract $2w$)
$l = 20$ (divide by $2$)	$l = \dfrac{P - 2w}{2}$ (divide by $2$)

The steps are exactly the same. The only difference is that the answer is an expression instead of a number.

Some formulas you’ll use

Formula	Meaning	Rearranged examples
$A = lw$	area of a rectangle	$l = \dfrac{A}{w}$
$P = 2l + 2w$	perimeter of a rectangle	$w = \dfrac{P - 2l}{2}$
$d = st$	distance = speed $\times$ time	$t = \dfrac{d}{s}$, $\;s = \dfrac{d}{t}$
$C = 2\pi r$	circumference of a circle	$r = \dfrac{C}{2\pi}$
$F = \dfrac{9}{5}C + 32$	Celsius to Fahrenheit	$C = \dfrac{5}{9}(F - 32)$
$y = mx + b$	a linear relation	$m = \dfrac{y - b}{x}$, $\;x = \dfrac{y - b}{m}$

(The Grade 9 curriculum also writes linear relations as $y = ax + b$. It means the same thing: the rate of change times $x$, plus the initial value.)

Substituting values

When you substitute, put each value in brackets, especially negatives and fractions. That keeps the signs right:

$$y = mx + b \text{ with } m = -\tfrac{2}{3},\ x = 9,\ b = 4: \quad y = \left(-\tfrac{2}{3}\right)(9) + 4 = -6 + 4 = -2$$

Two ways to find an unknown

If you know all but one of the values, you can either:

• substitute first, then solve the equation you get, or
• rearrange first, then substitute.

Both give the same answer. Rearranging first is better when you need to do the same calculation many times (for example, a table of values), because you only rearrange once.

Worked examples

Example 1: Distance, speed and time

Rearrange $d = st$ to make $t$ the subject. Then find how long a 540 km drive takes at an average speed of 90 km/h.

Solution. $t$ is multiplied by $s$, so divide both sides by $s$:

$$d = st \quad\Rightarrow\quad \frac{d}{s} = t \quad\Rightarrow\quad t = \frac{d}{s}$$

Substitute $d = 540$ and $s = 90$:

$$t = \frac{540}{90} = 6$$

The drive takes $6$ hours.

Check: $90 \times 6 = 540$ km. ✓

Example 2: Perimeter of a rectangle

A rectangular poster has a perimeter of 50 cm and a length of $14\tfrac{1}{2}$ cm. Rearrange $P = 2l + 2w$ to find the width.

Solution. Undo the addition first, then the multiplication:

$$\begin{aligned} P &= 2l + 2w \\ P - 2l &= 2w && \text{subtract } 2l \\ \frac{P - 2l}{2} &= w && \text{divide by } 2 \end{aligned}$$

So $w = \dfrac{P - 2l}{2}$. Substitute $P = 50$ and $l = 14\tfrac{1}{2} = 14.5$:

$$w = \frac{50 - 2(14.5)}{2} = \frac{50 - 29}{2} = \frac{21}{2} = 10.5$$

The poster is $10.5$ cm (or $10\tfrac{1}{2}$ cm) wide.

Check: $2(14.5) + 2(10.5) = 29 + 21 = 50$. ✓

Example 3: Temperature

The formula $F = \dfrac{9}{5}C + 32$ changes a Celsius temperature $C$ into Fahrenheit $F$. Make $C$ the subject. Then convert $-4\,^\circ\text{F}$ and $98.6\,^\circ\text{F}$ to Celsius.

Solution.

$$\begin{aligned} F &= \frac{9}{5}C + 32 \\ F - 32 &= \frac{9}{5}C && \text{subtract } 32 \\ \frac{5}{9}(F - 32) &= C && \text{multiply by } \tfrac{5}{9} \end{aligned}$$

To undo “multiply by $\tfrac{9}{5}$”, multiply by its reciprocal, $\tfrac{5}{9}$. So $C = \dfrac{5}{9}(F - 32)$.

For $F = -4$:

$$C = \frac{5}{9}(-4 - 32) = \frac{5}{9}(-36) = -20$$

For $F = 98.6$:

$$C = \frac{5}{9}(98.6 - 32) = \frac{5}{9}(66.6) = 37$$

So $-4\,^\circ\text{F}$ is $-20\,^\circ\text{C}$ (a cold winter day), and $98.6\,^\circ\text{F}$ is $37\,^\circ\text{C}$ (normal body temperature).

Check: $\dfrac{9}{5}(-20) + 32 = -36 + 32 = -4$. ✓

Example 4: Fractions in a linear relation

The point with $x = \dfrac{7}{4}$ and $y = 3$ lies on the line $y = mx - \dfrac{1}{2}$. Find the slope $m$.

Solution. Rearrange $y = mx + b$ for $m$:

$$\begin{aligned} y &= mx + b \\ y - b &= mx && \text{subtract } b \\ \frac{y - b}{x} &= m && \text{divide by } x \end{aligned}$$

Substitute $y = 3$, $b = -\dfrac{1}{2}$ and $x = \dfrac{7}{4}$. Brackets around the negative fraction keep the signs right:

$$m = \frac{3 - \left(-\frac{1}{2}\right)}{\frac{7}{4}} = \frac{\frac{7}{2}}{\frac{7}{4}} = \frac{7}{2} \times \frac{4}{7} = 2$$

The slope is $m = 2$.

Check: $2 \times \dfrac{7}{4} - \dfrac{1}{2} = \dfrac{7}{2} - \dfrac{1}{2} = 3$. ✓

Common mistakes

Undoing operations in the wrong order. For $P = 2l + 2w$, you must subtract $2l$ before dividing by 2. Dividing first means dividing every term: $\dfrac{P}{2} = l + w$, which also works, but only if you divide all the terms.

Dividing only part of an expression. $w = \dfrac{P - 2l}{2}$ is not the same as $w = P - \dfrac{2l}{2}$. The fraction bar works like brackets: the whole top is divided by 2.

Forgetting brackets when substituting negatives. With $F = -4$, write $\dfrac{5}{9}(-4 - 32)$. And $x^2$ with $x = -3$ is $(-3)^2 = 9$, not $-9$.

Using the reciprocal the wrong way round. To undo multiplying by $\dfrac{9}{5}$, multiply by $\dfrac{5}{9}$, not by $\dfrac{9}{5}$ again.

Mixing up units. In $d = st$, if the speed is in km/h, the time must be in hours. 30 minutes is $0.5$ h, not $30$ h.

Practice

1. (Warm-up) Rearrange $A = lw$ to make $l$ the subject. Then find the length of a rectangular room with area $36$ m² and width $4.5$ m.

Solution

Divide both sides by $w$: $l = \dfrac{A}{w}$.

$$l = \frac{36}{4.5} = 8$$

The room is $8$ m long. Check: $8 \times 4.5 = 36$. ✓

2. (Warm-up) Rearrange $C = 2\pi r$ to make $r$ the subject. Then find the radius of a circle with circumference $50$ cm, to one decimal place.

Solution

$r$ is multiplied by $2\pi$, so divide both sides by $2\pi$: $r = \dfrac{C}{2\pi}$.

$$r = \frac{50}{2\pi} \approx 7.96 \approx 8.0$$

The radius is about $8.0$ cm. Check: $2\pi(7.96) \approx 50.0$. ✓

3. (Warm-up) Use $y = mx + b$ to find $y$ when $m = -\dfrac{2}{3}$, $x = 9$ and $b = 4$.

Solution

$$y = \left(-\frac{2}{3}\right)(9) + 4 = -6 + 4 = -2$$

4. (Core) Rearrange $y = mx + b$ to make $x$ the subject. Then find $x$ when $y = -5$, $m = -\dfrac{3}{4}$ and $b = 1$.

Solution

$$\begin{aligned} y &= mx + b \\ y - b &= mx && \text{subtract } b \\ x &= \frac{y - b}{m} && \text{divide by } m \end{aligned}$$

Substitute:

$$x = \frac{-5 - 1}{-\frac{3}{4}} = (-6) \times \left(-\frac{4}{3}\right) = 8$$

Check: $\left(-\dfrac{3}{4}\right)(8) + 1 = -6 + 1 = -5$. ✓

5. (Core) A rectangular picture frame has a perimeter of $2.4$ m and a width of $0.5$ m. Use $P = 2l + 2w$ to find its length.

Solution

Make $l$ the subject: subtract $2w$, then divide by 2.

$$l = \frac{P - 2w}{2} = \frac{2.4 - 2(0.5)}{2} = \frac{2.4 - 1}{2} = \frac{1.4}{2} = 0.7$$

The frame is $0.7$ m long. Check: $2(0.7) + 2(0.5) = 1.4 + 1 = 2.4$. ✓

6. (Core) Use $d = st$.

• (a) A cyclist rides at $18$ km/h. How long does it take to ride $27$ km? Give your answer in hours and minutes.
• (b) A hiker walks $10.5$ km in $2\tfrac{1}{2}$ hours. What is her average speed?

Solution

(a) $t = \dfrac{d}{s} = \dfrac{27}{18} = 1.5$ hours, which is 1 hour 30 minutes.

(b) Divide both sides of $d = st$ by $t$: $s = \dfrac{d}{t}$. With $t = 2\tfrac{1}{2} = \dfrac{5}{2}$:

$$s = \frac{10.5}{\frac{5}{2}} = 10.5 \times \frac{2}{5} = \frac{21}{5} = 4.2$$

Her average speed is $4.2$ km/h. Check: $4.2 \times 2.5 = 10.5$. ✓

7. (Core) On a very cold morning in Winnipeg, the temperature is $-31\,^\circ\text{F}$. Use $C = \dfrac{5}{9}(F - 32)$ to convert it to Celsius.

Solution

$$C = \frac{5}{9}(-31 - 32) = \frac{5}{9}(-63) = -35$$

It’s $-35\,^\circ\text{C}$. Check: $\dfrac{9}{5}(-35) + 32 = -63 + 32 = -31$. ✓

8. (Challenge) Is there a temperature that is the same number in Celsius and Fahrenheit? Use $F = \dfrac{9}{5}C + 32$ to find it.

Solution

If the two numbers are the same, then $F = C$. Replace $F$ with $C$:

$$\begin{aligned} C &= \frac{9}{5}C + 32 \\ 5C &= 9C + 160 && \text{multiply every term by } 5 \\ -4C &= 160 && \text{subtract } 9C \\ C &= -40 \end{aligned}$$

$-40\,^\circ\text{C}$ is the same as $-40\,^\circ\text{F}$. Check: $\dfrac{9}{5}(-40) + 32 = -72 + 32 = -40$. ✓

9. (Challenge) The area of a trapezoid is $A = \dfrac{(a + b)h}{2}$, where $a$ and $b$ are the parallel sides and $h$ is the height. Make $b$ the subject. Then find $b$ when $A = 45$ cm², $h = 6$ cm and $a = 8$ cm.

Solution

$$\begin{aligned} A &= \frac{(a + b)h}{2} \\ 2A &= (a + b)h && \text{multiply by } 2 \\ \frac{2A}{h} &= a + b && \text{divide by } h \\ b &= \frac{2A}{h} - a && \text{subtract } a \end{aligned}$$

Substitute:

$$b = \frac{2(45)}{6} - 8 = \frac{90}{6} - 8 = 15 - 8 = 7$$

The other parallel side is $7$ cm. Check: $\dfrac{(8 + 7)(6)}{2} = \dfrac{90}{2} = 45$. ✓