Changing the Subject When It Appears Twice

How to change the subject of a formula in which the subject appears in more than one term: clear fractions, collect its terms on one side, take it out as a common factor, and divide by a bracket that must not be zero.

Before you start

Lesson 4.6: to change the subject, do the same to both sides, undoing what was done to the subject in reverse order. A fraction is cleared by multiplying both sides by its denominator.

Chapter 3: a common factor can be taken out of a sum of terms, as in $ax - bx = x{(a - b)}$ and $x + xy = x{(1 + y)}$ .

Collect, Factorise, Divide

If the new subject appears in two terms, you cannot undo the operations one by one: there are two places to free it from. Bring the two terms together and take the subject out as a common factor. Then only one bracket stands beside it, and dividing by that bracket leaves the subject alone.

A subject that appears twice

1. Clear any fractions and expand any brackets that contain the subject.

2. Collect every term with the subject on one side, and every other term on the other side.

3. Factorise: take the subject out as a common factor.

4. Divide both sides by the bracket, and state that the bracket must not be zero.

Example 1: Factorise Straight Away

A shop sells an item for $S$ dollars: its cost of $C$ dollars plus a profit of $p\%$ of the cost, so $S = C + \dfrac{pC}{100}$ . Make $C$ the subject.

Solution

Clear the fraction: multiply every term by $100$ : $100S = 100C + pC$ .

Both terms with $C$ are already on one side. Factorise: $100S = C{(100 + p)}$ .

Divide by the bracket: $C = \dfrac{100S}{100 + p}$ , for $p \neq -100$ . A profit is positive, so the bracket is never zero.

Check with $C = 80$ , $p = 25$ : $S = 80 + \dfrac{2000}{100} = 100\text{,}$ and $\dfrac{100 \times 100}{125} = 80\text{.}$

$C = \dfrac{100S}{100 + p}$ , for $p \neq -100$

Example 2: The Subject on Both Sides

Make $x$ the subject of $ax + b = cx + d$ .

Solution

Collect the $x$ terms on the left: subtract $cx$ from both sides, and subtract $b$ : $ax - cx = d - b$ .

Factorise: $x{(a - c)} = d - b$ .

Divide by ${(a - c)}$ : $x = \dfrac{d - b}{a - c}$ , for $a \neq c$ .

Check with numbers: $5x + 2 = 3x + 10$ has solution $x = 4$ ( $22 = 22$ ), and $\dfrac{10 - 2}{5 - 3} = 4$ .

$x = \dfrac{d - b}{a - c}$ , for $a \neq c$

Example 3: Expand, Collect, Factorise

Make $x$ the subject of $ax = b{(x + c)}$ .

Solution

Expand the bracket that contains $x$ : $ax = bx + bc$ .

Collect the $x$ terms: $ax - bx = bc$ .

Factorise: $x{(a - b)} = bc$ . Divide: $x = \dfrac{bc}{a - b}$ , for $a \neq b$ .

Check with $a = 5$ , $b = 2$ , $c = 3$ : $\dfrac{6}{3} = 2$ , and $5 \times 2 = 10 = 2{(2 + 3)}\text{.}$

$x = \dfrac{bc}{a - b}$ , for $a \neq b$

Example 4: Choosing the Side

Make $x$ the subject of $a - bx = c{(x + 1)}$ .

Solution

Expand: $a - bx = cx + c$ .

The $x$ terms can be collected on either side. On the right they are both positive: add $bx$ to both sides and subtract $c$ : $a - c = cx + bx$ .

Factorise: $a - c = x{(c + b)}$ . Divide: $x = \dfrac{a - c}{b + c}$ , for $b + c \neq 0$ .

Collecting on the left instead gives $-bx - cx = c - a$ , so $x = \dfrac{c - a}{-b - c}$ : the same answer, with a minus sign on top and below.

Check with $a = 10$ , $b = 2$ , $c = 3$ : $x = \dfrac{7}{5} = 1.4$ ; then $10 - 2.8 = 7.2$ and $3 \times 2.4 = 7.2$ .

$x = \dfrac{a - c}{b + c}$ , for $b + c \neq 0$

Fractions That Contain the Subject

If the subject is in a numerator and a denominator, multiplying by the denominator clears the fraction but leaves the subject on both sides. Expand, then collect and factorise as before.

Example 5: Clear the Fraction First

Make $x$ the subject of $y = \dfrac{x + 1}{x - 1}$ .

Solution

Multiply both sides by ${(x - 1)}$ : $y{(x - 1)} = x + 1$ .

Expand: $xy - y = x + 1$ . Now $x$ is in two terms.

Collect: $xy - x = y + 1$ .

Factorise: $x{(y - 1)} = y + 1$ . Divide: $x = \dfrac{y + 1}{y - 1}$ , for $y \neq 1$ .

Check with $x = 3$ : $y = \dfrac{4}{2} = 2$ , and $\dfrac{2 + 1}{2 - 1} = 3$ .

$x = \dfrac{y + 1}{y - 1}$ , for $y \neq 1$

Example 6: Two Fractions with the Subject

Make $x$ the subject of $\dfrac{x}{a} + \dfrac{x}{b} = c$ .

Solution

Multiply every term by $ab$ : $bx + ax = abc$ .

Factorise: $x{(a + b)} = abc$ . Divide: $x = \dfrac{abc}{a + b}$ , for $a + b \neq 0$ .

Check with $a = 2$ , $b = 3$ , $c = 5$ : $x = \dfrac{30}{5} = 6$ , and $\dfrac{6}{2} + \dfrac{6}{3} = 3 + 2 = 5\text{.}$

$x = \dfrac{abc}{a + b}$ , for $a + b \neq 0$

Common Mistakes

Stopping with the subject on both sides: from $ax = b + cx$ , writing $x = \dfrac{b + cx}{a}$ is not finished. Collect and factorise: $x = \dfrac{b}{a - c}$ .
Losing the $1$ when factorising: $ax - x = x{(a - 1)}$ , not $xa$ . The term $x$ is $1 \times x$ .
Collecting before the fraction is cleared or the bracket expanded: in $y{(x - 1)} = x + 1$ , the $x$ inside the bracket must come out first.
Dividing by a bracket that could be zero without saying so: $x = \dfrac{d - b}{a - c}$ needs $a \neq c$ .

Practice

Make the letter in brackets the subject, and state the value the bracket you divide by must not take. Check each answer with numbers.

Question 1

(a) $S = ar + a$ $(a)$ (b) $T = 3m + mn$ $(m)$

ShowHide solution

(a) Factorise: $S = a{(r + 1)}$ , so $a = \dfrac{S}{r + 1}$ , for $r \neq -1$ .

(b) Factorise: $T = m{(3 + n)}$ , so $m = \dfrac{T}{3 + n}$ , for $n \neq -3$ .

Check (b) with $m = 2$ , $n = 4$ : $T = 6 + 8 = 14$ , and $\dfrac{14}{7} = 2$ .

(a) $a = \dfrac{S}{r + 1}$ , for $r \neq -1$ (b) $m = \dfrac{T}{3 + n}$ , for $n \neq -3$

Question 2

(a) $ax + 3 = bx + 7$ $(x)$ (b) $px - q = rx + s$ $(x)$

ShowHide solution

(a) $ax - bx = 7 - 3$ , so $x{(a - b)} = 4$ and $x = \dfrac{4}{a - b}$ , for $a \neq b$ .

(b) $px - rx = s + q$ , so $x{(p - r)} = s + q$ and $x = \dfrac{q + s}{p - r}$ , for $p \neq r$ .

Check (a) with $a = 5$ , $b = 3$ : $x = 2$ , and $10 + 3 = 13 = 6 + 7$ .

(a) $x = \dfrac{4}{a - b}$ , for $a \neq b$ (b) $x = \dfrac{q + s}{p - r}$ , for $p \neq r$

Question 3

(a) $a{(x - b)} = cx$ $(x)$ (b) $p{(x + q)} = r{(x - q)}$ $(x)$

ShowHide solution

(a) Expand: $ax - ab = cx$ . Collect: $ax - cx = ab$ . Factorise: $x{(a - c)} = ab$ . So $x = \dfrac{ab}{a - c}$ , for $a \neq c$ .

Check (a) with $a = 3$ , $b = 2$ , $c = 1$ : $x = \dfrac{6}{2} = 3$ , and $3{(3 - 2)} = 3 = 1 \times 3$ .

(b) Expand: $px + pq = rx - rq$ . Collect: $px - rx = -rq - pq$ . Factorise both sides: $x{(p - r)} = -q{(p + r)}\text{.}$ So $x = \dfrac{-q{(p + r)}}{p - r} = \dfrac{q{(p + r)}}{r - p}\text{,}$ for $p \neq r$ .

Check (b) with $p = 3$ , $q = 2$ , $r = 1$ : $x = \dfrac{2 \times 4}{-2} = -4\text{,}$ and $3{(-4 + 2)} = -6 = 1 \times {(-4 - 2)}\text{.}$

(a) $x = \dfrac{ab}{a - c}$ , for $a \neq c$ (b) $x = \dfrac{q{(p + r)}}{r - p}$ , for $p \neq r$

Question 4

(a) $5 - ax = bx$ $(x)$ (b) $k - 2x = m{(x - 1)}$ $(x)$

ShowHide solution

(a) Add $ax$ to both sides: $5 = bx + ax = x{(a + b)}\text{.}$ So $x = \dfrac{5}{a + b}$ , for $a + b \neq 0$ .

(b) Expand: $k - 2x = mx - m$ . Collect on the right, where both $x$ terms are positive: $k + m = mx + 2x = x{(m + 2)}\text{.}$ So $x = \dfrac{k + m}{m + 2}$ , for $m \neq -2$ .

Check (b) with $k = 4$ , $m = 3$ : $x = \dfrac{7}{5} = 1.4$ ; then $4 - 2.8 = 1.2$ and $3 \times 0.4 = 1.2$ .

(a) $x = \dfrac{5}{a + b}$ , for $a + b \neq 0$ (b) $x = \dfrac{k + m}{m + 2}$ , for $m \neq -2$

Question 5

(a) $y = \dfrac{2x + 3}{x - 4}$ $(x)$ (b) $y = \dfrac{x}{x + a}$ $(x)$

ShowHide solution

(a) $y{(x - 4)} = 2x + 3$ , so $xy - 4y = 2x + 3$ . Collect: $xy - 2x = 4y + 3$ . Factorise: $x{(y - 2)} = 4y + 3$ . So $x = \dfrac{4y + 3}{y - 2}$ , for $y \neq 2$ .

Check (a) with $x = 5$ : $y = 13$ , and $\dfrac{55}{11} = 5$ .

(b) $y{(x + a)} = x$ , so $xy + ay = x$ . Collect on the right: $ay = x - xy = x{(1 - y)}\text{.}$ So $x = \dfrac{ay}{1 - y}$ , for $y \neq 1$ .

Check (b) with $a = 2$ , $x = 2$ : $y = \dfrac{2}{4} = \dfrac{1}{2}$ , and $\dfrac{2 \times \frac{1}{2}}{\frac{1}{2}} = 2$ .

(a) $x = \dfrac{4y + 3}{y - 2}$ , for $y \neq 2$ (b) $x = \dfrac{ay}{1 - y}$ , for $y \neq 1$

Question 6

$\dfrac{x + a}{b} = \dfrac{x - b}{a}$ $(x)$

ShowHide solution

Multiply both sides by $ab$ : $a{(x + a)} = b{(x - b)}$ , so $ax + a^2 = bx - b^2$ .

Collect: $ax - bx = -a^2 - b^2$ . So $x{(a - b)} = -{(a^2 + b^2)}$ and $x = \dfrac{a^2 + b^2}{b - a}$ , for $a \neq b$ .

Check with $a = 1$ , $b = 2$ : $x = 5$ , and $\dfrac{6}{2} = 3 = \dfrac{3}{1}$ .

$x = \dfrac{a^2 + b^2}{b - a}$ , for $a \neq b$

Question 7

$\dfrac{x}{2} - \dfrac{x}{a} = b$ $(x)$

ShowHide solution

Multiply every term by $2a$ : $ax - 2x = 2ab$ .

Factorise: $x{(a - 2)} = 2ab$ . So $x = \dfrac{2ab}{a - 2}$ , for $a \neq 2$ .

Check with $a = 4$ , $b = 3$ : $x = \dfrac{24}{2} = 12$ , and $6 - 3 = 3$ .

$x = \dfrac{2ab}{a - 2}$ , for $a \neq 2$

Question 8

Siti has $x$ stickers. She gives $\dfrac{1}{a}$ of them to her brother and $4$ to a friend, and has $k$ left, so $x - \dfrac{x}{a} - 4 = k$ . (a) Make $x$ the subject. (b) How many stickers did she have if she gave one third of them to her brother and has $10$ left?

ShowHide solution

(a) Multiply every term by $a$ : $ax - x - 4a = ak$ .

Collect the $x$ terms: $ax - x = ak + 4a$ . Factorise both sides: $x{(a - 1)} = a{(k + 4)}$ .

So $x = \dfrac{a{(k + 4)}}{a - 1}$ , for $a \neq 1$ . (If $a$ were $1$ , she would give all her stickers to her brother.)

(b) $a = 3$ , $k = 10$ : $x = \dfrac{3 \times 14}{2} = 21$ . Check: $21 - 7 - 4 = 10$ .

(a) $x = \dfrac{a{(k + 4)}}{a - 1}$ , for $a \neq 1$ (b) $21$ stickers

Question 9

A photo frame is $b$ cm wide and $kb$ cm long, so its perimeter is $P = 2b + 2kb$ cm. (a) Make $b$ the subject. (b) A frame with $k = 1.5$ has a perimeter of $60$ cm. How wide and how long is it?

ShowHide solution

(a) Factorise: $P = b{(2 + 2k)}$ , so $b = \dfrac{P}{2 + 2k}$ , for $k \neq -1$ . Here $k$ is positive, so the bracket is never zero.

(b) $b = \dfrac{60}{2 + 3} = 12$ , and the length is $1.5 \times 12 = 18$ .

Check: $2 \times 12 + 2 \times 18 = 24 + 36 = 60\text{.}$

(a) $b = \dfrac{P}{2 + 2k}$ , for $k \neq -1$ (b) $12$ cm wide and $18$ cm long

Question 10

A drink is made by mixing $s$ ml of syrup with $w$ ml of water. The fraction of the drink that is syrup is $c = \dfrac{s}{s + w}$ . (a) Make $s$ the subject. (b) How much syrup must be mixed with $300$ ml of water so that one quarter of the drink is syrup?

ShowHide solution

(a) Multiply by ${(s + w)}$ : $c{(s + w)} = s$ , so $cs + cw = s$ .

Collect on the right: $cw = s - cs = s{(1 - c)}\text{.}$ So $s = \dfrac{cw}{1 - c}$ , for $c \neq 1$ . The drink has some water in it, so $c$ is less than $1$ and the bracket is not zero.

(b) $s = \dfrac{\frac{1}{4} \times 300}{1 - \frac{1}{4}} = \dfrac{75}{\frac{3}{4}} = 100\text{.}$

Check: $\dfrac{100}{100 + 300} = \dfrac{100}{400} = \dfrac{1}{4}\text{.}$

(a) $s = \dfrac{cw}{1 - c}$ , for $c \neq 1$ (b) $100$ ml

Question 11

Find and correct the error. (a) Lina makes $x$ the subject of $x + a = bx$ and writes $x = bx - a$ . (b) Wei makes $x$ the subject of $ax - b = cx$ and writes $x = \dfrac{b}{a + c}$ .

ShowHide solution

(a) Her $x$ is still on both sides. Collect: $a = bx - x = x{(b - 1)}\text{,}$ so $x = \dfrac{a}{b - 1}$ , for $b \neq 1$ .

Check (a) with $a = 2$ , $b = 3$ : $x = 1$ , and $1 + 2 = 3 = 3 \times 1$ .

(b) Test $a = 5$ , $b = 6$ , $c = 2$ : $5x - 6 = 2x$ gives $x = 2$ , but his formula gives $\dfrac{6}{7}$ . Moving $cx$ to the left makes it $-cx$ : $ax - cx = b$ , so $x{(a - c)} = b$ and $x = \dfrac{b}{a - c}$ , for $a \neq c$ . With the numbers: $\dfrac{6}{3} = 2$ .

(a) $x = \dfrac{a}{b - 1}$ , for $b \neq 1$ (b) $x = \dfrac{b}{a - c}$ , for $a \neq c$

Key Points

If the subject appears in more than one term, it cannot be freed one step at a time: collect its terms and factorise.
First clear any fractions and expand any brackets that contain the subject.
Collect every term with the subject on one side and the rest on the other; the side where those terms are positive saves minus signs.
Take the subject out as a common factor, keeping the $1$ : $ax - x = x{(a - 1)}$ .
Divide by the bracket, and state the values for which it would be zero.
Check by putting numbers into both forms.