Finding an Unknown in a Formula

How to find the value of any letter in a formula from the values of the others: substitute and solve the equation, or rearrange first and then substitute, choosing the sign of a root from the context and checking the answer.

Before you start

In Secondary 1 you substituted numbers into a formula to find its subject: if $v = u + at$ , $u = 5$ , $a = 3$ and $t = 4$ , then $v = 5 + 12 = 17$ .

This lesson finds any letter, not just the subject. The step after substituting is solving an equation, sometimes one with fractions. Multiplying every term by the LCM of the denominators clears the fractions:

Reminder: An Equation with Fractions

Solve $\dfrac{a - 2}{5} + \dfrac{a - 1}{3} = 1\text{.}$

Solution

The LCM of $5$ and $3$ is $15$ . Multiply every term by $15$ : $3{(a - 2)} + 5{(a - 1)} = 15\text{.}$

Expand: $3a - 6 + 5a - 5 = 15$ , so $8a - 11 = 15$ , $8a = 26$ and $a = \dfrac{26}{8} = \dfrac{13}{4}$ .

Check: $\dfrac{\frac{13}{4} - 2}{5} + \dfrac{\frac{13}{4} - 1}{3} = \dfrac{1}{4} + \dfrac{3}{4} = 1\text{.}$

$a = 3\tfrac{1}{4}$

Substitute, Then Solve

Finding an unknown in a formula

1. Write down the formula.

2. Substitute every value you know. Put brackets round negative numbers.

3. Solve the equation that is left for the unknown letter.

4. Check: put your answer back into the formula, and make sure no denominator is zero. Give the unit, and round only at the end.

Example 1: A Linear Formula

A car speeds up steadily. Its velocity after $t$ seconds is $v = u + at$ , where $u$ m/s is its starting velocity and $a$ m/s² its acceleration. Find $t$ when $v = 20$ , $u = 5$ and $a = 3$ .

Solution

Substitute: $20 = 5 + 3t$ .

Solve: $3t = 15$ , so $t = 5$ .

Check: $5 + 3 \times 5 = 20$ .

$t = 5$ : the car reaches $20$ m/s after $5$ seconds.

Example 2: The Unknown inside a Bracket

The area of a trapezium is $A = \dfrac{1}{2}{(a + b)}h$ . Find $b$ when $A = 30$ , $a = 4$ and $h = 5$ (lengths in cm).

Solution

Substitute: $30 = \dfrac{1}{2}{(4 + b)} \times 5\text{.}$

Multiply both sides by $2$ : $60 = 5{(4 + b)}$ . Divide by $5$ : $12 = 4 + b$ . So $b = 8$ .

Check: $\dfrac{1}{2} \times 12 \times 5 = 30$ .

$b = 8$ cm

Example 3: Negative Values

(a) $C = \dfrac{5}{9}{(F - 32)}$ . Find $F$ when $C = -40$ . (b) $s = ut + \dfrac{1}{2}at^2$ . Find $u$ when $s = 18$ , $t = 3$ and $a = -4$ .

Solution

(a) $-40 = \dfrac{5}{9}{(F - 32)}\text{.}$ Multiply by $\dfrac{9}{5}$ : $-72 = F - 32$ . Add $32$ : $F = -40$ . At $-40$ degrees the two scales agree.

(b) $18 = 3u + \dfrac{1}{2}{(-4)}(3)^2 = 3u + \dfrac{1}{2}{(-4)}(9) = 3u - 18\text{.}$

So $3u = 36$ and $u = 12$ .

Check (b): $12 \times 3 + \dfrac{1}{2}{(-4)}(9) = 36 - 18 = 18\text{.}$

(a) $F = -40$ (b) $u = 12$

The Unknown in a Denominator

If the unknown is in a denominator, the equation has fractions. Clear them, or combine the known fractions first. Then check that your answer does not make any denominator zero: such a value can never be an answer.

Example 4: Reciprocals

For a lens, $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ . Find $v$ when $f = 10$ and $u = 15$ .

Solution

Substitute: $\dfrac{1}{10} = \dfrac{1}{15} + \dfrac{1}{v}\text{.}$

Subtract $\dfrac{1}{15}$ : $\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3}{30} - \dfrac{2}{30} = \dfrac{1}{30}\text{.}$

Turn both sides over: $v = 30$ .

Check: $\dfrac{1}{15} + \dfrac{1}{30} = \dfrac{2}{30} + \dfrac{1}{30} = \dfrac{3}{30} = \dfrac{1}{10}\text{.}$

$v = 30$

Example 5: The Unknown Appears Twice

Given $y = \dfrac{x + 3}{x - 2}$ , find $x$ when $y = 2$ .

Solution

Substitute: $2 = \dfrac{x + 3}{x - 2}$ .

Multiply both sides by ${(x - 2)}$ : $2{(x - 2)} = x + 3$ , so $2x - 4 = x + 3$ .

Collect: $2x - x = 3 + 4$ , so $x = 7$ .

Check: the denominator $x - 2 = 5$ is not zero, and $\dfrac{7 + 3}{7 - 2} = \dfrac{10}{5} = 2\text{.}$

$x = 7$

The Unknown Squared or under a Root

If the unknown is squared, it has two possible values, one positive and one negative ( $\pm$ ), because a positive number has two square roots. Keep both unless the context rules one out: a length, a time, a speed or a mass is positive.

If the unknown is under a square root, square both sides once the root is on its own. Use the $\pi$ key of your calculator, and round only at the end, usually to $3$ significant figures.

Example 6: Choosing the Sign

For an object moving in a straight line, $v^2 = u^2 + 2as$ . (a) Find the possible values of $u$ when $v = 10$ , $a = 2$ and $s = 16$ . (b) Which value is the answer if $u$ is the speed at the start?

Solution

(a) Substitute: $100 = u^2 + 2 \times 2 \times 16 = u^2 + 64\text{.}$

So $u^2 = 36$ , and $u = \pm 6$ . Both $6^2$ and ${(-6)}^2$ are $36$ .

(b) A speed is not negative, so $u = 6$ .

Check: $6^2 + 64 = 100 = 10^2$ .

(a) $u = 6$ or $u = -6$ (b) $u = 6$

Example 7: A Radius from an Area

A circle has an area of $50$ cm². Use $A = \pi r^2$ to find its radius, correct to $3$ significant figures.

Solution

Substitute: $50 = \pi r^2$ , so $r^2 = \dfrac{50}{\pi}$ .

A radius is positive: $r = \sqrt{\dfrac{50}{\pi}} = 3.9894\ldots$

Correct to $3$ significant figures, $r = 3.99$ cm.

Check: $\pi \times 3.9894^2$ is $50.0$ to $3$ significant figures.

$r = 3.99$ cm (3 s.f.)

Example 8: The Unknown under a Root

A pendulum of length $l$ m takes $t$ seconds for one swing, where $t = 2\pi\sqrt{\dfrac{l}{g}}$ and $g = 9.8$ . Find the length of a pendulum that takes $2$ seconds, correct to $3$ significant figures.

Solution

Substitute: $2 = 2\pi\sqrt{\dfrac{l}{9.8}}$ .

Divide by $2\pi$ : $\sqrt{\dfrac{l}{9.8}} = \dfrac{1}{\pi}$ .

Square both sides: $\dfrac{l}{9.8} = \dfrac{1}{\pi^2}$ , so $l = \dfrac{9.8}{\pi^2} = 0.99294\ldots$

Check: $2\pi\sqrt{\dfrac{0.99294}{9.8}}$ is $2.00$ to $3$ significant figures.

$l = 0.993$ m (3 s.f.)

Rearrange First, or Substitute First?

There are two ways to find an unknown. Substitute first, then solve, as above. Or rearrange first: make the unknown the subject (lessons 4.6 to 4.8), then substitute.

Both give the same answer. Substituting first is usually quicker for one value, because the numbers keep the equation short. Rearranging first is better when you need the same letter for many sets of values, or when the question asks for the formula.

Example 9: Both Ways

Using $v^2 = u^2 + 2as$ , find $s$ when $v = 14$ , $u = 6$ and $a = 4$ , (a) by substituting first, (b) by rearranging first.

Solution

(a) $196 = 36 + 8s$ , so $8s = 160$ and $s = 20$ .

(b) $2as = v^2 - u^2$ , so $s = \dfrac{v^2 - u^2}{2a} = \dfrac{196 - 36}{8} = \dfrac{160}{8} = 20\text{.}$

$s = 20$ either way

Example 10: Many Values from One Formula

The simple interest $I$ dollars on $P$ dollars at $R$ % per year for $T$ years is $I = \dfrac{PRT}{100}$ . Find the rate $R$ for each loan: A, $I = 90$ , $P = 600$ , $T = 3$ ; B, $I = 54$ , $P = 900$ , $T = 2$ ; C, $I = 120$ , $P = 1500$ , $T = 4$ .

Solution

Three values of $R$ are needed, so rearrange once: $100I = PRT$ , so $R = \dfrac{100I}{PT}$ .

A: $R = \dfrac{9000}{1800} = 5$ . B: $R = \dfrac{5400}{1800} = 3$ . C: $R = \dfrac{12000}{6000} = 2\text{.}$

Check A: $\dfrac{600 \times 5 \times 3}{100} = 90\text{.}$

A: $5$ % B: $3$ % C: $2$ % per year

Common Mistakes

Forgetting brackets round a negative number: with $a = -4$ , $\dfrac{1}{2}at^2 = \dfrac{1}{2}{(-4)}(9) = -18\text{.}$
Squaring too much: in $\dfrac{1}{2}at^2$ only $t$ is squared. With $a = -4$ , $t = 3$ : $\dfrac{1}{2} \times {(-4)} \times 3^2 = -18\text{.}$ Squaring the whole product $\dfrac{1}{2} \times {(-4)} \times 3 = -6$ would give $36$ .
Turning over each fraction separately: from $\dfrac{1}{10} = \dfrac{1}{15} + \dfrac{1}{v}\text{,}$ $v$ is not $10 - 15$ . Find $\dfrac{1}{v}$ first.
Giving one root where both are possible, or a negative root for a length, a time or a speed.
Rounding in the middle of the working. Keep the full calculator value and round at the end.
Not checking: put the answer back into the formula, and make sure no denominator is zero.

Practice

Use a calculator where needed, with its $\pi$ key. Give answers that are not exact correct to 3 significant figures, and check each answer in the formula.

Question 1

Given $v = u + at$ , find $a$ when $v = 30$ , $u = 6$ and $t = 4$ .

ShowHide solution

$30 = 6 + 4a$ , so $4a = 24$ and $a = 6$ . Check: $6 + 24 = 30$ .

$a = 6$

Question 2

(a) The perimeter of a rectangle is $P = 2{(l + b)}$ . Find $b$ when $P = 46$ and $l = 15$ . (b) The angles of an $n$ -sided polygon add up to $S = 180{(n - 2)}$ degrees. Find $n$ when $S = 1080$ .

ShowHide solution

(a) $46 = 2{(15 + b)}$ , so $23 = 15 + b$ and $b = 8$ .

(b) $1080 = 180{(n - 2)}$ , so $6 = n - 2$ and $n = 8$ : an octagon.

(a) $b = 8$ (b) $n = 8$

Question 3

(a) Given $F = \dfrac{9C}{5} + 32$ , find $C$ when $F = 14$ . (b) Given $y = mx + c$ , find $x$ when $y = -7$ , $m = -3$ and $c = 5$ .

ShowHide solution

(a) $14 = \dfrac{9C}{5} + 32$ , so $\dfrac{9C}{5} = -18$ , $9C = -90$ and $C = -10$ .

(b) $-7 = {(-3)}x + 5$ , so $-3x = -12$ and $x = 4$ . Check: $-12 + 5 = -7$ .

(a) $C = -10$ (b) $x = 4$

Question 4

(a) Given $\dfrac{1}{R} = \dfrac{1}{a} + \dfrac{1}{b}$ , find $b$ when $R = 4$ and $a = 12$ . (b) Given $T = \dfrac{k}{x - 2}$ , find $x$ when $T = 4$ and $k = 10$ .

ShowHide solution

(a) $\dfrac{1}{b} = \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{3}{12} - \dfrac{1}{12} = \dfrac{2}{12} = \dfrac{1}{6}\text{,}$ so $b = 6$ . Check: $\dfrac{1}{12} + \dfrac{1}{6} = \dfrac{3}{12} = \dfrac{1}{4}\text{.}$

(b) $4 = \dfrac{10}{x - 2}$ , so $4{(x - 2)} = 10$ , $x - 2 = 2.5$ and $x = 4.5$ . The denominator $2.5$ is not zero.

(a) $b = 6$ (b) $x = 4.5$

Question 5

Given $y = \dfrac{2x + 1}{x + 3}$ , find $x$ when $y = 3$ .

ShowHide solution

$3{(x + 3)} = 2x + 1$ , so $3x + 9 = 2x + 1$ and $x = -8$ .

Check: $x + 3 = -5$ is not zero, and $\dfrac{2 \times {(-8)} + 1}{-8 + 3} = \dfrac{-15}{-5} = 3\text{.}$

$x = -8$

Question 6

(a) An object of mass $m$ kg moving at $v$ m/s has kinetic energy $E = \dfrac{1}{2}mv^2$ joules. Find its speed when $E = 100$ and $m = 8$ . (b) Given $y = {(x - 1)}^2 + 3$ , find the possible values of $x$ when $y = 19$ .

ShowHide solution

(a) $100 = 4v^2$ , so $v^2 = 25$ . A speed is positive: $v = 5$ .

(b) ${(x - 1)}^2 = 16$ , so $x - 1 = \pm 4$ and $x = 5$ or $x = -3$ . Check: $4^2 + 3 = 19$ and ${(-4)}^2 + 3 = 19$ .

(a) $5$ m/s (b) $x = 5$ or $x = -3$

Question 7

(a) Using $t = 2\pi\sqrt{\dfrac{l}{g}}$ with $g = 9.8$ , find the length $l$ m of a pendulum that takes $3$ seconds for a swing. (b) A sphere has volume $500$ cm³. Use $V = \dfrac{4}{3}\pi r^3$ to find its radius.

ShowHide solution

(a) $\sqrt{\dfrac{l}{9.8}} = \dfrac{3}{2\pi}$ . Square: $\dfrac{l}{9.8} = \dfrac{9}{4\pi^2}$ , so $l = \dfrac{9.8 \times 9}{4\pi^2} = 2.2341\ldots$

(b) $r^3 = \dfrac{3 \times 500}{4\pi} = 119.36\ldots\text{,}$ so $r = \sqrt[3]{119.36\ldots} = 4.9237\ldots$

(a) $2.23$ m (b) $4.92$ cm

Question 8

Given $s = ut + \dfrac{1}{2}at^2$ , find (a) $a$ when $s = 50$ , $u = 0$ and $t = 5$ (b) $u$ when $s = 40$ , $t = 4$ and $a = 3$ .

ShowHide solution

(a) $50 = 0 + \dfrac{1}{2}a \times 25$ , so $12.5a = 50$ and $a = 4$ .

(b) $40 = 4u + \dfrac{1}{2} \times 3 \times 16 = 4u + 24\text{,}$ so $4u = 16$ and $u = 4$ .

(a) $a = 4$ (b) $u = 4$

Question 9

Using $F = \dfrac{9C}{5} + 32$ , change $50$ °F, $77$ °F and $104$ °F to degrees Celsius. Rearrange the formula first.

ShowHide solution

Rearrange: $F - 32 = \dfrac{9C}{5}$ , so $C = \dfrac{5(F - 32)}{9}$ .

$50$ °F: $\dfrac{5 \times 18}{9} = 10$ . $77$ °F: $\dfrac{5 \times 45}{9} = 25$ . $104$ °F: $\dfrac{5 \times 72}{9} = 40$ .

$10$ °C, $25$ °C and $40$ °C

Question 10

Mrs Tan puts $2400 in a savings plan that pays simple interest at $2.5$ % per year. Using $I = \dfrac{PRT}{100}$ , find how many years it takes to earn $300 in interest.

ShowHide solution

$300 = \dfrac{2400 \times 2.5 \times T}{100} = 60T\text{,}$ so $T = 5$ .

Check: $\dfrac{2400 \times 2.5 \times 5}{100} = 300\text{.}$

$5$ years

Question 11

A garden plot in the shape of a trapezium has an area of $54$ m², a height of $6$ m and one parallel side of $7$ m. Use $A = \dfrac{1}{2}{(a + b)}h$ to find the other parallel side.

ShowHide solution

$54 = \dfrac{1}{2}{(7 + b)} \times 6 = 3{(7 + b)}\text{,}$ so $7 + b = 18$ and $b = 11$ .

Check: $\dfrac{1}{2} \times 18 \times 6 = 54$ .

$11$ m

Question 12

Using $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ with $f = 6$ and $u = 10$ , Ahmad writes $\dfrac{1}{v} = \dfrac{1}{6} - \dfrac{1}{10} = \dfrac{1}{6 - 10} = -\dfrac{1}{4}\text{,}$ so $v = -4$ . Check his answer, and find $v$ correctly.

ShowHide solution

Check: $\dfrac{1}{10} + \dfrac{1}{-4} = 0.1 - 0.25 = -0.15\text{,}$ but $\dfrac{1}{6}$ is about $0.167$ . So $v = -4$ is wrong.

He subtracted the denominators. Correct: $\dfrac{1}{6} - \dfrac{1}{10} = \dfrac{5}{30} - \dfrac{3}{30} = \dfrac{2}{30} = \dfrac{1}{15}\text{,}$ so $v = 15$ .

Check: $\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3}{30} + \dfrac{2}{30} = \dfrac{5}{30} = \dfrac{1}{6}\text{.}$

$v = 15$

Key Points

To find an unknown, substitute the known values (with brackets round negatives), then solve the equation.
Or rearrange first to make the unknown the subject, then substitute. This is better when you need many values.
With fractions, clear them or combine the known ones, and check that no denominator is zero.
If the unknown is squared there are two roots, $\pm$ ; keep only the positive one for a length, a time, a speed or a mass.
If the unknown is under a root, get the root alone, then square both sides.
Round only at the end, and check the answer in the formula.