Back to: MATHEMATICS JSS3

**Welcome to Class !!**

*We are eager to have you join us !!*

*In today’s Mathematics class, We will be looking at Simple Fractional Equations. We hope you enjoy the class!*

**SOLVING EQUATION EXPRESSIONS**

**WORD PROBLEMS**

**Worked Examples:**

- Find 1/4 of the positive difference between 29 & 11
- The product of a certain number and 5 is equal to twice the number subtracted from 20. Find the number
- The sum of 35 and a certain number is divided by 4 the result is equal to double the number. Find the number.

Solutions:

- Positive Difference 29 – 11 = 18

$\frac{1}{4}of18=4\frac{2}{5}$ $$

- Let the number be x

x X 5 = 20 – 2x

5x = 20 – 2x

5x + 2x = 20

7x = 20

x = 20/7 = 2

- Let the number be n

sum of 35 and n = n + 35

$divided4=\frac{n+35}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}result=2\times n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}therefore\frac{n+35}{4}=2n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}n+35=8n\phantom{\rule{0ex}{0ex}}8n\u2013n=35\phantom{\rule{0ex}{0ex}}7n=35\phantom{\rule{0ex}{0ex}}n=\frac{35}{7}\phantom{\rule{0ex}{0ex}}n=5$

**EVALUATION**

- From 50 subtract the sum of 3 & 5 then divide the result by 6
- The sum of 8 and a certain number is equal to the product of the number and 3 find the number

** **

**SOLVING EQUATION EXPRESSIONS WITH FRACTION**

Always clear fractions before beginning to solve an equation: –

To clear fractions, multiply each term in the equation by the LCM of the denominations of the fractions.

Examples:

Solve the following

__$1)\frac{x}{9}=2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2)\frac{x+9}{5}+\frac{2+x}{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3)2x=\frac{5x+1}{7}+\frac{3x\u20135}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Solutions:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1)\frac{x}{9}=2\phantom{\rule{0ex}{0ex}}crossmultiply\phantom{\rule{0ex}{0ex}}x=18\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2)\frac{x+9}{5}+\frac{2+x}{2}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}MultiplybytheLCm(10)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}10\left(\frac{x+9}{5}\right)+10\left(\frac{2+x}{2}\right)=10(0\left)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\right(x+9)+5(2+x)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2x+18+10+5x=0\phantom{\rule{0ex}{0ex}}2x+5x+28=0\phantom{\rule{0ex}{0ex}}7x+28=0\phantom{\rule{0ex}{0ex}}7x=\u201328\phantom{\rule{0ex}{0ex}}x=\frac{\u201328}{7}\phantom{\rule{0ex}{0ex}}x=\u20134\phantom{\rule{0ex}{0ex}}$__

__2x + 18 + 10 + 5x = 0__

- 2x =

$\frac{5x+1}{7}+\frac{3x\u20135}{2}$

Multiply by the LCM (14)

$14X2x=14\left(\frac{5x+1}{7}\right)+14\left(\frac{3x\u20135}{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}28x=2(5x+1)+7(3x\u20135)\phantom{\rule{0ex}{0ex}}28x=10x+2+21x\u201335\phantom{\rule{0ex}{0ex}}$

28x = 31x – 33

28x – 31x = -33

-3x = -33

x = 33/3 = 11

** **

**EVALUATION**

Solve the following equations.

$1)\frac{7}{3c}=\frac{21}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2)\frac{6}{y+3}=\frac{11}{y\u20132}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3)\frac{3}{2b\u20135}\u2013\frac{4}{b\u20133}=0$

**Furthermore, we can consider the word equations or expressions into:**

- Sum & Differences
- Products
- Expressions with fractions & equations

** **

**SUM & DIFFERENCES**

The sum of a set of numbers is a result obtained when the numbers are added together. The difference between two numbers is a result of subtracting one number from the other.

Worked Examples:

- Find the sum of -2 & -3.4
- Find the positive difference between 19 & 8
- The difference between two numbers is 7. If the smaller number is 7 find the other.
- The difference between -3 and a number is 8, find the two possible values for the number.
- Find the three consecutive numbers whose sum is 63.

** **

**Solutions:**

- -2 + -3.4 = -5.4
- 19 – 8 = +11
- let the number be Y i.e. Y -7 = 7

i.e. Y = 7 + 7 = 14

- Let M represent the number

M – (-3) = 8

m + 3 = 8

m = 8 – 3

m = +5

also -3 – m = 8

-m = 8 + 3

-m = 11

m = -11

the possible values are +5 & -11

- Consecutive numbers are 1, 2, 3,4,5,6, Consecutive odd numbers are

1, 3,5,7,9……….. consecutive even numbers are 2, 4, 6, 8,10……….

Representing in terms of X, we have 2X, 2X + 2, 2X + 4, 2X + 6, 2X + 8, 2X + 10…………

for consecutive even numbers, we have X, X + 2, X + 4, X + 6…….

for consecutive odd numbers, we have X + 1, X + 2, X + 3, X + 4…

for consecutive numbers.

let the first number be x,

let the second number be x + 1

let the third number be x + 2

Therefore x + x + 1 + x + 2 = 63

3x + 3 = 63

3x = 63 – 3

3x = 60

x = 60 /3

= 20

The numbers are 20, 21, and 22.

** **

**EVALUATION**

- Find the sum of all odd numbers between 10 and 20
- The sum of four consecutive odd numbers is 80 find the numbers

3. The difference between 2 numbers is 9, the largest number is 32 find the numbers.

** **

**PRODUCTS**

The product of two or more numbers is the result obtained when the numbers are multiplied together.

Worked Examples:

- Find the product of – 6, 0.7, &
- The product of two numbers is 8. If one of the numbers is 1/4 find the other.
- Find the product of the sum of -2 & 9 and the difference between -8 & -5.

Solutions

- Products -6 x 0.7 x 20/3

-6 x 7/10 x 20/3 =

$\frac{\u20136\times 7\times 20}{10\times 3}$= -2 x 7 x 2 = -28

- Let the number be x

$\frac{1}{4}$

X x = 8 multiply both sides by 4

x = 8 x 4 = 33

- Sum = -2 + 9 = 7

Difference = -5-(-8) = -5 + 8 = 3

Products= 7 x 3 = 21

** **

**EVALUATION**

- The product of three numbers is 0.084 if two numbers are 0.7 & 0.2 find the third number
- Find the product of the difference between 2 & 7 and the sum of 2 & 7
- From 50 subtract the sum of 3 & 5 then divide the result by 6
- The sum of 8 and a certain number is equal to the product of the number and 3 find the number

*We have come to the end of this class. We do hope you enjoyed the class?*

*Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.*

*In our next class, we will be talking about Change Of Subject Formulae. We are very much eager to meet you there.*

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