Question: Make $n$ the subject of $P = 2m + 3n - 5$.

Step 1. Move every term that doesn't contain $n$ to the other side.

$$\begin{aligned} P &= 2m + 3n - 5 \\ P - 2m + 5 &= 3n \end{aligned}$$

You subtract $2m$ and add $5$ to both sides to leave the $n$-term alone.

Step 2. Divide both sides by the coefficient of $n$.

$$\begin{aligned} \frac{P - 2m + 5}{3} &= n \end{aligned}$$

Solution:

$$n = \frac{P - 2m + 5}{3}$$

Check: Substitute back. Let $P = 16,\; m = 3,\; n = ?$

$$n = \frac{16 - 2(3) + 5}{3} = \frac{16 - 6 + 5}{3} = \frac{15}{3} = 5$$

Verify: $2(3) + 3(5) - 5 = 6 + 15 - 5 = 16 = P$ ✔

⚠ Watch out: When transposing, every term moves — don't forget the $-5$ becomes $+5$.

Question: Make $x$ the subject of $y = a + \sqrt{x + 1}$.

Step 1. Isolate the radical:

$$y - a = \sqrt{x + 1}$$

Step 2. Square both sides:

$$(y - a)^2 = x + 1$$

Step 3. Subtract 1:

$$x = (y - a)^2 - 1$$

Note: No $\pm$ here, because the original equation defines $\sqrt{x + 1}$ as the principal (non-negative) square root. Squaring doesn't undo a square — it follows from the definition.

Tip: Always isolate the radical BEFORE squaring. Squaring $y = a + \sqrt{x+1}$ directly gives $y^2 = a^2 + 2a\sqrt{x+1} + (x+1)$, which is worse.

Question: Make $y$ the subject of $3y + 5x = 2(y - 4) + 7x$

Step 1. Expand the bracket on the right:

$$3y + 5x = 2y - 8 + 7x$$

Step 2. Collect all $y$-terms on one side — subtract $2y$ from both sides:

$$3y - 2y + 5x = -8 + 7x$$ $$y + 5x = 7x - 8$$

Step 3. Isolate $y$ — subtract $5x$ from both sides:

$$y = 7x - 5x - 8$$ $$\boxed{y = 2x - 8}$$

Check: Let $x = 3$, so $y = 2(3) - 8 = -2$.

$$\begin{aligned} \text{LHS} &= 3(-2) + 5(3) = -6 + 15 = 9 \\ \text{RHS} &= 2(-2 - 4) + 7(3) = -12 + 21 = 9 \;\checkmark \end{aligned}$$

⚠ Watch out: When $y$ appears on both sides, you must move all $y$-terms to one side before simplifying. Don't try to "cancel" $y$ across the equals sign — collect and combine instead.

Question: Make $r$ the subject of $V = \dfrac{4}{3}\pi r^3$.

Step 1. Multiply both sides by 3:

$$3V = 4\pi r^3$$

Step 2. Divide by $4\pi$:

$$r^3 = \dfrac{3V}{4\pi}$$

Step 3. Take cube root:

$$r = \sqrt[3]{\dfrac{3V}{4\pi}}$$

Note: No $\pm$ on the cube root. The principal cube root of any real number is unique.

Tip: This is the volume of a sphere, rearranged. The same pattern applies to any $V = k \cdot r^n$ formula: isolate $r^n$, then take the $n$-th root.

Question: Given $\dfrac{k}{3} = \dfrac{x}{7z - 6w}$, make $z$ the subject.

Note: $z$ appears once, linearly. Cross-multiply, then isolate $z$.

Step 1. Cross-multiply:

$$k(7z - 6w) = 3x$$

Step 2. Expand:

$$7kz - 6kw = 3x$$

Step 3. Isolate $z$:

$$7kz = 3x + 6kw$$ $$z = \dfrac{3x + 6kw}{7k}$$

Answer: $z = \dfrac{3x + 6kw}{7k}$.

Tip: Because $z$ appears once, no factoring is required after expanding. If you ever see two $z$-terms on the same side, the technique is "subject on both sides" instead.

CHIJ St Nicholas 2022 P2 Q2

(a)(i) Express $x^2 - 6x + 7$ in the form $(x + q)^2 + p$. [2m]

Step 1. Complete the square:

$$\begin{aligned} x^2 - 6x + 7 &= (x^2 - 6x + 9) - 9 + 7 \\ &= (x - 3)^2 - 2 \end{aligned}$$

Comparing with $(x + q)^2 + p$, you get $q = -3$ and $p = -2$.

Solution: $(x - 3)^2 - 2$, so $q = -3,\; p = -2$.

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(a)(ii) Explain why the minimum value of the graph of $y = x^2 - 6x + 7$ is $p$. [1m]

Since $(x - 3)^2 \geq 0$ for all real $x$, the smallest value of $(x-3)^2$ is $0$ (when $x = 3$).

Therefore the minimum value of $y = (x-3)^2 + p$ is $0 + p = p = -2$.

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(b) Given that $2r = \sqrt{\dfrac{p + r^3}{r}}$

(b)(i) Find the value of $p$ when $r = 2$. [2m]

Step 1. Square both sides to remove the square root:

$$4r^2 = \frac{p + r^3}{r}$$

Step 2. Multiply both sides by $r$:

$$4r^3 = p + r^3$$

Step 3. Substitute $r = 2$:

$$\begin{aligned} 4(2)^3 &= p + (2)^3 \\ 32 &= p + 8 \\ p &= 24 \end{aligned}$$

Solution: $p = 24$

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(b)(ii) Express $r$ in terms of $p$. [2m]

Step 1. Start from the squared form obtained above:

$$4r^3 = p + r^3$$

Step 2. Collect all $r$ terms on one side:

$$4r^3 - r^3 = p$$ $$3r^3 = p$$

Step 3. Divide by 3, then cube-root:

$$r^3 = \frac{p}{3}$$

Solution: $r = \sqrt[3]{\dfrac{p}{3}}$

Question: Given $\dfrac{k - x}{k + x} = \dfrac{m}{n}$, make $x$ the subject.

Step 1. Cross-multiply:

$$n(k - x) = m(k + x)$$

Step 2. Expand both sides:

$$nk - nx = mk + mx$$

Step 3. Group $x$-terms on one side, everything else on the other:

$$-nx - mx = mk - nk$$ $$-(n + m)x = k(m - n)$$

Step 4. Factor out $x$ and divide:

$$x = \dfrac{k(m - n)}{-(n + m)} = \dfrac{k(n - m)}{n + m}$$

Answer: $x = \dfrac{k(n - m)}{n + m}$.

 

Tip: Multiplying numerator and denominator by $-1$ at the end is just a tidying step to avoid the leading negative. Either form is mathematically correct.

Question: Given $\dfrac{1}{m} = \dfrac{1}{n} + \dfrac{1}{q}$, make $q$ the subject.

Note: $q$ appears inside a reciprocal fraction. Combine fractions first, then cross-multiply.

Step 1. Combine the right-hand side over a common denominator $nq$:

$$\dfrac{1}{m} = \dfrac{q + n}{nq}$$

Step 2. Cross-multiply:

$$nq = m(q + n)$$ $$nq = mq + mn$$

Step 3. Group $q$-terms:

$$nq - mq = mn$$

Step 4. Factor out $q$ and divide:

$$q(n - m) = mn$$ $$q = \dfrac{mn}{n - m}$$

Tip: When you have $\dfrac{1}{\text{subject}}$ appearing, combining all reciprocal fractions into a single fraction is almost always the first move. Then cross-multiply to get back to a polynomial equation.

The classic mistake: trying to divide while the variable still appears in multiple places.

Wrong ❌:

$$nk - nx = mk + mx$$ $$- nx = mk + mx - nk$$ $$x = \dfrac{mk + mx - nk}{-n}$$ $$x \text{ appears in both left and right side, which is incorrect}$$

Right ✅:

$$-nx - mx = mk - nk \quad \text{(group } x \text{-terms)}$$ $$x(-n - m) = mk - nk \quad \text{(factorize)}$$ $$x = \dfrac{mk - nk}{-(n + m)} = \dfrac{nk - mk}{n + m}$$

Rule: Never divide until the target variable appears in exactly one bracketed factor. Always factorize first.

Question: Given $wy^2 = \sqrt{\dfrac{1 - rt}{t}}$, make $t$ the subject.

Note: $t$ is inside the radical and also in the denominator. Square first, then clear the denominator.

Step 1. Square both sides to remove the radical:

$$(wy^2)^2 = \dfrac{1 - rt}{t}$$ $$w^2 y^4 = \dfrac{1 - rt}{t}$$

Step 2. Multiply both sides by $t$:

$$w^2 y^4 \cdot t = 1 - rt$$

Step 3. Group $t$-terms:

$$w^2 y^4 \cdot t + rt = 1$$ $$t(w^2 y^4 + r) = 1$$

Step 4. Divide:

$$t = \dfrac{1}{w^2 y^4 + r}$$

Tip: Always isolate the radical BEFORE squaring. Here the radical was already alone on one side. If it weren't, you'd transpose the non-radical terms first.

Question: Given $y = \dfrac{k - x^2}{\pi x^2 + 1}$, make $x$ the subject.

Note: $x^2$ appears in both numerator and denominator.

Step 1. Cross-multiply:

$$y(\pi x^2 + 1) = k - x^2$$

Step 2. Expand:

$$y\pi x^2 + y = k - x^2$$

Step 3. Group $x^2$-terms on one side:

$$y\pi x^2 + x^2 = k - y$$

Step 4. Factor out $x^2$:

$$x^2(y\pi + 1) = k - y$$

Step 5. Divide, then take $\pm\sqrt{\;}$:

$$x^2 = \dfrac{k - y}{y\pi + 1}$$ $$x = \pm\sqrt{\dfrac{k - y}{y\pi + 1}}$$

Tip: The $\pm$ is mandatory because both positive and negative values of $x$ would give the same $x^2$. Save the square root for the very last step.

Question: Make $x$ the subject of $y = \sqrt[3]{\dfrac{5x + 4k^2}{6 + x}}$.

Step 1. Cube both sides:

$$y^3 = \dfrac{5x + 4k^2}{6 + x}$$

Step 2. Cross multiply:

$$y^3(6 + x) = 5x + 4k^2$$ $$6y^3 + xy^3 = 5x + 4k^2$$

Step 3. Group $x$-terms:

$$xy^3 - 5x = 4k^2 - 6y^3$$

Step 4. Factorize:

$$x(y^3 - 5) = 4k^2 - 6y^3$$

Step 5. Divide:

$$x = \dfrac{4k^2 - 6y^3}{y^3 - 5}$$

Tip: Cube both sides first to remove the radical, THEN treat $y^3$ as a constant during the rearrangement. The cube root has no $\pm$ but you don't even need to take it again — $y^3$ is already an expression in $y$.

The ± rule: When you square-root both sides to remove a squared term, you MUST include $\pm$

$$x^2 = k \implies x = \pm\sqrt{k}$$

Writing only $x = \sqrt{k}$ loses the negative root and costs a mark.

Exception — DON'T add $\pm$:

• Cube roots: $x^3 = k \implies x = \sqrt[3]{k}$ — single-valued, no $\pm$.
• Squaring to undo a square root: $\sqrt{A} = B \implies A = B^2$ — no $\pm$ here because squaring is the operation, not the inverse.

Mnemonic: "$\pm$ when going FROM $x^2$ TO $x$." Going from $x$ to $x^2$ doesn't need $\pm$.

Question: Given $V = \dfrac{4}{3}\pi r^3$, make $r$ the subject.

Note: Volume of a sphere. The subject is $r^3$; we cube-root at the end (no $\pm$).

Step 1. Multiply both sides by $3$ to remove the fraction:

$$3V = 4\pi r^3$$

Step 2. Divide both sides by $4\pi$:

$$\dfrac{3V}{4\pi} = r^3$$

Step 3. Take cube root of both sides:

$$r = \sqrt[3]{\dfrac{3V}{4\pi}}$$

Tip: No $\pm$ on a cube root. The principal cube root of any real number is unique — unlike the square root which has two values. Students who write $r = \pm\sqrt[3]{\ldots}$ lose a mark.

Question: Given $3a - 2b = 3b + 5a$, find $\dfrac{b}{a}$.

 

Step 1. Collect like terms to link the two variables:

$$\begin{aligned} 3a - 5a &= 3b + 2b \\ -2a &= 5b \end{aligned}$$

Step 2. Divide both sides by $a$ (and by $5$) to match the target expression:

$$\dfrac{b}{a} = -\dfrac{2}{5}$$

Key takeaway: You don't need the individual values of $a$ and $b$ — just the relationship between them. Whenever the question asks for a ratio or compound expression, one rearrangement step is usually enough. Read the target expression first so you know exactly what form to aim for.

Watch for hidden perfect squares when rearranging:

$$a^2 + 2ab + b^2 = (a + b)^2$$ $$a^2 - 2ab + b^2 = (a - b)^2$$ $$4p^2 - 12pq + 9q^2 = (2p - 3q)^2$$

Why it matters: If you spot $(a \pm b)^2 = c^2$, skip expansion and go straight to:

$$a \pm b = \pm c$$

Example:

$$x^2 - y^2 = 2x - 1$$ $$x^2 - 2x + 1 = y^2$$ $$(x - 1)^2 = y^2$$ $$x = 1 \pm y$$

Spotting the perfect square saves several lines of algebra.