- Read the question twice; identify the quantity you are asked for.
- Define a letter for it, with units.
- Translate each sentence into an equation.
- Solve, choosing the appropriate method.
- Check the answer against the original words, and reject impossible solutions.
At this level linear equations rarely appear on their own — they arrive wrapped in a context, with fractions or brackets.
Solve $\dfrac{2x - 1}{3} = \dfrac{x + 4}{5}$.
A taxi charges a fixed £$3.50$ plus £$1.20$ per mile. A journey cost £$15.50$. How long was it?
| Situation | Best method |
|---|---|
| Factorises easily over the integers | Factorise |
| No middle term, e.g. $x^2 = 7$ | Square-root both sides ($\pm$!) |
| Exact answer wanted, or vertex needed | Complete the square |
| Nothing obvious, or ugly numbers | The formula |
$\Delta > 0$: two roots · $\Delta = 0$: one repeated root · $\Delta < 0$: no real roots
Solve $x(x + 3) = 4x + 6$, giving exact answers.
Find the values of $k$ for which $x^2 + kx + 4 = 0$ has exactly one solution.
Cubics are new at this level. The route is always the same: find one root by the factor theorem, divide out, then solve the quadratic that remains.
- Rearrange to $\mathrm{f}(x) = 0$.
- Test factors of the constant term until $\mathrm{f}(a) = 0$.
- Divide $\mathrm{f}(x)$ by $(x-a)$.
- Solve the resulting quadratic.
A cuboid has height $x$ cm, width $(x+1)$ cm and length $(x+4)$ cm. Its volume is $84$ cm³. Find $x$.
Solve $x^3 + 2x^2 - 13x + 10 = 0$.
Solve $3x + 4y = 18$ and $5x - 2y = 4$.
Solve $y = 2x - 1$ and $x^2 + y^2 = 25$. Interpret the answer geometrically.
Define first
Write "let $x$ be…" with units.
Quadratics
Everything to one side, $=0$, then choose a method.
Never divide by $x$
You lose the root $x=0$.
Discriminant
$b^2-4ac$ counts the real roots.
Cubics
Factor theorem, divide, solve the quadratic.
Two linear
Eliminate by scaling and adding.
Linear + quadratic
Substitute the linear one in.
Pair solutions
Each $x$ with its matching $y$.
Tangency
One repeated solution $\Rightarrow$ tangent.
Reject nonsense
Negative lengths, times before zero.
Solve $\dfrac{3x+1}{4} = \dfrac{x-2}{2}$.
▶ Show solution
$2(3x+1) = 4(x-2)$
$6x + 2 = 4x - 8$
$2x = -10$, so $x = -5$.
Solve $2x^2 + 7x - 15 = 0$.
▶ Show solution
$ac = -30$; two numbers multiplying to $-30$ and adding to $7$ are $10$ and $-3$.
$2x^2 + 10x - 3x - 15 = 2x(x+5) - 3(x+5) = (2x-3)(x+5)$
$x = \tfrac32$ or $x = -5$.
Solve $3x^2 - 5x - 1 = 0$, giving answers to 3 significant figures.
▶ Show solution
$\Delta = 25 + 12 = 37$
$x = \dfrac{5 \pm \sqrt{37}}{6}$
$x = \dfrac{5 + 6.0828}{6} = 1.85$ or $x = \dfrac{5 - 6.0828}{6} = -0.180$
Solve $x^3 - 6x^2 + 11x - 6 = 0$.
▶ Show solution
$\mathrm{f}(1) = 1 - 6 + 11 - 6 = 0$ ✓
Dividing by $(x-1)$: $x^2 - 5x + 6 = (x-2)(x-3)$
$x = 1$, $2$ or $3$.
Solve $4x - 3y = 11$ and $2x + y = 8$.
▶ Show solution
From the second, $y = 8 - 2x$.
$4x - 3(8-2x) = 11$
$4x - 24 + 6x = 11$, so $10x = 35$ and $x = 3.5$.
$y = 8 - 7 = 1$.
Solve $y = x + 2$ and $y = x^2 - 4$.
▶ Show solution
$x + 2 = x^2 - 4$
$x^2 - x - 6 = 0$, so $(x-3)(x+2) = 0$.
$x = 3 \Rightarrow y = 5$; $x = -2 \Rightarrow y = 0$.
Solutions: $(3, 5)$ and $(-2, 0)$.
Find the values of $k$ for which $2x^2 + kx + 8 = 0$ has no real solutions.
▶ Show solution
No real solutions means $\Delta < 0$:
$k^2 - 4(2)(8) < 0 \;\Rightarrow\; k^2 < 64$
$-8 < k < 8$
The sum of two numbers is $13$ and the sum of their squares is $97$. Find them.
▶ Show solution
Let them be $x$ and $y$. Then $x + y = 13$ and $x^2 + y^2 = 97$.
From the first, $y = 13 - x$:
$x^2 + (13-x)^2 = 97$
$x^2 + 169 - 26x + x^2 = 97$
$2x^2 - 26x + 72 = 0 \;\Rightarrow\; x^2 - 13x + 36 = 0$
$(x-4)(x-9) = 0$, so the numbers are $4$ and $9$.
Show that the line $y = 2x + 1$ is a tangent to the curve $y = x^2 + 4x + 2$, and find the point of contact.
▶ Show solution
Set them equal: $2x + 1 = x^2 + 4x + 2$
$x^2 + 2x + 1 = 0$
$\Delta = 4 - 4 = 0$, so there is exactly one solution — the line meets the curve once, i.e. it is a tangent.
$(x+1)^2 = 0$ gives $x = -1$, and then $y = 2(-1) + 1 = -1$.
The point of contact is $(-1,\, -1)$.
A cinema sells adult tickets at £$9$ and child tickets at £$5$. On one evening $240$ tickets were sold for a total of £$1720$.
(a) Set up and solve simultaneous equations to find how many of each were sold. (b) The next evening the same total number of tickets raised £$1960$. Explain whether this is possible. (c) The manager says "if we sell $x$ adult tickets out of $240$, income is $I = 1200 + 4x$." Verify this formula and use it to answer (b) directly.
▶ Show solution
(a) Let $a$ = adult tickets and $c$ = child tickets.
$a + c = 240$ and $9a + 5c = 1720$
From the first, $c = 240 - a$:
$9a + 5(240 - a) = 1720$
$9a + 1200 - 5a = 1720 \;\Rightarrow\; 4a = 520 \;\Rightarrow\; a = 130$
So $130$ adult and $110$ child tickets. Check: $9(130) + 5(110) = 1170 + 550 = 1720$ ✓
(b) $9a + 5(240-a) = 1960$ gives $4a = 760$, so $a = 190$.
That is a whole number and $0 \leqslant 190 \leqslant 240$, so yes, it is possible — $190$ adult and $50$ child tickets.
(c) $I = 9x + 5(240 - x) = 9x + 1200 - 5x = 1200 + 4x$ ✓
Setting $1200 + 4x = 1960$ gives $x = 190$ immediately, agreeing with (b).
The formula also shows the income must lie between $I(0) = £1200$ and $I(240) = £2160$, and must be a multiple of $4$ above $1200$ — a quick test for whether any claimed total is achievable.