Corrigé

Algebra

Pack A — Réponses

#	Question	Réponse
1	Simplify $5x + 3x - 2x$.	$6x$
2	Simplify $5a + 3b + 2a - b$.	$7a + 2b$
3	Expand $3(x + 4)$.	$3x + 12$
4	Expand $-2(x - 5)$.	$-2x + 10$
5	Factorise $6x + 9$.	$3(2x + 3)$
6	Factorise $4x^2 + 6x$.	$2x(2x + 3)$
7	Simplify $(4x)^2$.	$16 x^2$
8	Simplify $\sqrt{12}$.	$2\sqrt{3}$
9	Find the value of $3x + 2y$ when $x = 4$ and $y = 5$.	22
10	Solve $3x + 5 = 17$.	$x = 4$
11	Expand $(x + 3)(x + 5)$.	$x^2 + 8x + 15$
12	Expand $(x - 4)(x + 3)$.	$x^2 - x - 12$
13	Expand $(x + 5)^2$.	$x^2 + 10x + 25$
14	Expand $(x + 6)(x - 6)$.	$x^2 - 36$
15	Factorise $x^2 + 7x + 12$.	$(x + 3)(x + 4)$
16	Factorise $x^2 + 3x - 10$.	$(x + 5)(x - 2)$
17	Simplify $\sqrt{6} \times \sqrt{24}$.	12
18	Simplify $\sqrt{8} + \sqrt{18}$.	$5\sqrt{2}$
19	Solve $3(x - 4) = 9$.	$x = 7$
20	When $a = 3$ and $b = -2$, find the value of $a^2 - 3ab + 2b$.	23
21	Factorise $x^2 - 49$.	$(x + 7)(x - 7)$
22	Factorise $2x^2 + 7x + 3$.	$(2x + 1)(x + 3)$
23	Simplify $(3 + \sqrt{5})(3 - \sqrt{5})$.	4
24	Rationalise the denominator of $\dfrac{6}{\sqrt{3}}$.	$2\sqrt{3}$
25	Solve the system: $\begin{cases} y = 2x + 1 \\ y = -x + 7 \end{cases}$	$x = 2$, $y = 5$
26	Solve the system: $\begin{cases} 2x + 3y = 15 \\ 2x + 1y = 9 \end{cases}$	$x = 3$, $y = 3$
27	Factorise fully $2x^2 - 18$.	$2(x+3)(x-3)$
28	Solve $\dfrac{x + 3}{4} = 5$.	$x = 17$
29	Expand and simplify $(x + 2)(x + 5) - (x + 3)^2$.	$x + 1$
30	The sum of two numbers is $30$ and their difference is $6$. Find both numbers.	18 and 12
31	Solve the system: $\begin{cases} 3x + 2y = 12 \\ 2x - 5y = -11 \end{cases}$	$x = 2$, $y = 3$
32	Factorise fully $6x^2 - 5x - 4$.	$(3x - 4)(2x + 1)$
33	Rationalise the denominator of $\dfrac{2}{3 + \sqrt{5}}$.	$\dfrac{2(3 - \sqrt{5})}{4} = \dfrac{3 - \sqrt{5}}{2}$
34	If $x + \dfrac{1}{x} = 3$ (with $x > 0$), find $x^2 + \dfrac{1}{x^2}$.	7
35	A cinema sells adult tickets at £$12$ and child tickets at £$8$. One evening it sold $80$ tickets for £$832$. How many adult and child tickets were sold?	48 adults, 32 children
36	Solve $x^2 + 7x + 12 = 0$.	$x = -3$ or $x = -4$
37	Simplify $\dfrac{x^2 - 25}{x + 5}$.	$x - 5$
38	Solve simultaneously: $y = x^2$ and $y = 3x + 4$.	$x = 4, y = 16$ or $x = -1, y = 1$
39	Simplify $(\sqrt{3} + \sqrt{12})^2$.	27
40	Find $x$ such that $x^2 = 50$, giving exact answers in surd form.	$x = \pm 5\sqrt{2}$

Pack B — Réponses

#	Question	Réponse
1	Simplify $7y + 2y - 4y$.	$5y$
2	Simplify $5a + 3b + 2a - b$.	$4p + 7q$
3	Expand $5(x + 2)$.	$5x + 10$
4	Expand $-4(x - 3)$.	$-4x + 12$
5	Factorise $8x + 12$.	$4(2x + 3)$
6	Factorise $9x^2 + 12x$.	$3x(3x + 4)$
7	Simplify $(7x)^2$.	$49 x^2$
8	Simplify $\sqrt{18}$.	$3\sqrt{2}$
9	Find the value of $3x + 2y$ when $x = 6$ and $y = 3$.	24
10	Solve $4x + 2 = 18$.	$x = 4$
11	Expand $(x + 4)(x + 7)$.	$x^2 + 11x + 28$
12	Expand $(x - 5)(x + 2)$.	$x^2 - 3x - 10$
13	Expand $(x + 6)^2$.	$x^2 + 12x + 36$
14	Expand $(x + 8)(x - 8)$.	$x^2 - 64$
15	Factorise $x^2 + 9x + 20$.	$(x + 4)(x + 5)$
16	Factorise $x^2 + 2x - 15$.	$(x + 5)(x - 3)$
17	Simplify $\sqrt{5} \times \sqrt{20}$.	10
18	Simplify $\sqrt{12} + \sqrt{27}$.	$5\sqrt{3}$
19	Solve $5(x - 2) = 25$.	$x = 7$
20	When $a = 3$ and $b = -2$, find the value of $a^2 - 3ab + 2b$.	21
21	Factorise $x^2 - 100$.	$(x + 10)(x - 10)$
22	Factorise $3x^2 + 8x + 4$.	$(3x + 2)(x + 2)$
23	Simplify $(4 + \sqrt{3})(4 - \sqrt{3})$.	13
24	Rationalise the denominator of $\dfrac{10}{\sqrt{5}}$.	$2\sqrt{5}$
25	Solve the system: $\begin{cases} y = 3x + 2 \\ y = -x + 14 \end{cases}$	$x = 3$, $y = 11$
26	Solve the system: $\begin{cases} 3x + 4y = 26 \\ 3x + 2y = 16 \end{cases}$	$x = 2$, $y = 5$
27	Factorise fully $2x^2 - 50$.	$2(x+5)(x-5)$
28	Solve $\dfrac{x + 2}{3} = 7$.	$x = 19$
29	Expand and simplify $(x + 1)(x + 6) - (x + 2)^2$.	$3x + 2$
30	The sum of two numbers is $50$ and their difference is $10$. Find both numbers.	30 and 20
31	Solve the system: $\begin{cases} 2x + 3y = 12 \\ 5x - 2y = 11 \end{cases}$	$x = 3$, $y = 2$
32	Factorise fully $4x^2 - 4x - 3$.	$(2x - 3)(2x + 1)$
33	Rationalise the denominator of $\dfrac{4}{5 + \sqrt{3}}$.	$\dfrac{4(5 - \sqrt{3})}{22} = \dfrac{2(5 - \sqrt{3})}{11}$
34	If $x + \dfrac{1}{x} = 4$ (with $x > 0$), find $x^2 + \dfrac{1}{x^2}$.	14
35	A cinema sells adult tickets at £$15$ and child tickets at £$10$. One evening it sold $60$ tickets for £$750$. How many adult and child tickets were sold?	30 adults, 30 children
36	Solve $x^2 + 8x + 15 = 0$.	$x = -3$ or $x = -5$
37	Simplify $\dfrac{x^2 - 36}{x + 6}$.	$x - 6$
38	Solve simultaneously: $y = x^2$ and $y = 5x + 6$.	$x = 6, y = 36$ or $x = -1, y = 1$
39	Simplify $(\sqrt{2} + \sqrt{8})^2$.	18
40	Find $x$ such that $x^2 = 72$, giving exact answers in surd form.	$x = \pm 6\sqrt{2}$

Problèmes — Solutions détaillées

**Rectangle dimensions.** A rectangle has length $(x + 3)$ cm and width $(x + 1)$ cm. Its area is $35$ cm². (a) Form an equation in $x$ and expand the brackets. (b) Solve the equation to find $x$. (c) State the dimensions of the rectangle.

Réponse

(a) $x^2 + 4x + 3 = 35$ → $x^2 + 4x - 32 = 0$ (b) $x = 4$ (reject $x = -8$) (c) 7 cm × 5 cm

(a) Area: $(x+3)(x+1) = x^2 + 4x + 3 = 35$. Rearrange: $x^2 + 4x - 32 = 0$. (b) Factorise: need two numbers multiplying to $-32$ and summing to $+4$: $8$ and $-4$. So $(x+8)(x-4) = 0$, giving $x = -8$ or $x = 4$. Reject $x = -8$ (lengths can't be negative). So $x = 4$. (c) Length = $4 + 3 = 7$ cm; Width = $4 + 1 = 5$ cm. Check: $7 \times 5 = 35$ ✓.

**Coins.** Aoife has a mix of 50p and 20p coins. She has 15 coins in total, with a total value of £4.80. How many of each coin does she have?

Réponse

6 fifty-pence and 9 twenty-pence coins

Let $f$ = number of 50p coins, $t$ = number of 20p coins. $$f + t = 15$$ $$0.50 f + 0.20 t = 4.80$$ Multiply second eq by 10: $5f + 2t = 48$. From first: $t = 15 - f$. Substitute: $5f + 2(15 - f) = 48$, so $5f + 30 - 2f = 48$, $3f = 18$, $f = 6$. Then $t = 9$. Check: 6 × £0.50 + 9 × £0.20 = £3.00 + £1.80 = £4.80 ✓.

**Surds in geometry.** A right-angled triangle has legs of length $\sqrt{3}$ cm and $\sqrt{12}$ cm. (a) Find the hypotenuse, giving the exact answer in simplest surd form. (b) Find the area of the triangle. (c) Find the perimeter, giving an exact answer in simplest surd form.

Réponse

(a) $\sqrt{15}$ cm (b) 3 cm² (c) $3\sqrt{3} + \sqrt{15}$ cm

(a) Hypotenuse: $h = \sqrt{(\sqrt{3})^2 + (\sqrt{12})^2} = \sqrt{3 + 12} = \sqrt{15}$ cm. (b) Area = $\frac{1}{2}(\sqrt{3})(\sqrt{12}) = \frac{1}{2}\sqrt{36} = \frac{1}{2}(6) = 3$ cm². (c) Perimeter = $\sqrt{3} + \sqrt{12} + \sqrt{15}$. Simplify $\sqrt{12} = 2\sqrt{3}$. So perimeter $= \sqrt{3} + 2\sqrt{3} + \sqrt{15} = 3\sqrt{3} + \sqrt{15}$ cm.

**Algebraic identities.** (a) Show that $(x+y)^2 - (x-y)^2 = 4xy$. (b) Hence, without a calculator, find the value of $103^2 - 97^2$. (c) Generalise: $a^2 - b^2 = (a+b)(a-b)$. Use this to find $215^2 - 185^2$.

Réponse

(a) See working (b) 1200 (c) 12 000

(a) Expand: $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)^2 = x^2 - 2xy + y^2$. Subtract: $(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 4xy$ ✓. (b) Treat 103 and 97 as $100 + 3$ and $100 - 3$. Using (a): $(100+3)^2 - (100-3)^2 = 4(100)(3) = 1200$. (c) $215^2 - 185^2 = (215+185)(215-185) = 400 \times 30 = 12{,}000$.

**Consecutive integers.** Three consecutive integers have a sum of 102. (a) Set up an equation using $n$ as the smallest integer. (b) Find the three integers. (c) Show that for any three consecutive integers, their sum is divisible by 3.

Réponse

(a) $n + (n+1) + (n+2) = 102$ (b) 33, 34, 35 (c) See working

(a) Let smallest be $n$. Then $n + (n+1) + (n+2) = 102$. (b) Simplify: $3n + 3 = 102$, so $3n = 99$, $n = 33$. Integers: 33, 34, 35. Check: $33 + 34 + 35 = 102$ ✓. (c) Sum of any three consecutive integers $n, n+1, n+2$ is $3n + 3 = 3(n+1)$. This is $3 \times$ (an integer), so it is always divisible by 3.

**Mixed factorising.** Factorise each fully. (a) $4x^2 - 25$ (b) $x^3 - 9x$ (c) $2x^2 + 8x + 6$ (d) $x^2 + 4x + 4 - y^2$ *(tricky)*

Réponse

(a) $(2x+5)(2x-5)$ (b) $x(x+3)(x-3)$ (c) $2(x+1)(x+3)$ (d) $(x+2+y)(x+2-y)$

(a) Difference of squares: $4x^2 - 25 = (2x)^2 - 5^2 = (2x+5)(2x-5)$. (b) Common factor $x$: $x(x^2 - 9) = x(x+3)(x-3)$. (c) HCF 2: $2(x^2 + 4x + 3) = 2(x+1)(x+3)$. (d) Recognise $x^2 + 4x + 4 = (x+2)^2$. So expression is $(x+2)^2 - y^2$, a difference of squares: $((x+2)+y)((x+2)-y) = (x+2+y)(x+2-y)$.

**Two unknown coefficients.** A quadratic $x^2 + bx + c$ has roots 2 and 7. (a) Use the factorised form to find $b$ and $c$. (b) Verify by substituting $x = 2$ into $x^2 + bx + c$.

Réponse

(a) $b = -9$, $c = 14$ (b) See working

(a) If roots are 2 and 7, factorised form is $(x-2)(x-7) = x^2 - 9x + 14$. So $b = -9$, $c = 14$. (b) Check $x = 2$: $2^2 + (-9)(2) + 14 = 4 - 18 + 14 = 0$ ✓.

**Rationalising and simplifying.** (a) Simplify $\sqrt{50} + \sqrt{18} - \sqrt{8}$. (b) Rationalise $\dfrac{4}{\sqrt{2} + 1}$. (c) Hence calculate $(\sqrt{2} + 1)\left(\dfrac{4}{\sqrt{2}+1}\right)$ — verify your answer makes sense.

Réponse

(a) $6\sqrt{2}$ (b) $4(\sqrt{2} - 1) = 4\sqrt{2} - 4$ (c) 4 ✓

(a) $\sqrt{50} = 5\sqrt{2}$; $\sqrt{18} = 3\sqrt{2}$; $\sqrt{8} = 2\sqrt{2}$. Sum: $5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2} = 6\sqrt{2}$. (b) Multiply top and bottom by conjugate: $\frac{4}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{4(\sqrt{2}-1)}{2 - 1} = 4(\sqrt{2} - 1)$. (c) Product: $(\sqrt{2}+1) \cdot 4(\sqrt{2}-1) = 4((\sqrt{2})^2 - 1^2) = 4(2 - 1) = 4$. This confirms the rationalising step: the original was indeed $\frac{4}{\sqrt{2}+1}$ rewritten without surd in denominator.

**Triangle perimeter system.** Two sides of an isosceles triangle have length $x + 3$ and the third (the base) has length $2x - 1$. The perimeter is $20$ cm. (a) Form an equation in $x$. (b) Find $x$ and state the side lengths. (c) Could this triangle exist if the perimeter were instead 5 cm? Explain.

Réponse

(a) $4x + 5 = 20$ (b) $x = 3.75$; sides $6.75, 6.75, 6.5$ cm (c) No — sides would be negative or violate triangle inequality

(a) Perimeter: $(x+3) + (x+3) + (2x-1) = 4x + 5 = 20$. (b) $4x = 15$, $x = 3.75$. Sides: $x + 3 = 6.75$ cm (twice), $2x - 1 = 6.5$ cm. Total $= 6.75+6.75+6.5 = 20$ ✓. (c) If perimeter $= 5$: $4x + 5 = 5$, so $x = 0$. Sides: 3, 3, $-1$ cm. A negative side length is impossible. Also even with $x$ slightly positive ($x = 0.5$), sides $3.5, 3.5, 0$ — degenerate. The triangle does **not** exist.

**Solving a quadratic.** Solve $x^2 - 5x = 6$.

Réponse

$x = 6$ or $x = -1$

Rearrange to standard form: $x^2 - 5x - 6 = 0$. Factorise: two numbers $\times (-6)$ and sum $-5$: $-6$ and $1$. So $(x-6)(x+1) = 0$. Therefore $x = 6$ or $x = -1$. Check: $36 - 30 = 6$ ✓; $1 + 5 = 6$ ✓.

**Algebraic system in context.** A youth-club hires a hall. The cost is a fixed fee plus an hourly rate. - 3 hours costs £45. - 5 hours costs £67. (a) Set up two equations using $f$ (fixed fee) and $r$ (hourly rate). (b) Solve to find $f$ and $r$. (c) Predict the cost of an 8-hour booking.

Réponse

(a) $f + 3r = 45$, $f + 5r = 67$ (b) $f = 12$, $r = 11$ (c) £100

(a) Cost = fixed + (rate × hours). So $f + 3r = 45$ and $f + 5r = 67$. (b) Subtract: $2r = 22$, so $r = 11$. Then $f + 33 = 45$, so $f = 12$. (c) 8 hours: $12 + 8 \times 11 = 12 + 88 = \mathbf{\pounds 100}$.

**Surds and Pythagoras.** A square has diagonal length 8 cm. (a) Find the exact side length of the square, simplifying any surds. (b) Find the exact area of the square. (c) Find the perimeter, giving an exact answer.

Réponse

(a) $4\sqrt{2}$ cm (b) 32 cm² (c) $16\sqrt{2}$ cm

(a) Let side be $s$. Diagonal of square: $d = s\sqrt{2}$. So $s\sqrt{2} = 8$, giving $s = \frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$ cm. (b) Area = $s^2 = (4\sqrt{2})^2 = 16 \times 2 = 32$ cm². (c) Perimeter = $4s = 4 \times 4\sqrt{2} = 16\sqrt{2}$ cm.