## MATHEMATICS FORM ONE TOPIC 7-ALGEBRA

MATHEMATICS FORM ONE TOPIC 7-ALGEBRA

ALGEBRA

Algebra is a study which deals with situations whereby some values unknown. Normally these unknown are represented by letters. Those letters are also referred to as variables.

Algebraic expression

An expression – is a mathematical statement which consists of several variables. An expression can only be simplified, that is we cannot find values of the variables (s) on it.

Examples

1. a + 2
2. x + 3y + 9z

3.16p – qp

1. a + b + c + d
2. 40

An equation

An equation is formed when two expression are joined by an equal sign

E.g

1. i) 2x – y = 16
2. ii) x + 2 = 6 – 5

iii) 3y + xy = 9

Each member of an equation or expression is called Term

Coefficient

When a number is multiplied by a variable (s) that number is called coefficient of that variable

Example

What is the coefficient of the variables in the following?

1. a) 6x – 8p + y
2. b)– k + 3d
3. c) 2a + 3b – c

Solutions

Coefficients of a) x is 6

y is 1

P = -8

1. b) K = -1

d = 3,
= 1

1. c) a = 2

b = 3

c = -1

Addition and subtraction of algebraic expression

Addition and subtraction of algebraic expression can be done by adding or subtracting like term.

Like terms are those terms which has identical (same) variables

Examples

1. 2a + 4a = 6a
2. 5a + 16a = 21a
3. 2x + 10x – 3x = 9x

Examples: simplify the expression

3n – 7n + 12n

Solution

-4n + 12n

12n – 4n

= 8n

Examples: simplify

6m – 4 – 2m + 15

Soln

6m – 2m – 4n + 15

= 4m + 11

Example: simplify 4x + 6y – 3x + 5y

Solution

4x – 3x + 6y + 5y

= x + 11y

Coefficient: y = 11

x = 1

Number of terms = two

Exercise 7.1

1. Simplify each of the following expressions and after simplifying state
2. the number of terms
3. the coefficient of each of the terms
4. i) n + n + n + n + n + k + k + k + x + x = 5n + 3k + 2x

Solution

(a) There are3 terms

(b) Coefficient of “n” is 5

Coefficients of “y” is 3

Coefficients of “x” is 2

1. ii) 3x + 4y – 7z + 3x – 7y + 2z

Solution
a) There are 3 terms

6x – 3y – 5z
b) coefficients of x is 6

Coefficients of y is -3

Coefficients of z is -5

iii) 3 x + 7x –  x =

Solution

1. a) There is 1 term
b) Coefficient of x is 10

Simplify each of the expressions in numbers 2 – 6

1. 12m + 13m

12m
+ 13m

25m = 25m

1. 5y + 7y – 4y

12y – 4y

= 8y

1. 24w – 28w

-28w + 24w

= -4w

1. 15n – 9n

15n – 9n

= 6n

1. 4k – k + 3k

3k + 3k

= 6k

1. 8y – 3 – 7y + 4

8y – 7y – 3 + 4

y + 4 – 3

y + 1

= y+ 1

1. 14x + 8 – 3x + 2

14x – 3x + 8 + 2

=11x + 10

1. 3a – 5b – 7a + 6c + 7a + 8b

3a – 7a + 7a – 5b + 8b + 6c

=3a + 3b + 6c

1. 4x – 6y + 7x + 2y

4x + 7x – 6y + 2y

=11x – 4y

1. 3x + 4 + 8x – 4 – 11x

3x + 8x – 11x + 4 – 4

11x – 11x + 4 – 4

0 + 0

= 0

1. 8m + 0.4m – 2 – 6m + 8

8m + 0.4m – 6m – 2 + 8

8.4m – 6m + 6

= 2.4m + 6

Multiplication and division of algebraic expression

Example:1) Multiply a – 2b + 6ab by 12xy

Solution

(a – 2b + 6ab) x 12xy

= 12axy – 24bxy + 72abxy

Examples:2) Re – write without brackets

– 16a (-2mn + 9xb – 3kbc)

Solution

-16a (-2mn+9xb-3kbc) = (-16ax-2mn) + (-16a x 9xb) + (-16a x -3kbc)
= 32amn + -144axb + 48abck
= 32amn – 144axb +48abc

Example:3) divide 36xyz – 48xwz – 24xz by 12z

Solution
(36xyz – 48xwz – 24xz) ÷ 12z

Exercise 7.2

1. Complete the following

60xy – 30y + 90z = 30 ( )
Solution

60xy – 30y + 90z = 30 (2xy – y +3x)

2. Simplify i) xy + yz + 2xy – 3zy

1. ii) 8m ÷ 2 + 3mn ÷ n

Solution

1. i) xy + yz + 2xy – zyxy+yz +2xy – zy = xy + 2xy +yz – 3zy
= 3xy – 2yz
2. ii) 8m ÷ 2 + 3mn ÷ n

4m + 3m

(4 + 3) m

=7m

1. Simplify the following
2. i) 5mn – 3mn

= 2mn

1. ii) xyz + 3xy + 4zx – zyx

= xyz – zyx + 3xy + 4zx

= 0 + 3xy + 4zx

= 3xy + 4zx

iii) 3 (2n + 3) + 4 (5n – 3)

Solution

6n + 9 + 20n – 12

6n + 20n + 9 – 12

=26n – 3

1. iv) abc + bac – cab

Solution

abc + abc – abc

abc – abc + abc

abc + 0

= abc

1. v) 2 (5x + 3y) + 3(3x + 2y)

Solution

10x + 6y + 9x + 6y

10x + 9x + 6y + 6y

= 19x + 12y

1. vi) m (2n + 3) + n (3m + 4)

Solution

2nm + 3m + 3mn + 4n

2nm + 3mn + 3m + 4n

= 5mn + 3m + 4n

vii) x (y – 5) + y (x + 2)

Solution

xy – 5x + yx + 2y

xy + yx – 5x + 2y

= 2xy – 5x + 2y

1. ix) Pq -2qp + 3pq – 2qp

Solution

Pq + 3pq – 2qp – 2qp

4pq – 4pq

= 0

1. x) (4x + 8y) ÷ 2 + (9xw + 4xy) ÷ w

solution

1. xi) Multiply 6a – 5b by 3x

Solution

3x (6a – 5b) = 3x x 6a – 3x x 5b
= 18ax – 15bx

= 18ax – 15bx

Equations

An equations is a mathematical statement which involves two expression connected or joined by an equal sign
So we define an equation also as statement of equality e.g. 2y – 6 = 3x + 12
The values of variables can be found in equation if the number of equations is equal to the number of unknown.

FORMULATION OF AN EQUATION

There are three steps to follow when formulating an equation which are;

1. i)Understand the problem/question, what it is asking for
2. ii) Let the unknown be represented by a variable

iii) Formulate the equation using the given information

Signs, words or phrase used when formulating an equation:-

+ Addition, sum of, increase by, greater than, plus, taller than, more than

– Difference, subtract, decrease, less than, shorter than.

× Multiplication, times, products.

÷ Division, divided, Quotient.

= Equals, is, given, result.

Example 01

1. The age of the father is equal to the sum of the ages of his son and daughter. If the son’s age is thrice the age his sister, formulate an equation.

Solution

Let y be the father age

And x be the age of the daughter

The age of son = 3x

y = 3x + x

y = 4x

1. The sum of two numbers is 20. If one of thenumber is 8 formulate an equation.

Solution

Let one of the number be x

And the other number = 8

x + 8 = 20

1. A girl is 14 years old, how old will she be in x years time

Solution

A girl = 14 years

Let “y” be a girls age in x years time.

In years time = + x

y = 14 + x

1. The difference between 24 and another number is 16, form an equation

Soln

Let another number = x

24 – x = 16

Exercise 7.3

Formulate equations for each of the following

1. Five times a number gives twenty

Soln

x= 20

5x = 20

1. The differencebetween123 and another number is 150

Solution

let another number = x

Then x – 123 = 150

x – 123 = 150

1. The sum of 21andanother number is 125

solution

let another number = y

Sum means (+)

21 + y = 125

21 + y = 125

1. When a certainnumberis increased by 15, the result is 88

solution

Let the number be x

Then x + 15 = 88

x + 15 = 88

1. When 99 is increased by a certainnumberthe result is 63

Solution

Let the number = y

Then 99 + y = 63

99 + y = 63

1. The product of 12andanother number is the same as two times the sum of 12 and the number

Solution

Let the number be x

Then 12 x x = 2 x (12 + x)

12x = 24 + 2x

7.A number is such that when it is double and 8 added to it, the result is the same as multiplying the number by 3 and subtracting 7.

solution

Let the number be x

Then x + x + 8 = x x 3 – 7

2x + 8 = 3x – 7

1. When 36 is added to a certainnumber, theresult is the same as multiplying the number by 5.

solution

Let x be the number

Then x + 36 = x x 5

x + 36 = 5x

1. If John is n years old and is 6 years older that James older, write an expression of the sum of their ages.

Solution

Let “J” be john, and “Q” be James and “N” be the year

Let Q = q years

J = n + 6 years

The sum of their age = q + n + 6

= q + n + 6

1. When the sum of nand(n + 3) is multiplied by 5 the result half the product of the two numbers.

Write the expression of this statement:-

Solution

(n + (n +3) x 5 = ½ (n + (n +3) )

(2n + 3) x 5 = ½(2n + 3)

SOLVING FOR EQUATIONS

Solving means finding the value of the unknown in the equation

Example 1

1. x + 5 = 8

Solution

x + 5 = 8

x + 5 – 5 = 8 – 5

x + 0 = 3

x = 3

1. x – 8 = 15

x – 8 + 8 = 15 + 8

x = 23

1. 3x – 5 = 7

3x – 5 + 5 = 7 + 5

3x = 12

x = 4

4. + 3 = 12

solution;

multiply 2 both side

1. (3x – 2) = 10

Solution

1. = 2

Solution

=

1 = (3x – 2)

8 = 6x – 4

8 + 4 = 6x – 4 + 4

12 = 6x

=

x = 2

1. –  = 4

Solution

2m = 4 x 15

2m = 60

m= 30

1.  = 5

Solution

10x = 5 x 8

10x = 40

x = 4

1. 2x – 5 = 3x – 8

Solution

3x-8=2x-5

3x-2x=8-5

x = 3

1. 4 – 3t = 0.3t – 5.9

Solution

4 + 5.9 = 0.3t + 3t

9.9 = 3.3t

9.9 = 3.3t
3.3 3.3

t = 3

Solution

Multiply by 8 both side

1.  –  solve for x

Solution

=  –

7

14x-7 = 9x

14x -9x =7

5x = 7

x =

### MATHEMATICS FORM ONE TOPIC 5-ALGEBRA

EXERCISE 7.4

Solve the following equations

1. x + 12 = 25

Solution

x = 25 – 12

x = 13

x = 5

1. 2x + 12 = 25

Solution

2x + 12 – 12 = 25 – 12

2x + 0 = 13

1. x – 8 = 8

Solution

x– 8 + 8 = 8 + 8

x = 16

1. x = 55

Solution

X = 5 5

X = 25

1. 2x – 8 = 8

Solution

2x =8+8

2x = 16

x = 8

1. 3x – 3 = 15

Solution

3x – 3 + 3 = 15 + 3

x =6

1. – 3 = 5

Solution

1. 0.2x + 7 = 9

Solution

0.2x + 7 – 7 = 9 – 7

x = 10

1. 0.6x – 5 = 7

Solution

0.6x – 5 + 5 = 7 + 5

x = 20

1. + 3 = 5

Solution

1. 4x – 7= 7

Solution

4x = 7 + 7

4x = 14

=

1. = 14

Solution

Solution

1. = 6

Solution

= 6 x 5

=

x = 10

Solution

3x = 25 + 1

=

1. = 10

5 = 10x

1. 18.10

Solution

5 x 1 = 10 (x + 1)

5 = 10x + 10

5 – 10 = 10x

solution

1 (x + 5) = 3 (x – 1)

x+ 5 = 3x – 3

3 + 5 = 3x – x

x = 4

Solution

1 (x + 5) = 5 (x – 1)

x + 5 = 5x – 5

5 + 5 = 5x – x

Solving word problems

E.g. 1

If John has hundred shillings, how many oranges can be buy if orange costs 50 shillings?

Solution

Let k be the number of oranges John can buy but one orange costs 50shs.

50 x k = 200

K = 4

John can buy 4 oranges

Example 2:

A father age is 4 times the age of his son. If the sum of there is fifty years Find the age of the son.

Solution

Let the age of father be y

Let the age of the son be x

Therefore the age of the father is y = 4x

Their sum = 4x + x = 50
5x = 50

The son’s age is 10years old

Example 3:

The sum of 2 consecutive numbers is 31. Find the smaller numbers

Solution

Let the smaller number be x

Let the bigger number be x + 1

x+ x + 1 = 31
2x + 1 = 31
2x = 31 – 1
2x = 30

The smaller number is 15

Exercise 7.5

1. If 4 is added to anumberand the sum is multiplied by 3 the result is 27. Find the number.

Solution

Let the number be ‘b’

(b + 4) x 3 = 27

12 + 3b = 27

3b = 15

b= 5

1. Okwi’s age is six times uli’s age.15 yearshenceOkwi will be three times as old as Uli. Find their ages.

Solution

Let the age of Uli be x

Okwi = 6x

Okwi Uli

6x x

6x + 15 x + 5

6x + 15 = 3x + 45

fifteen years to come
6x + 15 15 + x
Then 6x + 15 = 3(x + 15)
6x + 15 = 3x + 45
6x – 3x = 45 – 15
3x = 30

x = 10

Okwi = 60 years

Uli = 10 years

3 . The sum of two consecutive odd numbers is 88. Find the numbers

Solution

Let the number be n

n + 2, n + 4

n + 2 + n + 4 = 88

2n + 6 = 88

n = 41

The smaller number = 41 + 2 = 43

The bigger number = 41 + 4 = 45

1. Obi’s age is twice Oba’s age. 4 years ago Obi was three times as old as Oba. Find their ages.

Solution

Oba’s age let it be x

Obi Oba

2x x

2x – 4 x – 4

2x – 4 = (x – 4) 3

2x – 4 = 3x – 12

8 = x

Obi = 16 years old.

Oba = 8 years old.

Inequalities in one unknown

The following rules are useful when solving inequalities

1. i) Adding or subtracting equal amounts from each side does not change the inequalities sign

Example : solve x – 2 ≤ 4

Solution

X – 2 + 2 ≤ 4 + 2

X ≤ 6

Example 2: 2x + 4 ≥ 16

Solution

2x + 4 – 4 ≥ 16 – 4

≥

X ≥ 6

1. ii) Multiplying or dividing by samepositive number each side change the inequality sign

Example: solve 3y + 16 < 50

Solution

3y + 16 – 16 < 50 – 16

,

Example 2: (2x – 4) ≥ 9

(2x – 4) ≥ 9 x 3

2x – 4 + 4 ≥ 29 x 3

≥

X ≥ 3

iii) Multiplying or dividing each side by negative number CHANGES the inequality sign.

Example. Solve the inequality

((4 – 3x) < 4

Solution

2 (4 – 3x) < 4 x 3

2 () <

4 – 3x < 6

-3x < 6 – 4

The sign changes

Examples 1: Solve -4x + 3≥

-4x + 3≥
-4x ≥ -3

-4x ≥ –

÷ -4

x ≤

Examples 2. solve

Find their L.C.M

>  x 4

3 (2x – 6) 4 (3 – 2x)

6x – 18 > 12 – 8x

6x + 8x > 12 + 18

X >

BINARY OPERATIONS

Is an operation denoted by *, which describe the formula of a given variables.

if P * q = 5pq – p: Find

1. i) 2 * 3 =

p = 2 and q = 3

2 * 3 = 5 (2) (3) – 2

= 30 – 2 = 28

2 * 3 = 28

1. ii) (1* 2) * 3

Solution

(1 * 2) = p = 1 and q = 2

In (1 * 2) = 5 (1) (2) -1 = 9
=10-1
=9
12=9

9 * 3 = p * q

9 * 3 = 5 (9) (3) – 9

= 135 – 9

= 126

(1 * 2) * 3 = 126

1. iv) (2 * 1) * (3 * 2)

Solution

2 * 1 = p = 2 and q = 1

5 (2) (1) – 2

2 * 1 = 10 – 2

= 8

3 * 2 = p = 3 and q = 2

5 (3) (2) – 3

15 x 2 – 3

3 * 2 = 30 – 3

32= 27

2 * 1 = 8

3 * 2 = 27

8 * 27

8 * 27 = p = 8 and q = 27

5 (8) (27) – 8

40 x 27 – 8

1080 – 8

Then: 8 * 27 = 1072

(2 * 1) * (3 8 2) = 1072

1. if (t * 5) = 50 find t

Solution

t * 5 = p = t and q = 5

t * 5 = 5 (t) (5) – t

15t – t

24t=50

t * 5 = 24t

=

t =

## MATHEMATICS FORM ONE TOPIC 10- COORDINATE GEOMETRY

Coordinate geometry is the study of geometric figures by plotting them in the coordinate axes. Figures such as straight lines, curves, circles, ellipse, hyperbola, polygons, can be easily drawn and presented to scale in the coordinate axes. Further coordinate geometry helps to work algebraically and study the properties of geometric figures with the help of the coordinate system

## MATHEMATICS FORM ONE TOPIC 4- COORDINATE GEOMETRY

COORDINATE GEOMETRY

Introduction

The position of points on a line found by using a number line, that is

When two number lines one vertical and another one horizontal are considered one kept at 90o and intersecting at their zero marks, The result is called xy – plane or Cartesian plane. The horizontal one is called x – axis and the vertical is called y – axis.
Origin is the where the two axes that is x – axis and y axis (intersect)

Coordinates of a point

The position of a point in the xy – plane is given by a pair of in the form of ordered pair. Thus ordered pair is called coordinate. The coordinate of the point is therefore written in the form of (a, b), Where the first number ‘’a’’ is the value in the horizontal axis i.e x – axis b is the value in the y – axis
The value in the x – axis is also referred to as abscissa and y – axis is called ordinate. All distance in the xy – plane are measured the origin.

Examples write the coordinates of the following point A, B, C, D, E, F

Solutions
The coordinates of the points are
A = (0,5)
B = (5,0)
C = (0,4)
D = (-5,5)

Exercise 10.1

1. a) Write down the coordinates of each of the labeled points in figure 9.2
b) State the quadrant in which each of these points F, H, V and I belong

2. Draw axes on a graph paper and plot the points given below. Join in the order given with straight lines forming polygonal figures shape have you drawn in each case.

a) (1, 1), (3, 1), (3, 3), (1, 3)
b) (-2, 1), (2, 5), (2, -2)
c) (3, 1), (5.4, 1) (4.3, 2), (3.3, 2)
d) (5.5, 3.4), (6.5, 3.4), (6.8, 4.3), (6.0, 4.9), (5.2, 4.3)
e) (1.5, -3), (6, 3), (1.5, 3), (-6, -3)
f) (-1, 0), (-2, 2), (0, 1), (2, 2), (1, 0), (2, -2), (0, -1)

SLOP/GRADIENT OF A LINE

Slope / gradient is the change in the vertical axis to the change in the horizontal axis.

Example: Find the gradients of the lines joining

(a) A (2, 4) and B (-2, 6)
(b ) A (-2, -2) and B (2, -4)
(c) A (0, -1) and B (2, 3)
Solution
a. Let (x1, y1) be (2, 4) and
(x2, y2) be (-2, 6)
M=
b. let (x1, y1) be (-2, -2)
(x2, y2) be (2, -4)
M =

m=
c. A (0, -1) and B (2, 3)
Solution
Let (x1, y1) be (0, -1)
(x2, y2) be (2, 3)
M =
M = 2

Exercise 10.2

1.Plot pair of the following points on a graph paper and join them by straight line. For each pair, calculate the gradient of the line and state whether it is positive, negative, zero or undefined.
(a) (0,3), (2,5)
(b) (5,8), (4,1)
(c) (1,5), (4,7)
(d) (2,6), (5,3)
(e) (1,6), (3,-1)
(f) (3,6), (-2,-1)
(g) (0,2), (6,2)
(h) (2,3), (-1,-3)
(i) (2,10), (2,0)
(j) (), , 2)
(k) (-2,1), (4,3)
(l) (-4,4),(-3,3)
(m) (0,0), (-3,4)
(n) (99,6), (119,1)
(o) (0.64,-1.62, (1.36,-0.62))

EQUATION OF A LINE

We have already discussed how to find the gradient of a line for example the gradient of the line joining points (2, – 4) and (5,0) is given as.
Since the two points are collinear.
we can find the equation of the line having any point on a line say (x,y) and any point,
then from

let (x1,y1) = (x,y), and (x2,y2) = (5,0)

4(5-x) = 3x-y
20-4x = -3y

∴y =  –
In general the equation of a straight line is written as y = mx + c. Where m – Is the slope of the line and c is ordinate of the y. called y- intercept
The point on the line (x,y) is called arbitrary point

Example: – find the equation of line passing through the points.
(12,-6) and (2, 6)
Solution

let (x,y)=(12,-6),
(x2-y2)= (2,6)

5(y-6)=-6(x-2)
5y-30=-6x+12
5y=-6+12+30
5y=-6x+42

(2) Give that y = – + 6 find the gradient of this line

The gradient is

Example: find the equation of the line passing through the point (4, 6) and having a slope -1/2

Solution
(x,y) , (4,6) , M =
=
2(y-6) = x – 4
2y – 12 = x – 4
2y = x + 8
X-intercept and y – intercept.
X-intercept is the point where a line meets (cuts) the x-axis, at the value of y (ordinate) is equal to zero.
That is to say the x-intercept is found by substituting y = 0 in the equation. Therefore for the equation y = mx + c.
Y = 0, 0 = mx + c
Mx + c = 0
Mx = -c
x= -c/m.

Therefore the coordinate of x-intercept is (-c/m, 0).
y- Intercept is the point where the line and the y- axis meet. All this point the abscissa is normally equal to zero. The x- intercept is found by setting. x=0
i.e y = mx + c
x=0
y = m(0) + c
y = c
The intercept (0,c)
The coordinate of the Y-intercept is (0,c)
Example : – a line L is passing through the points A(5 – 2) and B(1,4).
Find
i. The equation of the line in the form of
Y= mx +c and ax + by + c = o
ii. The x and y intercept.
Solution:
i.

(x,y) = (1,4)
=
-3 + 3x = 8 – 2y
2y = 11- 3x
Y =

Y=

then -3 + 3x = 8 – 2y

-3 + 3x = 8 – 2y
-3 + 3x – 8 + 2y = 0
3x – 11 + 2y = 0
3x + 2y – 11 = 0

(ii) Y – intercept
Let x = 0
3x + 2y – 11 = 0
y =
Y – Intercept =0,
The coordinate of Y-intercept is (0,11/2)
∴ find the x-intercept
Let y=0
mx + c = y
0 = + 11

x =

The coordinate of X-intercept is (22/3,0)

(iii) If ax + by = 12 goes through points (1,-2) and (4, 1) find the value a and b-

solution
+

let the two collinear point be (x,y),(4,1) and gradient 1
then from

x-4 = y-1
x-y=–1+4
x-y=3
if the equation is multiple by 4 both side we have
4(x-y)=4×3
4x-4y=12
compare the equations.
x-y=3 and ax+by=12
ax+by=4(3)

x-y=3

then

a=1

b=-4
The value of a=1 and b=-4

EXERCISE 10.3

3. Find the equations of lines through points
a) (2,1) with gradient 2.
b) (0,5) with gradient -2
c) (1,-3) with gradient-3
d) (-2, -4) with gradient
e) (0, 0 ) with gradient -3
f) (-3 , -3) and y- intercept
g) ( 6, 2) and y intercept -2
h) (-1 , -1 ) and y – intercept –
i) ( 1 , 2 ) and y- intercept = 2
j) (5 , 5) and y intercept 0

2. Find the equations of the following lines
a. (i) y – intercept – 2 , gradient 1
b. (ii) y – intercept 7, gradient
(iii) y- intercept -16, gradient 4
d. (iv) y -intercept 2, gradient – 10
e. (v) y- intercept 0.4, gradient -0.7
(3) Rewrite the following equations in the form y = mx + c, and then determine the gradient and the y- intercept of each.
a) (i) 7x + 4y = 11
(ii) 14x + 3y = 12
c) (iii) 2x = 5 + y
d) (iv) 4x + 5y = 40
(v) 8x – ()y = 0(vi) 6x = 5 – 2y
(vii) = 9
Qn. 4. Find the area of the shaded region in the following figure. If the equation line AB is 5x + 6y-60 =0

SIMULTANEOUS EQUATIONS

Are equations of more than one variable which can be solved at the same time. There are two ways of solving simultaneous equations.

1. (i) Elimination method
(ii) Substitution method.
(iii) Graphical Method

The principle of solving equations is that the number of equations should be equal to the number of unknowns
Example of simultaneous equation
(a) =

(b) =

These are examples of simultaneous equations with two unknowns.
1. Elimination method
Is the method of omitting one variable and solve the remaining variables.
How to eliminate
i. Check if there are equal coefficients
ii. If there are equal coefficients of same variables in the both equations subtract.
iii. If there are equal and opposite to coefficients of same variable in both equations, add.
2. If all coefficients are different modify the equations
Example 1. 2x + 3y = 6
3x + 2y = 4
Modification
Omitting: x

here we can now omit x by subtracting.

13y=10
Omitting y

13x=24

=
x=
Let find the value of y by take one equation
2x+ 3y=6

Example 2:
Solution

3x=6
x =2
Let find the value of y by take one equation
6x+y=15
2 + y = 15
12 + y = 15
y = 15 – 12
y = 3
∴x =2 and y = 3
Example 03:
2x + y = 10
3x – 2y = 1
Solution. by Eliminate

7x= 21

x=3
Let find the value of y by taken one equation
3x-2y=1
3
9-2y=1
-2y=1-9

y= 4

y=4 and x = 3
Solve the following simultaneous equations by elimination method
1. x + y = 7
5x + 12y = 7
Solution
Modify

-7y = -28

y= 4

Let find the value of x by taken one equation
x + y=7
x + 4 =7
x=7- 4
x=3Therefore: x=3 and y = 4
2. x + 8y =19
2x + 11y = 28
Solution
=

5y=10

y= 2
Let find the value of x by take one equation
x + 8y = 19
x + 8
x=19-16
x = 3
Therefore: x=3 and y = 2
3. 8x + 5y = 9
3x + 2y = 4
Solution
–
y=5

Let find the value of x by taken one equation

8x + 5y = 9

8x + 5(5) =9
8x + 25 = 9
8x = 9 – 25
8x = -16

x = -2
Therefore: x = -2 and y = 5
4. 2x- 3y = 7
15x + y = 9
Solution

17x = 34

=
x = 2
Let find the value of y by taken one equation
2y – 3y = 7

2(2) – 3y =7

-3y=7-4, -3y=3

4 – 3y = 7

y=-1
Therefore: x = 2, y = -1

5. 2x + 3y = 8
2x = 2 + 3y
solution
2x + 3y = 8
2x -3y = 2

6y=6
y=1
Let find the value of x by taken one equation
2x + 3y = 8
2x + 3(1) = 8
2x = 8-3
x=5
= =
x =
Therefore: x=  and y = 1
6. 3x – 4y = 20
x + 2y = 5
Solution

–

2x = 20

x= 10

Let find the value of y by taken the one equation
x + 2y =5
10+ 2y = 5
2y = 5 – 10

Therefore: y =  and x = 10
7. 6x = 7y + 7
7y – x = 8
Solution

= –

5x= 15

x=3
Let find the value of y by taken one equation
6x – 7y =7
6 (3) -7y = 7
Therefore :
8. y = 4x – 7
16x – 5y = 25
solution
-4x + y = -7
16x – 5y = 25

-4x=-10
x = 2.5 or 5/2

Let find the value of y by taken equation one
y – 4x = -7
y = – 4 (2.5)= -7
y = -7 +10
y= 3
Therefore x = 2.5 and y = 3
9. 2x + 7y = 39
3x + 5y = 31
solution
–
y= 5
Let find the value of x by taken one equation
3x +5y = 31
3x+ 5(5) = 31
3x=31- 25

x = 2

Therefore: x = 2 and y = 5
10. 15x – 8y = 29
17x + 12y = 75
Solution
x =3

Let find the value of y by taken one equation
15x – 8y = 29
15(3) – 8y = 29
Therefore: y = 2 and x = 3

### 2: SUBSTITUTION

Example: 01 solve
6x + y = 15
3x + y = 9 by substitution method.
solution
6x + y = 15 ……… (i)

3x + y = 9……….. (ii)

from equation 1
6x+y =15
y=15-6x……………..(iii)

Substitute equation (iii) into (ii)
3x + y = 9
3x + 15 – 6x = 9
3x – 6x + 15 = 9

-3x = 9 – 15

-3x=-6

x=2
y = 15 – 6x
y = 15 – 6 x 2
y= 15 – 12 = 3
x= 2 and y = 3
Example 02
2x + y = 10……………..(i)
3x – 2y = 1………………(ii)
From (i)
2x + y = 10
2x – 2x + y = 10 – 2x
y = 10 – 2x ……………..(iii)
Put (iii)into (ii)
3x – 2 (10 – 2x) = 1
3x – 20 + 4x = 1
3x + 4x – 20 = 1
∴x = 3
y = 10 – 2x
y = 10 – 2 (3)
y = 10 – 6
y = 4
∴x = 3 and y = 4

Solve:
3x + 2y = 8………. (i)
2x + 3y = 12…… (ii)
solution
from (i)
3x + 2y = 8
2y=8-3x
Y =  ……. (iii)
Put (iii) into (ii)
3x + 2y = 8
3(0) + 2y = 8
2y=8
∴Y = 4
X = 0 and y = 4
Exercise 10 .4
1. Solve the simultaneous equations by using elimination method.
(i). 2x + y = 5
4x – y = 7
(ii) 3x + y = 6
5x + y = 8
(iii) 5x – 2y = 16
x + 2y = 8
(iv) 8x+5y=40
9x + 5y = 5
(v) 7x – 4y = 17
5x – 4y = 11
(vi) 0.7x – 0.5y = 2.5
0.7x – 0.3y = 2.9

2. Solve the following simultaneous equations by using substitution method(i). 3x – 2y =
2x + y = 8
(ii) 5x + y = 23
3x- 2y = 6
(iii) x – 3y = 2
4x + 2y = 36
(iv) 7x – y = 14
8x – 2y = 16
(v) 7x + y = 14

8x – 2y = 16

3. Solve the following by using any method
(i) 3y – x =
y + 2x = 6
(ii) 8m – n = 38
m – 3n = -1
(iii) 5x – 2y = 10
-x + 3y = 24

1. 4.Solve the following simultaneous equations by substitution method
(i) x – y = -3
2x – y = -5
(ii) X – 2y = 6
X + 2y = 2
3. (iii) 3x – 4y = -11
2x + 3y = 1
4. (iv) 2x – 3y =32
3x – 4y = 30
(v) 5a – 5b = 7
2a – 4b = 2

5. Solve the following system of simultaneous equations by elimination.
6. (i) 10u + 3v – 4 = 0
6u + 2v – 2 =0
(ii) x – y = 1
4x + 3y = (iii) 3x + 3y = 15
2x + 5y =14
(iv) 7x – 3y = 15
5x- 2y = 19
(v) x + y = 5
x – y = 16. Solve the following by any method
Solving word problems leading to simultaneous equations.
1. A fathers age is four times the age of his son. If the sum of their ages is 60 years. Find the age of the son and that of the father.
Solution.
Let x be the age of the son
Let y be the age of father
y = 4x……………….. (i)
x+ y = 60 …………(ii)
By substitution method
x + y = 6, but y=4x
then
x + 4x = 60
5x=60
x = 12
By solve value of y you can take one equation
y= 4x
y = 4 (12) = 48
The age of the father is 48 years and that of the son is 12 years
2. 2. The sum of two number is 12 and their difference is 2 find the number
Solution
Let the first number be = x
Let the second number be= y
=
2x + 0 = 14
2x=14
x = 7
Let find the value of y
7 + y = 12
y = 12 – 7
y = 5
The numbers are 5 and 7

Example 03. If the numerator of a fraction is decreased by 1 its value become2/3 but if it denominator is increased by 5 its value becomes ½ , what the fraction?
solution
let the fraction be a/b
…………….(i)
+ 5 =  …………..(ii)
from (i)
3(a-1) =2b
3a – 3 =2b
3a – 2b=3………………(iii)
from (ii)

Example 4

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number is increased by 3. Equal to 4 times the original number
Find the original number
Let the number be x and y
x + y = 7……………… (i)
The meaning of x, y = 10x + y
Similarly y, x = 10y + x
y, x + 3 = 4 (x, y)
10y+x+3=4(10x+y)
10y + x + 3 = 40x + 4y
6y+3=39x
=
13 x – 2y = 1……. (2)

= + 15x + 0 = 15
15x = 15
x = 1

Let find the value of y
x + y = 7
1+ y = 7 the number was x ,y is 1,6
y = 6

The original number is 16

Exercise 10.5

1. The sum of two number is 109 and the difference of the same numbers 29. find the numbers

2. Two number are such that the first number plus three times the second number is 1. And the first minus three times the second is 1/7. Find the two numbers

3. The sum of the number of boys and girls in a class is 36. If twice the number of girls exceeds the number of boys by 12, find the number of girls and that of boys in the class.

4. Twice the length of a rectangle exceeds three times the width of the rectangle by one centimeter and if one – third of the difference of the length and the width is one centimeter find the dimensions of the rectangle.
5. The cost of 4 pencils and five pens together is 6000 shillings while the cost of 6 pencils and 8 pens is 940 shillings, calculate the cost of one pencil and one pe
6. Half of Paul’s money plus one – fifth of John’s money is 1400 shilling John’s money is 2650 shillings. How much has each?
7. A farmer buys 3 sheep and 4 goats for shs 290,Another buys sheep and goats from the some market for shs 170.What price did they pay for (a) 1 goat (b) 1 sheep

SOLUTIONS.

EXERCISE 10.1

1. (a) Their coordinates are
A (2, 7)
J (0, 4)
F (3, 4)
N (4, 1.5)
M (6, 1)
V (2.5, -2)
D (6, -2)
K (3, -5)
C (-3, -5)
I (-2, -2)
P (-4.5, -2.5)
H (-4, 4)
G (-2, 6)
L (-3, 1)
E (0, 0)

(b) F belongs to quadrant I
H belongs to quadrant II
V belongs to quadrant IV

I belong to quadrant III

2 (a) Square

(b) Triangle

(c) Trapezium

(d) Parallelogram

(e) Octagon

EXERCISE 10.2

1. a) (0, 3), (2, 5)

let (x1,y1) be (0,3)
(x2,y2) be (2,5)

M = 1 it is positive gradient.

(b) (5, 8), (4, 1)

Let (x1, y1) be (5, 8)
(x2, y2) be (4, 1)

m= 7 Slope is positive

(c) (1, 5), (4, 7)

Let (x1, y1) be (1, 5)
(x2, y2) be (4, 7)