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.

On the SAT

Desmos can’t rearrange a formula for you, and on the SAT these questions usually ask “which equation gives $r$ in terms of $A$?”, so the answer is an expression, not a number. The quickest check is to pick easy numbers: choose values for the other variables, work out the subject from the original formula, then see which answer choice gives the same value (Desmos is handy for that arithmetic). Doing the rearranging by hand, one inverse operation at a time, is usually fastest of all. See using Desmos on the SAT.

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: $\dfrac{(8 + 7)(6)}{2} = \dfrac{90}{2} = 45$. ✓