Changing the Subject of a Formula

Solution

(a) $l$ is multiplied by $b$. Divide both sides by $b$: $\dfrac{A}{b} = l$, so $l = \dfrac{A}{b}$.

(b) $b$ has $a$ added to it. Subtract $a$ from both sides: $P - a = b$, so $b = P - a$.

(c) $r$ is multiplied by $2\pi$. Divide both sides by $2\pi$: $r = \dfrac{C}{2\pi}$.

Solution

How is $v$ built from $t$? First $t$ is multiplied by $a$, giving $at$; then $u$ is added, giving $u + at$.

The last step was "add $u$", so undo it first. Subtract $u$ from both sides: $v - u = at$.

Then undo "multiply by $a$". Divide both sides by $a$: $\dfrac{v - u}{a} = t$.

So $t = \dfrac{v - u}{a}$. The whole of $v - u$ is divided by $a$, so it sits above the fraction bar.

Check with $u = 1$, $a = 2$, $t = 3$: the first formula gives $v = 1 + 6 = 7$, and the new one gives $t = \dfrac{7 - 1}{2} = 3$.

Solution

How is $C$ built from $F$? First $32$ is subtracted from $F$; then the result is multiplied by $\dfrac{5}{9}$.

The last step was "multiply by $\dfrac{5}{9}$", so undo it first. Multiply both sides by $\dfrac{9}{5}$: $\dfrac{9}{5}C = F - 32$.

Then undo "subtract $32$". Add $32$ to both sides: $\dfrac{9}{5}C + 32 = F$.

So $F = \dfrac{9C}{5} + 32$.

Check: water boils at $100$ °C and $212$ °F. The new formula gives $\dfrac{900}{5} + 32 = 180 + 32 = 212\text{.}$

Solution

From $a$: add $b$, then multiply by $\dfrac{1}{2}$ and by $h$, which is the same as multiplying by $\dfrac{h}{2}$.

Undo the multiplying first. Multiply both sides by $2$: $2A = {(a + b)}h$. Divide both sides by $h$: $\dfrac{2A}{h} = a + b$.

Then subtract $b$: $a = \dfrac{2A}{h} - b$.

Check with $a = 3$, $b = 5$, $h = 2$: $A = \dfrac{1}{2} \times 8 \times 2 = 8$, and $\dfrac{2 \times 8}{2} - 5 = 3$.

Solution

The term with $x$ has a minus sign. It is easiest to add $2x$ to both sides first, so that it becomes positive: $y + 2x = 5$.

Subtract $y$ from both sides: $2x = 5 - y$.

Divide both sides by $2$: $x = \dfrac{5 - y}{2}$.

Undoing in reverse order also works: subtract $5$, $y - 5 = -2x$, then divide by $-2$: $x = \dfrac{y - 5}{-2} = \dfrac{5 - y}{2}\text{,}$ the same answer. Adding $2x$ first just avoids dividing by a negative number.

Check with $x = 1$: $y = 5 - 2 = 3$, and $\dfrac{5 - 3}{2} = 1$.

Solution

(a) Multiply both sides by $4$: $4y = 3x - 1$. Add $1$: $4y + 1 = 3x$. Divide by $3$: $x = \dfrac{4y + 1}{3}$.

(b) Multiply both sides by $3$: $3V = \pi r^2 h$. Divide by $\pi r^2$: $h = \dfrac{3V}{\pi r^2}$.

Check (a) with $x = 3$: $y = \dfrac{8}{4} = 2$, and $\dfrac{8 + 1}{3} = 3$.

Solution

(a) Multiply both sides by $t$: $st = d$. Divide by $s$: $t = \dfrac{d}{s}$. This needs $s \neq 0$; here $s$ is a speed, which is not $0$ on a journey.

(b) Multiply both sides by ${(x - 2)}$: $T{(x - 2)} = k$. Divide by $T$: $x - 2 = \dfrac{k}{T}$. Add $2$: $x = 2 + \dfrac{k}{T}$, for $T \neq 0$.

This can also be written as one fraction: $x = \dfrac{2T + k}{T}$.

Check (b) with $k = 6$, $x = 5$: $T = \dfrac{6}{3} = 2$, and $2 + \dfrac{6}{2} = 5$.

Solution

Subtract $\dfrac{1}{u}$ from both sides: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$.

Write the right side as one fraction (lesson 4.4), over $fu$: $\dfrac{1}{v} = \dfrac{u}{fu} - \dfrac{f}{fu} = \dfrac{u - f}{fu}\text{.}$

If two numbers are equal, so are their reciprocals. Turn both sides over: $v = \dfrac{fu}{u - f}$, for $u \neq f$.

Do not turn over each fraction one by one: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ does not give $v = f - u$.

Check with $f = 2$, $u = 3$: $v = \dfrac{6}{1} = 6$, and $\dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}$.

Solution

Take $F = 212$. The original formula gives $C = \dfrac{5}{9} \times 180 = 100\text{.}$

Mei's formula gives $F = \dfrac{9}{5} \times 132 = 237.6\text{,}$ not $212$. So her formula is wrong.

She undid the steps in the wrong order: she added $32$ before multiplying by $\dfrac{9}{5}$. Correct: $F = \dfrac{9}{5}C + 32$, which gives $180 + 32 = 212$.