Hard Algebraic Equations

Very hard questions for GCSE Maths

Conquer GCSE/IGCSE/GCE(OL) Linear Algebra Equations with Ease!
Algebra linear equations might seem like a daunting hurdle on your GCSE/IGCSE/GCE(OL) math journey, but fear not! This comprehensive tutorial is here to equip you with the knowledge and skills to tackle even the most challenging problems. We'll break down complex concepts into clear, step-by-step explanations, making linear equations a breeze.
By the end of this tutorial, you'll be able to:
So, whether you're a GCSE, IGCSE, or GCE(OL) student looking to solidify your foundation or someone aiming to ace your exams, this tutorial is your perfect companion. Let's dive in and conquer those linear equations!

Solving equations

 

An equation is almost a sort of seesaw: you add something to the left, lose the balance and are forced to do the same to the right; you divide and multiply by something, once again, the same must be done to the other side; if you subtract something, there is no exception. Therefore, solving equation means, getting rid of everything around x by seesaw method.

E.g. 1

2(x + 5) = 18
:- 2 => 2(x + 5) :- 2 = 18 :- 2
x + 5 = 9
- 5 => x + 5 - 5 = 9 - 5
x = 4

E.g. 2

5(x - 2) = 2(x - 3)
5x - 10 = 2x - 6
+10 => 5x - 10 + 10 = 2x - 6 + 10
5x = 2x + 4
-2x => 5x - 2x = 2x - 2x + 4
3x = 4
:-3 => 3x / 3 = 4 /3
x = 1.3

 

 

E.g.3

4(x + 4) + 3(x -3) = 2(x -3) + 12
4x + 16 + 3x - 9 = 2x - 6 + 12
7x + 7 = 2x + 6
- 7 => 7x + 7 - 7 = 2x + 6 - 7
7x = 2x - 1
-2x => 7x - 2x = 2x - 2x -1
5x = -1
:-5 => 5x / 5 = -1 / 5
x = -0.2

E.g. 4

(x + 5) / 4 = (x -3) / 2
X 4 => 4 X (x + 5) /4 = 4 X (x- 3) / 2
(x + 5) = 2 (x -3)
x + 5 = 2x - 6
- 5 => x +5 -5 = 2x - 6 - 5
x = 2x - 11
-2x => x - 2x = 2x - 2x -11
-x = -11
-1 X x = 11

E.g. 5

3 + 2(x + 5) = 3 - (2x - 1)
3 + 2x + 10 = 3 -2x + 1
13 + 2x = 4 - 2x
-13 => 2x + 13 - 13 = 4 - 2x - 13
2x = -2x - 9
+2x => 2x + 2x = 2x - 2x - 9
4x = -9
:- 4 => 4x / 4 = -9 / 4
x = -2.25

Hard Equation Generator

With this simple programme, you can generate questions at random, along with answers - unlimited number of questions. Generate the question first, work out the solution and then check with the answer shown below the question.

 

 

Word Problem Solving

E.g. 6

Ted gave £1400 to share between his children, Berty, Clare and Damion in such a way that Berty gets twice as much as Clare and Damion gets half of what Clare gets. How much does Clare get?

Berty	Clare	Damion
2x	x	x/2

2x + x + x/2 = 1400
4x + 2x + x = 2800
7x = 2800
x = £400
Clare gets $400.

E.g. 7

The numerator of a fraction is 2 less than its denominator. If both numerator and denominator are increased by 2, the fraction becomes 7/9. Find the original fraction.

Let the original fraction be x/(x+ 2)
x +2 /(x+ 4) = 7/9
9x + 18 = 7x + 28
2x = 10
x = 5
Original fraction = 5/7

E.g. 8

A courier driver has 72 parcels in his van. At a certain store, he drops 2 parcels and x were collected. At the next store, he drops a fifth of the parcels, before collecting 4. If he has 64 parcels in his van at this stage, find x.

At the first store,
Number of parcels = 72 - 2 + x = 70 + x
At the second store,
Number of parcels = (70+x)/5 + 4
4(70+x)/5 + 4 = 64
4(70+x)/5 = 60
70 + x = 300/4 = 75
x = 5

E.g. 9

The mean of the numbers, 4, 5, 10, 11 , Y, placed in ascending order, is the same as the median. Find Y.

10 = (4 + 5 + 10 + 11 + Y)/5
30 + Y = 50
Y = 20

E.g. 10

The first 5 terms and the last term of a linear sequence are as follows:
5, 9, 13,......, 805
a) Find the n^th term of the sequence.
b) Find the 42^nd term of the sequence.
c) Can 107 be a term of this sequence? d) How many terms are there in the sequence?

a) The difference between the successive terms = 4
Zero^th term = 1 = 4 x 0 + 1
So n^th term, N = 4n + 1
b) n = 42 → N = 4(42) + 1 = 169
42^nd term = 169
c) If N = 107,
4n + 1 = 107 → 4n = 106
n = 26.5
The term number of a sequence has to be an integer, not a fraction. Therefore, 107 does not belong to this sequence.
d) The last term, N, is 805
4n + 1 = 805
4n = 804
n = 201
There are 201 terms in the sequence.

E.g. 11

The product of the first and last of the four consecutive, positive integers is 154. Find the sum of the second and third integer.

Let the numbers be x, x+1, x+ 2 and x+3.
x(x + 3) = 154
x² + 3x - 154 = 0
(x + 13)(x - 11) = 0
x = 11 or x = -13
Since the numbers are positive, x = 11.
The four consecutive numbers are 11, 12, 13, and 14.
The sum of the second and third = 25.

 

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E.g. 12

The gradient of a straight line is 5. It passes through the point, (4, -6). Find its equation.

y = mx + c
m = 5; x = 4; y = -6
-6 = 5(4) + c
c = -26
The equation of the line is, y = 5x - 26.

E.g. 13

The gradient of a straight line is 2/3. It passes through the point, (6, 5). Find its equation and express in the form, ay + bx = c.

y = mx + c
m = 2/3; x = 6; y = 5
5 = (2/3)(6) + c
c = 1
y = (2/3)x + 1
Multiplying by 3, 3y = 2x + 3
3y - 2x = 3
The equation of the line is, 3y - 2x = 3.

E.g. 14

A straight line passes through two points. The coordinates of the points are (2,5) and (4, 11) respectively. Find the equation of the line.

m = change in y / change in x
m = (11 - 5)/(4 - 2) = 6/2 = 3
Taking the coordinates of the first point into account,
y = mx + c
5 = 3(2) + c
c = -1
y = 3x - 1

 

You can practise a few questions as in the example 14 with the following interactive applet: just click the button to get two random points along with their coordinates; then, find the gradient and y-intercept of the line as shown above; finally, you can check the answer by clicking on the checkbox.

 

 

Practice is the key to mastering maths; please visit this page, for more worksheets.