CAIE IGCSE Mathematics Extended Equation Questions

Practise rearranging formulae and solving quadratic, algebraic-fraction and linear–quadratic simultaneous equations with exact or rounded roots.

Exam points

clear fractions or rearrange formulae systematically while keeping all operations balanced
solve quadratics by factorisation or the formula and round roots only to the stated accuracy
substitute a linear equation into a quadratic one and recover both coordinate pairs

Covered syllabus points

E2.5.1 Construct expressions, equations and formulas E2.5.2 Solve linear equations E2.5.3 Solve fractional equations E2.5.4 Solve simultaneous linear equations E2.5.5 Solve linear and non-linear simultaneous equations E2.5.6 Solve quadratic equations E2.5.7 Change the subject of formulas

[Maximum number: 7]

(a)

Antonio travels 100 km at an average speed of $x \mathrm{~km} / \mathrm{h}$ .

He then travels a further 150 km at an average speed of (x+10) km / h.
The time taken for the whole journey is 4 hours 20 minutes.

[ 7 ]

(i)

Show that $13x^2-620x-3000=0$ .

[ 4 ]

$\frac{100}{x}+\frac{150}{x+10}=4\frac13$ oe or $150=\left(\frac{13}{3}-\frac{100}{x}\right)(x+10)$
M1
$\frac{100(x+10)+150x}{x(x+10)}$ or better
M1
$300x+3000+450x=13x^2+130x$ oe or better
B1
Allow correct multiples of $13x^2-620x-3000=0$

(ii)

Solve $13x^2-620x-3000=0$ to find the speed Antonio travels for the first 100 km of the journey.
You must show all your working and give your answer correct to 1 decimal place.

[ 3 ]

$\frac{620\pm\sqrt{(-620)^2-4(13)(-3000)}}{2(13)}$
or equivalent quadratic formula
M2
M1 for $(-620)^2-4\times13\times(-3000)$ oe
x=52.1 to 52.2 final answer
B1
Reject negative speed

[Maximum number: 5]

(a)

Solve.
4x+15=9x=

[ 2 ]

-1.5 or $-1\frac12$ or $-\frac32$
M1 for 4x=9-15 or $x+\frac{15}{4}=\frac94$

(b)

Solve by factorisation.
$4x^2+8x-5=0$ $x=\qquad$ or $x=\qquad$

[ 3 ]

(2x-1)(2x+5)=0 oe
B2
$x=\frac12$ and $x=-\frac52$
B1

[Maximum number: 7]

(a)

Solve the equations.

[ 7 ]

(i)

$\frac{3(3-x)-2(x+2)}{5}=1$ x=

[ 4 ]

$\frac{36}{17}$ oe
B3 for -15x-2x=5+4-45 or better; lower method marks as in markscheme

(ii)

$\frac{5}{x+3}=\frac{3}{x+5}$ x=

[ 3 ]

All question bank results loaded

Covered syllabus points

E2.5 Equations question 1

Question (a)

Question (i)

Question (ii)

E2.5 Equations question 2

Question (a)

Question (b)

E2.5 Equations question 3

Question (a)

Question (i)

Question (ii)