EQUATIONS WITH FRACTIONS

Subject:

MATHEMATICS

Term:

First Term

Week:

Week 5

Class:

JSS 3 / BASIC 9

Previous lesson: Pupils have previous knowledge of

FACTORIZATION: FACTORIZING EXPRESSION WITH A COMMON FACTOR BRACKET AND BY GROUPING

that was taught in their previous lesson

Topic:

EQUATIONS WITH FRACTIONS

Behavioural Objectives:

At the end of the lesson, learners will be able to

• Clearing fractions (revision)
• Algebraic fractions with: (i) unknown at the numerator.
• Unknown at the denominator (iii) binomial denominators.

 

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum
• Workbooks

 

CONTENT:

 

WEEK 5

TOPIC: EQUATIONS WITH FRACTIONS.

CONTENTS:

• Clearing fractions (revision)
• Algebraic fractions with: (i) unknown at the numerator.
• Unknown at the denominator (iii) binomial denominators.

• Change of subject of formulae and substitution.

 

CLEARING FRACTIONS/ ALGEBRAIC FRACTIONS UNKNOWN AT THE NUMERATOR

We will now discuss how to clear algebraic fractions when the unknown is at the numerator (top).

As we learned in a previous lesson, to clear an algebraic fraction, we need to find the Lowest Common Multiple (LCM) of the denominators.

For example, suppose we have the following equation:

frac 4x+1x-2=frac 32

To clear this equation, we first need to find the LCM of the denominators, which is 6. We can do this by listing the multiples of each denominator and finding their intersection:

– Multiples of 2: 2, 4, 6, 8, 10, 12, 14, …

– Multiples of 3: 3, 6, 9, 12, 15, 18, 21, …

– Multiples of x-2: x-2, 2(x-2), 3(x-2), 4(x-2), 5(x-2), …

From this list, we can see that the LCM of 2 and 3 is 6, and 6 is also a multiple of (x-2). Therefore, the LCM of all three denominators is 6.

Equations such as   etc. are equations involving fractions.

To solve any of these equations, we consider the L.C.M of the denominators and multiply each term of the equation by the L.C.M to clear the fractions and solve the equation as usual.

 

Example 1

Solve the equation:

 

Solution

 

The denominators of the fractions in this equation are; 5 and 2.

L.C.M of 5 and 2 is 10.

We multiply through by this L.C.M. i.e ) = 2  = 20 – 5

Opening brackets we have,

2  8 = 20 – 5

Collecting like terms we have,

5  + 2  = 20 +8

7  = 28.

Thus,  =  = 4

 

Example 2

 

Solve   +

 

Solution

The denominators are 3, 5 and 15.

Their L.C.M is 30

Thus, 30 x  + 30 x

 

 

10 x  + 6 x 4 = 2 x 17

+ 24 = 3;

;

CLASS ACTIVITY

Solve

• = 5    (b)    (c)

NOTE: Sometimes it is not necessary to find the L.C.M of the denominators. If the equation has a single denominator on both sides of the equation, we simply solve the equation by cross multiplying as illustrated in the next example below.

Change of subject of formulae and substitution.

In order to solve equations which contain fractions, we need to be able to clear them. This means getting rid of the fraction by multiplying both sides of the equation by the denominator.

For example, consider the equation:

4x – 3 = 2x + 5

We can see that there is a fraction on the left hand side, so we need to clear it. We do this by multiplying both sides of the equation by the denominator, which in this case is 3:

4x – 3 = 2x + 5

12x – 9 = 6x + 15

We can now see that there is no fraction on the left hand side, so we can continue solving the equation as normal.

It is important to note that when we clear a fraction, we must multiply both sides of the equation by the *same* number. If we do not, then the equation will no longer be true.

For example, consider the equation:

2x + 1 = 3

We can see that there is a fraction on the right hand side, so we need to clear it. We do this by multiplying both sides of the equation by the denominator, which in this case is 3:

2x + 1 = 3

6x + 3 = 9

We can now see that there is no fraction on the right hand side, so we can continue solving the equation as normal.

Now let’s consider an equation with a binomial denominator. For example:

4x – 3 = 2x + 5

We can see that there is a fraction on the left hand side, so we need to clear it. We do this by multiplying both sides of the equation by the denominator, which in this case is (4x + 3):

4x – 3 = 2x + 5

16x^2 – 12x – 9 = 8x^2 + 12x + 15

We can now see that there is no fraction on the left hand side, so we can continue solving the equation as normal.

It is important to note that when we clear a fraction with a binomial denominator, we must multiply both sides of the equation by the *same* number. If we do not, then the equation will no longer be true.

For example, consider the equation:

2x + 1 = 3

We can see that there is a fraction on the right hand side, so we need to clear it. We do this by multiplying both sides of the equation by the denominator, which in this case is (2x – 3):

2x + 1 = 3

4x^2 – 6x + 3 = 6x – 9

We can now see that there is no fraction on the right hand side, so we can continue solving the equation as normal.

FRACTIONS WITH UNKNOWNS IN THE DENOMINATOR

Example1 Solve  i.          ii.

Solution

cross multiplying, we have,

 

Rearranging the terms,

Cross multiplying, we have,

Example 2

Solve

 

Solution

Cross multiplying, we have,

2( )

Opening brackets, we have,

Collecting like terms

CLASS ACTIVITY

solve the following equations

3. 10 = 1 +

FRACTIONS WITH BINOMIAL DENOMINATORS

Expressions of the form , etc. are examples of algebraic fractions with binomials as  the denominator.

Example:

Solve the following:

1. a) b)

Solution:

1. a)

Multiply both sides by (x + 2) (x – 3)

x – 3 = 3(x + 2)

x – 3 = 3x + 6

-3-6 = 3x – x

-9 = 2x

x = -9/2

x = -4 ½

1. b)

Multiply both sides by (2x – 1) (3 – x)

3 – x = 2x – 1

3 + 1 = 2x + x

4 = 3x

4/3 = 3x/3

x = 1

CLASS ACTIVITY

Solve the following equations:

WORD PROBLEMS INVOLVING FRACTIONS.

When solving word problems identify the unknown and represent it by any letter of the alphabet. Form an equation in terms of the unknown based on the given information and solve the equation. Study the table below

 

S.N	word problem	unknown	mathematical form
1	a number added to triple the number	k	k + 3K
2	Three-quarters of a number subtracted from twice the number	r	2r
3	The product of 5 and one-quarter of a number added to ten	m	5
4	eight subtracted from three times a number is the same as twice the number	z	3z – 8 = 2z
5	Two-third of a number subtracted from thrice the number gives five	d	3d  = 5
6	The positive difference between nine and five	—–	9 – 5
7	one-quarter of w subtracted from twenty	w

 

CLASS DRILL

Write the following in mathematical form

1. The positive difference between and
2. One-third of a number subtracted from gives ten

iii.  Five times m plus twice n gives ten

1. The product of two-fifth of k and two
2. One-fourth of the product of eight and five
3. Three-fifth of the difference between four and

Example1

A fisherman had 30 fish in his net. He ate some of them and discovered that there were 19 fish left. How many did he eat?

Solution

Total number of fish in the net = 30.

Let number of fish eaten =

Number of fish left in the net = 19

Hence, 30 = 19

Or = 30 – 19 = 11.  Hence the man ate 11 fish

 

Example2

The sum of the ages of a man and his son is 56 years. In 8 years’, time, the ratio of their ages will be 13:5. (a) How old are they now  (b) find the difference between their ages.

Solution

(a)        Let the age of the man be  years.

Therefore, the son’s age will be (56 – ) years.

In 8 years’ time, their ages will be(8 + ) years (man)

and (8 +56 – m) years (son).

The ratio of their ages at this time is 13:5.

Hence,

Cross multiplying,

we have 5(8 + m) = 13( 64 – m)

Opening brackets,

40 + 5m =  832 – 13m

Collecting like terms,

5m + 13m = 832 – 40

18m = 792

Hence, the man’s age is 44 years and the son’s age is 56 – 44 = 12 years.

(b)The difference between their ages is 44 – 12 = 32

 

Example 3

When a certain number is subtracted from 56 and the result is divided by 5, it is the same as if 14 is added to the number and the result divided by 2. Find the number

Solution

Let the number be n

The number subtracted from 56 is 56 – n

The result divided by 5 is

14 added to the number is n + 14

The result divided by 2 is

Hence, according to the information given,

Cross multiplying, we have,

Opening brackets,

112 – 2n = 5n + 70

Collecting like terms,

2n – 5n = 70 – 112

n =

n = 14

CLASS ACTIVITY

1. Subtract 8 from 78, then find one-seventh of the result
2. If I add 8 to a certain number and I double the result, my final answer is 36. What is the number?
3. Subtract 8 from 78, then find one-seventh of the result. 
4. If I add 8 to a certain number and I double the result, my final answer is 36. What is the number?

    

5. I have a 4 digit number. When I subtract 1 from it, the result is 3 times the number reversed. What is my original number?

    

6. I take away 2 from a number and then multiply by 3. If my final answer is 18, what was my starting number?

    

7. I divide a certain number by 3 and then add 7 to the result. If my final answer is 12, what was the original number?

    

8. I divide a certain number by 4 and then add 5. If my final answer is 13, what was the original number?

    

9. I subtract 8 from 78, then find one-seventh of the result.

    

10. If I add 8 to a certain number and I double the result, my final answer is 36. What is the number?

     

11. I have a 4 digit number. When I subtract 1 from it, the result is 3 times the number reversed. What is my original number?

     

12. I take away 2 from a number and then multiply by 3. If my final answer is 18, what was my starting number?

     

13. I divide a certain number by 3 and then add 7 to the result. If my final answer is 12, what was the original number?

     

14. I divide a certain number by 4 and then add 5. If my final answer is 13, what was the original number?

FORMULAE AND SUBSTITUTION

Formulae are equations containing variables that stand for specific quantities. The varibles are usually represented by letters.

Substitution means to replace one letter or a set of letters in an algebraic expression or formula with their numerical values in order to evaluate it.

EXAMPLE:

Find:    a) 3y + x², when x = 2, and y = 4

1. b) , where n = 6 and m = 5

Solution:

1. a) 3y + x², when x = 2, and y = 4

= 3(4) + 2²

            = 12 + 4 = 16

1. b) , where n = 6 and m = 5

=  +  =  =

 

CHANGING SUBJECT OF A FORMULAR

Example1

Make h the subject of the formula V = πh(R + r)

Solution

To make h the subject of the formula means to make h stand alone.

Divide both sides by π(R + r)

 

h =

Example2: Given 3a – 2b = 8, express a in terms of b. Hence, find a when b = 11.

Solution:

3a – 2b = 8

3a = 8 + 2b

a =

When a = 11,

a =

a =

a = 30/3

a = 10.

 

PRACTICE EXERCISE

1. Make T the subject of the simple interest formula I = .

Hence, find how long it takes for a principal of ₦25 000 to earn an interest 0f ₦2 625 at 3% per annum.

2. If a stone is dropped, the distance d m, which it falls in t seconds is given by the formula d= 4.9t²
3. a) How far does it fall in 3 seconds?
4. b) How long does it take to fall 490m
5. c) How long does it fall in 122 ½ m?
6. d) How far does it fall in ?
7. Solve the equations:
8. a)
9. b)
10. c)

 

 

ASSIGNMENT

Solve the following equations:

 

PRESENTATION:

Step 1:

The subject teacher revises the previous topic

Step 2:

He or she introduces the new topic

 

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

 

CONCLUSION:

The subject goes round to mark the pupil’s notes. He does the necessary corrections