Algebra, equations & inequalities — Mathematics Grade 12

3 exam-style question sets on Algebra, equations & inequalities (Paper 1), each with a hint and a fully worked answer. The app holds 15 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Equations, inequalities & integer exponentsEquations & inequalities

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Question 1

Equations, inequalities & integer exponentsRoutine24 marksPaper 1

Answer ALL the parts below. Show ALL your working, and where you reject a value, say why.

aSolve for $x$ : $x^{2} - 5 x - 24 = 0$ (3)
bSolve for $x$ , correct to TWO decimal places: $3 x^{2} - 5 x = 4$ (4)
cSolve for $x$ : $2 x + 7 = x - 4$ (4)
dSolve for $x$ : $x^{2} - 3 x - 18 \leq 0$ (4)
eSolve simultaneously for $x$ and $y$ : $x - y = 4$ and $\frac{3}{x} + \frac{1}{y} = 1$ (5)
fThe integers $x$ and $y$ satisfy $2^{x + 2} + 2^{x + 1} + 2^{x} = 7^{y + 1} + 7^{y}$ . Determine the value of $x^{2} + y^{2}$ . (4)

Hint

Factorise wherever the numbers allow it (parts a, d and e); part (b) needs the quadratic formula after rearranging to standard form; in part (c) square both sides and then TEST both candidates in the original equation; in part (f) take out the smallest power on each side and compare the powers of each prime.

Worked answer

a) $x^{2} - 5 x - 24 = 0$
$(x - 8) (x + 3) = 0$ ✓
$x = 8$ ✓ or $x = - 3$ ✓

b) Standard form: $3 x^{2} - 5 x - 4 = 0$ ✓
$x = \frac{5 \pm ( - 5 ) ^{2} - 4 ( 3 ) ( - 4 )}{2 ( 3 )} = \frac{5 \pm 73}{6}$ ✓
$x = 2, 26$ ✓ or $x = - 0, 59$ ✓ (TWO decimal places)

c) $2 x + 7 = x - 4$ , so we need $x \geq 4$ . Square both sides: $2 x + 7 = x^{2} - 8 x + 16$ ✓
$x^{2} - 10 x + 9 = 0$ , so $(x - 9) (x - 1) = 0$ ✓ candidates $x = 9$ or $x = 1$ ✓
Test $x = 1$ : $9 = 3$ but $1 - 4 = - 3 \neq = 3$ , so reject $x = 1$ (introduced by squaring). Test $x = 9$ : $25 = 5 = 9 - 4$ , valid.
$∴ x = 9$ ✓

d) $x^{2} - 3 x - 18 \leq 0$
$(x - 6) (x + 3) \leq 0$ ✓ Critical values: $x = - 3$ and $x = 6$ ✓
The parabola opens upwards, so the expression is negative or zero between its roots ✓
$∴ - 3 \leq x \leq 6$ ✓

e) From $x - y = 4$ : $y = x - 4$ ✓
Substitute into $\frac{3}{x} + \frac{1}{y} = 1$ : $\frac{3}{x} + \frac{1}{x - 4} = 1$ . Multiply by $x (x - 4)$ : $3 (x - 4) + x = x (x - 4)$ ✓
$4 x - 12 = x^{2} - 4 x$ , so $x^{2} - 8 x + 12 = 0$ and $(x - 2) (x - 6) = 0$ ✓
$x = 2$ or $x = 6$ ✓ then $y = - 2$ or $y = 2$ ✓
Solutions: $(2; - 2)$ and $(6; 2)$

f) LHS: $2^{x + 2} + 2^{x + 1} + 2^{x} = 2^{x} (4 + 2 + 1) = 7 \cdot 2^{x}$ ✓
RHS: $7^{y + 1} + 7^{y} = 7^{y} (7 + 1) = 8 \cdot 7^{y} = 2^{3} \cdot 7^{y}$ ✓
So $7^{1} \cdot 2^{x} = 2^{3} \cdot 7^{y}$ . Matching the powers of each prime: $x = 3$ and $y = 1$ ✓
$∴ x^{2} + y^{2} = 9 + 1 = 10$ ✓

Modelled on Nov 2023 Paper 1, Question 1 — same skills, new scenario.

Question 2

Equations & inequalitiesComplex27 marksPaper 1

Answer ALL parts of this question. Show ALL your working.

aSolve for $x$ : $x (x + 7) = 0$ (2)
bSolve for $x$ , correct to TWO decimal places: $2 x^{2} = 6 x - 3$ (4)
cSolve for $x$ : $x^{2} + 3 x - 10 < 0$ (4)
dSolve for $x$ : $4^{x} - 6 \cdot 2^{x} - 16 = 0$ (5)
eSolve for $x$ : $x + 5 = x - 1$ (4)
fSolve simultaneously for $x$ and $y$ : $x - y = 3$ and $y^{2} + 2 x y = 45$ (5)
gThe product $P = (1 - \frac{1}{2 ^{2}}) (1 - \frac{1}{3 ^{2}}) (1 - \frac{1}{4 ^{2}}) \dots (1 - \frac{1}{n ^{2}})$ is formed for a natural number $n \geq 2$ . Determine the value of $n$ for which $P = \frac{51}{100}$ . (3)

Hint

Factorise first wherever you can; test BOTH candidates of the surd equation in the original equation, and in part (g) write each bracket as a single fraction and look for cancelling.

Worked answer

a) $x (x + 7) = 0$ , so $x = 0$ ✓ or $x = - 7$ ✓

b) Standard form: $2 x^{2} - 6 x + 3 = 0$ ✓
$x = \frac{6 \pm 36 - 4 ( 2 ) ( 3 )}{2 ( 2 )} = \frac{6 \pm 12}{4}$ ✓
$x = 2, 37$ ✓ or $x = 0, 63$ ✓ (TWO decimal places)

c) $(x + 5) (x - 2) < 0$ ✓ Critical values: $x = - 5$ and $x = 2$ ✓
The parabola opens upwards, so the expression is negative strictly between its roots ✓
$∴ - 5 < x < 2$ ✓

d) Let $k = 2^{x}$ , then $4^{x} = k^{2}$ : $k^{2} - 6 k - 16 = 0$ ✓
$(k - 8) (k + 2) = 0$ ✓ so $k = 8$ ✓ ( $k = - 2$ rejected, since $2^{x} > 0$ for all real $x$ ) ✓
$2^{x} = 8 = 2^{3}$ , so $x = 3$ ✓

e) Square both sides: $x + 5 = (x - 1)^{2} = x^{2} - 2 x + 1$ ✓
$x^{2} - 3 x - 4 = 0$ , so $(x - 4) (x + 1) = 0$ ✓ candidates $x = 4$ or $x = - 1$ .
Test $x = 4$ : $9 = 3$ and $4 - 1 = 3$ ✓ valid. Test $x = - 1$ : $4 = 2$ but $- 1 - 1 = - 2 \neq = 2$ , so reject $x = - 1$ (introduced by squaring).
$∴ x = 4$ ✓

f) From the linear equation: $x = y + 3$ ✓ Substitute: $y^{2} + 2 y (y + 3) = 45$ ✓
$3 y^{2} + 6 y - 45 = 0$ , so $y^{2} + 2 y - 15 = 0$ and $(y + 5) (y - 3) = 0$ ✓
$y = - 5$ or $y = 3$ ✓ then $x = - 2$ or $x = 6$ ✓ Solutions: $(- 2; - 5)$ and $(6; 3)$

g) Each factor: $1 - \frac{1}{k ^{2}} = \frac{( k - 1 ) ( k + 1 )}{k ^{2}}$ , so
$P = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2 \cdot 4}{3 \cdot 3} \cdot \frac{3 \cdot 5}{4 \cdot 4} \dots \frac{( n - 1 ) ( n + 1 )}{n \cdot n}$ ✓ which telescopes to $P = \frac{n + 1}{2 n}$ ✓
$\frac{n + 1}{2 n} = \frac{51}{100}$ gives $100 n + 100 = 102 n$ , so $n = 50$ ✓

Modelled on Nov 2024 Paper 1, Question 1 — same skills, new scenario.

Question 3

Equations & inequalitiesComplex25 marksPaper 1

Show ALL your working in each part of this question.

aSolve for $x$ : $(x + 4) (x - 9) = 0$ (2)
bSolve for $x$ , correct to TWO decimal places: $3 x^{2} + 2 = 8 x$ (4)
cSolve for $x$ : $3 x^{2} > 15 x$ (4)
dSolve for $x$ : $2 \cdot 4^{2 x} - 33 \cdot 4^{x} + 16 = 0$ (4)
eSolve for $x$ : $6 - \frac{1}{x} = \frac{1}{x} - 4$ (5)
fThe length $ℓ$ of a rectangular mural is $3$ m more than its width $w$ (both measured in metres). TWICE the area of the mural is equal to the square of its width, plus $135$ . Set up TWO equations and determine the dimensions of the mural. (6)

Hint

Take all terms to one side before factorising; in (d) substitute $y = 4^{x}$ and in (e) substitute $u = \frac{1}{x}$ , then test BOTH candidates in the original equation.

Worked answer

a) $(x + 4) (x - 9) = 0$ , so $x = - 4$ ✓ or $x = 9$ ✓

b) Standard form: $3 x^{2} - 8 x + 2 = 0$ ✓
$x = \frac{8 \pm 64 - 4 ( 3 ) ( 2 )}{2 ( 3 )} = \frac{8 \pm 40}{6}$ ✓
$x = 2, 39$ ✓ or $x = 0, 28$ ✓ (TWO decimal places)

c) $3 x^{2} - 15 x > 0$ ✓ so $3 x (x - 5) > 0$ ✓
Critical values: $x = 0$ and $x = 5$ ; the parabola opens upwards, so the product is positive on the outside of the roots ✓
$∴ x < 0$ or $x > 5$ ✓

d) Let $y = 4^{x}$ , then $4^{2 x} = y^{2}$ : $2 y^{2} - 33 y + 16 = 0$ ✓
$(2 y - 1) (y - 16) = 0$ ✓ so $y = 16$ or $y = \frac{1}{2}$ — BOTH valid, since $4^{x} > 0$ for every real $x$ .
$4^{x} = 16 = 4^{2}$ gives $x = 2$ ✓ and $4^{x} = \frac{1}{2} = 4^{- \frac{1}{2}}$ gives $x = - \frac{1}{2}$ ✓

e) Let $u = \frac{1}{x}$ (note $u > 0$ ): $6 - u = u - 4$ ✓
Square both sides: $6 - u = u^{2} - 8 u + 16$ ✓ so $u^{2} - 7 u + 10 = 0$ and $(u - 2) (u - 5) = 0$ ✓
$u = 5$ gives $x = \frac{1}{u ^{2}} = \frac{1}{25}$ ; $u = 2$ gives $x = \frac{1}{4}$ ✓
Test $x = \frac{1}{4}$ : LHS $= 6 - 2 = 2$ but RHS $= 2 - 4 = - 2$ , so reject $x = \frac{1}{4}$ (introduced by squaring). Test $x = \frac{1}{25}$ : $6 - 5 = 1$ and $5 - 4 = 1$ , valid.
$∴ x = \frac{1}{25}$ ✓

f) Equations: $ℓ = w + 3$ ✓ and $2 ℓ w = w^{2} + 135$ ✓
Substitute: $2 w (w + 3) = w^{2} + 135$ ✓ so $w^{2} + 6 w - 135 = 0$ and $(w + 15) (w - 9) = 0$ ✓
$w = 9$ ✓ (reject $w = - 15$ : a width CANNOT be negative) then $ℓ = 12$ ✓
The mural is $12$ m long and $9$ m wide.

Modelled on Nov 2025 Paper 1, Question 1 — same skills, new scenario.

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Grade 12 Mathematics: Algebra, equations & inequalities

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