All About Fraction

Fraction

Fraction can be defined as a part of a whole. If an orange is cut into two (2) equal parts, each part of the orange can be described as a fractional part of the whole orange.

It should be noted that "a whole" is represented with "1" in Mathematics or simply put, "a whole" means "1".

In the case of the orange used as example above and for Mathematics purpose, the whole orange can be represented with "1". If half $\left(\frac{1}{2}\right)$ of the orange is cut and we want to know what remains, it will be done like this:

The whole orange = 1
The part cut = $\frac{1}{2}$
The remaining orange = $1-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2}\right)$ Note that the LCM of "1" and "2" is "2".

N.B. You will notice that i introduced 1 as a denominator for 1 (which is a whole number) while trying to subtract $\frac{1}{2}$ from 1. That is the usual practice anytime we want to subtract a fraction from 1 or any whole number.

Description of Fraction

A fraction is usually expressed in the form of $\frac{a}{b}$, where "a" represents "Numerator" and "b" represents "Denominator". Examples of fractions are $\frac{1}{2}$, $\frac{3}{4}$, $\frac{7}{5}$, $\frac{9}{8}$ and so on.

It can be said, therefore, that the upper part of all fractions is known as a "Numerator" while the lower part of all fractions is known as a "Denominator". In the fraction $\frac{3}{4}$, "3" is known as a "Numerator" while "4" is a "Denominator".

Types of Fraction

There are three (3) types of fraction namely:

1. Proper Fractions
2. Improper Fractions
3. Mixed Numbers (This is not usually called a fraction.)

(1) A proper fraction is a fraction which has its denominator (i.e. the lower part of any fraction) greater than its numerator (i.e. the upper part of any fraction). Examples of proper fractions are $\frac{7}{9}$, $\frac{2}{3}$, $\frac{3}{4}$ and so on.

(2) An improper fraction is a fraction which has its numerator less than its denominator. Examples of improper fractions are $\frac{7}{5}$, $\frac{3}{2}$, $\frac{4}{3}$, $\frac{9}{4}$ and so on.

(3) A mixed number is defined as the combination (or coming together) of a $quotwhole number" and a "proper fraction". Examples of mixed numbers are $3\frac{1}{4}$, $7\frac{1}{3}$, $2\frac{1}{5}$ and so on.

All About Algebra

Definition of Algebra || Solving Algebraic Expression

Dividing by the Coefficient || Cross Multiplying Algebraic Expression

Algebraic Process for SSS 1 - 3 Students || Adding to Both Sides of an Equation

Subtracting from Both Sides of an Equation || Multiplying Both Sides of an Equation

Dividing Both Sides of an Equation || Word Problems in Linear Equation

Algebraic Expression

Algebra is simply the manipulation of alphabets and figures. Another term for algebra is equation. By equation, we mean that an expression will have a left hand side and a right hand side separated by an equal to (=) sign.

Basically there are two (2) types of equation (or algebra) and they are:

1. Linear equation
2. Quadratic equation

When the graph of a linear equation is drawn, a straight line is what we will get while a quadratic equation will give a curved line.

This is an example of a linear equation:

$2x+3=13$

This is an example of a quadratic equation:

$x^2+4x+8=0$

Solving Linear Equations

Solving linear equations is not a difficult task. It requires the following steps:

• Dividing Both Sides by the Coefficient
• Cross Multiplying
• Clearing Brackets
• Collecting Like Terms

In a very simple algebraic expression, there will be constant (or constants), coefficient and variable.

Lets us take the algebraic expression $2x+3=13$ into consideration.

The numbers '3' and '13' are constants.

The '2' preceding (or which comes before) 'x' is known as coefficient.

The 'x' that follows '2' is known as a variable.

A variable is an unknown value that we try to get by algebraic calculation in mathematics.

In a situation whereby the coefficient stands side-by-side with the variable, it will assist us to derive the value of the variable.

Let us solve some examples below to know how the steps involved in solving algebra operates in practice.

Dividing Both Sides by the Coefficient

Example 1

Solve the following equations:

(a) $4x=40$

(b) $7a=21$

(c) $10t=120$

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Solution to Example 1a

$4x=40$

divide both sides by the coefficient 4 to get:

$\frac{\stackrel{1}{\cancel{4}}x}{\cancel{4}}=\frac{\cancel{40}}{\cancel{4}}$

$x=10$

You can cross-check your answer like so:

$4x=40$

The relationship that exists between 4 and x is an invisible multiplication sign and $x=10$ from our calculation above

$4\left(10\right)=40$

$40=40$

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Solution to Example 1b

$7a=21$

divide both sides by the coefficient 7 to get:

$\frac{\cancel{7}a}{\cancel{7}}=\frac{\cancel{21}}{\cancel{7}}$

$a=3$

Cross check:

$7a=21$

$a=3$

$7\left(3\right)=21$

$21=21$

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Solution to Example 1c

$10t=120$

divide both sides by the coefficient 10 to get:

$\frac{\cancel{10}t}{\cancel{10}}=\frac{\cancel{120}}{\cancel{10}}$

$t=12$

Cross check:

$10t=120$

$t=12$

$10\left(12\right)=120$

$120=120$

You will notice that when we cross-check our answers by substituting them into the main algebraic questions, we have equal numbers on both sides of the equal to (=) signs.

The explanation above is why another term for algebra is equation.

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Cross Multiplying Algebraic Expressions

Let us solve some other examples to show how to cross multiply in Algebra to derive answers for given variables (unknown).

Example 2

Simplify the following equations:

(2a) $\frac{x}{8}=\frac{1}{4}$

(2b) $\frac{3x}{6}=3$

(2c) $\frac{2}{5a}=\frac{1}{10}$

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Solution to Example 2a

$\frac{x}{8}=\frac{1}{4}$

cross multiply to get:

$\overline{)xx4=\overline{)4x}/8x1=\overline{)8}}$

$4x=8$

divide both sides by the coefficient 4 to get:

$\frac{\cancel{4}x}{\cancel{4}}=\frac{\cancel{8}}{\cancel{4}}$

$x=2$

Cross-check your answer:

$\frac{x}{8}=\frac{1}{4}$

$x=2$

$\frac{2}{8}=\frac{1}{4}$

$\frac{\cancel{2}}{\cancel{8}}=\frac{\cancel{1}}{\cancel{4}}$
$\frac{1}{4}=\frac{1}{4}$

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Solution to Example 2b

$\frac{3x}{6}=3$

use 1 as a denominator for 3 to achieve easy cross-multiplying

$\frac{3x}{6}=\frac{3}{1}$

cross multiply to get:

$\overline{)3xx1=\overline{)3x}/6x3=\overline{)18}}$

$3x=18$

divide both sides by the cosfficient 3 to get:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{18}}{\cancel{3}}$

$x=6$

Cross-check your answer:

$\frac{3x}{6}=3$

$x=6$

$\frac{3x6}{6}=3$

$\frac{18}{6}=3$

$3=3$

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Solution to Example 2c

$\frac{2}{5a}=\frac{1}{10}$

cross multiply to get:

$\overline{)5ax1=\overline{)5a}/2x10=\overline{)20}}$

$5a=20$

divide both sides by the coefficient 5 to get:

$\frac{\cancel{5}a}{\cancel{5}}=\frac{\cancel{20}}{\cancel{5}}$

$a=4$

Cross-check your answer:

$\frac{2}{5a}=\frac{1}{10}$

$a=4$

$\frac{2}{5x4}=\frac{1}{10}$

$\frac{2}{20}=\frac{1}{10}$

$\frac{\cancel{2}}{\cancel{20}}=\frac{\cancel{1}}{\cancel{10}}$

$\frac{1}{10}=\frac{1}{10}$

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Algebraic Process for Senior Secondary School 1-3

Linear Equations

An equation is a mathematical statement which shows equality between two mathematical expressions.

There are different types of equations; one of which is linear equation.

A linear equation (or linear algebraic equation) is a mathematical statement which has only one variable combined with several constants. The left and the right hand sides of a linear equation are always equal to each other.

Examples of linear equations are:

$3y+7=19$

$2x+3=4x-13$

$4\left(a-1\right)=2\left(a+4\right)$

Apart from using the methods specified in Basic Algebra treated above (dividing both sides by the coefficient, cross multiplying and so on), Linear Equations can also be solved by using the following methods:

(a) Adding the Same Number to both Sides of the Equation

Example

Find the value of x in the following equation:

$2x-2=10$

Explanation and Solution

What we want to do is to derive the value of the variable 'x' which has a coefficient '2' preceding it.
Since we have the constant '-2', the only way we can derive the value of the variable 'x' is to introduce '+2' and add it to both sides of the expression like so:

$2x-2+2=10+2$

The '-2' and '+2' on the LHS of the expression above cancel out each other.

The explanation for the cancelling out of the '-2' and '+2' goes thus:

• See the '+2' as possession of two loaves of bread by Tom
• See the '-2' as the eating of the two loaves of bread by Tom
• If Tom has two loaves of bread and he has eaten the two loaves, the remainder will be '0'.

∴ −2+2 = 0

$2x=12$

divide both sides by the coefficient 2 to get:

$\frac{\cancel{2}x}{\cancel{2}}=\frac{\cancel{12}}{\cancel{2}}$

$x=6$

Verify your answer

$2x-2=10$

$x=6$

substitute 6 for x

$2\left(6\right)-2=10$

$12-2=10$

$10=10$

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(b) Subtracting the same Number from both Sides of the Equation

Example

Solve for y in the equation below:

$4y+3=19$

To solve for 'y' which has a coefficient '4' and a constant '+3' on the LHS of the expression above, introduce '−3' and subtract it from both sides of the expression like so:

$4y+3-3=19-3$

+3 and −3 on the LHS cancel out

$4y=19-3$

$4y=16$

divide both sides by the coefficient 4 to get:

$\frac{\cancel{4}y}{\cancel{4}}=\frac{\cancel{16}}{\cancel{4}}$

$x=4$

Verify your answer

$4y+3=19$

$x=4$

substitute 4 for y

$4\left(4\right)+3=19$

$16+3=19$

$19=19$

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(b) Multiplying both Sides of the Equation by the Same Number

Example

Solve for x in $\frac{3x}{4}=6$

To solve for 'x' which has a coefficient '3' and a divisor or a denominator '4' on the LHS of the expresson above, multiply both sides of the denominator '4' like so:

$\frac{3x}{4}x4=6x4$

$\frac{3x}{4}x\frac{4}{1}=\frac{6x4}{1}$

$\frac{3x}{1}=\frac{24}{1}$

$3x=24$

divide both sides by the coefficient 3 to get:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{24}}{\cancel{3}}$

$x=8$

Verify your answer

$\frac{3x}{4}=6$

$x=8$

substitute 8 for x

$\frac{3x8}{4}=6$

$\frac{24}{4}=6$

$6=6$

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(b) Dividing both Sides of the Equation by the Same Number

This is very similar to dividing both sides of an equation by the a coefficient.

Example

Find x in $20x=80$

Solution

$20x=80$

divide both sides by the coefficient 20 to get:

$\frac{\cancel{20}x}{\cancel{20}}=\frac{\cancel{80}}{\cancel{20}}$

$x=4$

Verify your answer:

$20x=80$

$x=4$

substitute 4 for x

$20\left(4\right)=80$

$80=80$

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Let us solve all the questions below to practise all the methods that been explained so far:

(1) $2x+3=3x+8$

(2) $\frac{y}{4-y}=7$

(3) $4\left(2a-5\right)=3\left(2a+8\right)$

(4) $\frac{3y}{4}-\frac{7}{3}=\frac{17}{12}$

(5) $\frac{2t-1}{3}-\frac{3t-1}{4}=1$

Solution to Question 1

$2x+3=3x+8$

collect like terms

$2x-3x=8-3$

$-x=5$

divide both sides by the coefficient −1

$\frac{\cancel{\mathrm{-}}x}{\cancel{-1}}=\frac{\cancel{5}}{\cancel{-1}}$

$x=-5$

Solution to Question 2

$\frac{y}{4-y}=7$

$\frac{y}{4-y}=\frac{7}{1}$

cross multiply

$y=7\left(4-y\right)$

$y=28-7y$

collect like terms

$y+7y=28$

$8y=28$

divide both sides by 8

$\frac{\cancel{8}y}{\cancel{8}}=\frac{\cancel{28}}{\cancel{8}}$

$y=3\frac{\cancel{4}}{\cancel{8}}$

$y=3\frac{1}{2}$

Solution to Question 3

$4\left(2a-5\right)=3\left(2a,+,8\right)$

clear brackets

$8a-20=6a+24$

$8a-20=6a+24$

collect like terms

$8a-6a=24+20$

$2a=44$

divide both sides by the coefficient 2 to get:

$\frac{\cancel{2}a}{\cancel{2}}=\frac{\cancel{44}}{\cancel{2}}$

$a=22$

Solution to Question 4

$\frac{3y}{4}-\frac{7}{3}=\frac{17}{12}$

This is similar to solving fraction and the LCM should be found.

The LCM, which is 12, will be used to multiply all terms on both sides like so:

$\cancel{12}x\frac{3y}{\cancel{4}}-\cancel{12}x\frac{7}{\cancel{3}}=\cancel{12}x\frac{17}{\cancel{12}}$

$3x3y-4x7=7$

$9y-28=7$

collect like terms

$9y=7+28$

$9y=35$

divide both sides by 9

$\frac{\cancel{9}y}{\cancel{9}}=\frac{\cancel{35}}{\cancel{9}}$

$y=3\frac{8}{9}$

Solution to Question 5

$\frac{2t-1}{3}-\frac{3t-1}{4}=1$

multiply all terms by 12

$12x\frac{\left(2t-1\right)}{3}-12x\frac{\left(3t-1\right)}{4}=12x1$

$4\left(2t-1\right)-3\left(3t-1\right)=12$

clear brackets

$8t-4-9t+3=12$

rearrange

$8t-9t+3-4=12$

$\mathrm{-t}-1=12$

collect like terms

$\mathrm{-t}=12+1$

$\mathrm{-t}=13$

divide both sides by the coefficient −1

$\frac{\cancel{\mathrm{-}}t}{\cancel{-1}}=\frac{\cancel{13}}{\cancel{-1}}$

$t=-13$

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Word Problems in Linear Equations

It is possible to change some statements to mathematical forms while still maintaining the meaning of the statement or there will not be any change in the meaning of the statement changed to a mathematical form.

The following rules should be strictly used in solving word problems:

Rule 1

Choose any letter (or alphabet) to represent the required number.

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Example

A number is added to 3 and the result is 7. What is the number?

The question above is very simple. It can be answered mentally without even writing on a paper at all.
We all know that the only number that can be added to '3' in order to get '7' is '4' but solving Word Problems expects us to represent the unknown number with any alphabet.
In this case, let us represent the unknown number with 'x'.

Rule 2

Convert each statement in the question to a mathematical statement containing the letter that has been chosen.

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Example

Half of a number added to one-third of itself is equal to 20.

Explanation

Half of a number means the following:

Half = $\frac{1}{2}$

of = $x$ (multiplication sign)

a number = you can use any alphabet. Let us use $t$ in this case.

∴ Half of a number = $\frac{1}{2}xt=\frac{1}{2}x\frac{t}{1}=\frac{t}{2}$

One-third of itself means one-third of the same number and it will be represented mathematically like so:

One-third = $\frac{1}{3}$

of = $x$ (multiplication sign)

the same number = let the unknown number be $t$

∴ One-third of a number = $\frac{1}{3}xt=\frac{1}{3}x\frac{t}{1}=\frac{t}{3}$

The last statement 'is equal to $20$' will be written like so:

$=20$

Rule 3

You will need to link all the different parts of the question to form an equation.

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Example

Considering the example for rule 2, the different mathematical statements formed from the question will then be put together as an equation like so:

$\frac{t}{2}+\frac{t}{3}=20$

Let us solve different examples to drive home our points on Word Problems in Linear Equations.

Example 1

Half of a number added to one-third of itself is equal to 20. What is the number?

Solution

Let the unknown number be t

Half of a number = $\frac{1}{2}xt=\frac{1}{2}x\frac{t}{1}=\frac{t}{2}$

One-third of the same number = $\frac{1}{3}xt=\frac{1}{3}x\frac{t}{1}=\frac{t}{3}$

$\frac{t}{2}+\frac{t}{3}=20$

look for the LCM of 2 and 3

multiply through with LCM 6

$\cancel{6}x\frac{t}{\cancel{2}}+\cancel{6}x\frac{t}{\cancel{3}}=6x120$

$3t+2t=120$

$5t=120$

divide both sides by the coefficient 5 to get:

$\frac{\cancel{5}t}{\cancel{5}}=\frac{\cancel{120}}{\cancel{5}}$

$t=24$

Verify your answer:

$\frac{t}{2}+\frac{t}{3}=20$

$t=24$

substitute 24 for $t$

$\frac{\cancel{24}}{\cancel{2}}+\frac{\cancel{24}}{\cancel{3}}=20$

$12+8=20$

$20=20$

Example 2

A mother is five times as old as her daughter. In four years time, she will be three times as old as the daughter. What are their ages now?

Solution

Let the daughter's age be $x$

The mother's age will then be $5xx=5x$

In four years' time, their ages will be:

daughter $=x+4$

mother $=5x+4$

but $5x+4$ is three times of $x+4$

∴ $5x+4=3\left(x+4\right)$

clear brackets and collect like terms

$5x+4=3x+12$

$5x-3x=12-4$

$2x=8$

divide both sides by the coefficient 2

$\frac{\cancel{2}x}{\cancel{2}}=\frac{\cancel{8}}{\cancel{2}}$

$x=4$

Their ages now are:

daughter $=x=4$ years

mother $=5x=5x4=20$ years

N.B. : It can be clearly seen from the ages of the daughter and the mother that 20 years is 5 times as old as 4 years.

Verify your answer based on what will happen in 4 years' time:

In 4 years' time, their ages will be:

Note that $x=4$

daughter $=x+4=4+4=8$

mother $=5x+4=5\left(4\right)+4=24$

From our verification above, it is evident that the mother will be 3 times as old as the daughter $\left(8x3=24\right)$ in 4 years' time.

Example 3

A father is three times as old as his son. In twelve years time, he will be twice as old. How old is the father now?

Solution

Let the son's age be $y$

The father's age will then be $3xy=3y$

In twelve years' time, their ages will be:

son $=y+12$

father $=3y+12$

but the father is twice as old as the son which means that $3y+12$ is twice as old as $y+12$

∴ $3y+12=2\left(y+12\right)$

clear brackets and collect like terms

$3y+12=2y+24$

$3y-2y=24-12$

$y=12$

The father's age now $=3y=3x12=36$

Verification can also be made for this solution by checking whether the father will be twice as old as the son in 12 years' time.

In 12 years' time, their ages will be:

Note that $y=12$

son $=y+12=12+12=24$

father $=3y+12=3\left(12\right)+12=48$

From our verification above, it is evident that the father will be twice as old as the son $\left(24x2=48\right)$ in 12 years' time.

Example 4

When 6 is subtracted from from four times a number and the remainder is divided by 4, the result becomes 7. What is the number?

Solution

$\frac{4x-6}{4}$

$\frac{4x-6}{4}=7$

$\frac{4x-6}{4}=\frac{7}{1}$

cross multiply to get:

$4x-6=28$

collect like terms

$4x=28+6$

$4x=34$

divide both sides by 4

$\frac{\cancel{4}x}{\cancel{4}}=\frac{\cancel{34}}{\cancel{4}}$

$y=8\frac{2}{4}$

$y=8\frac{1}{2}$

Example 5

Three times a certain number added to 14 is the same as when 2 times the number is subtracted from 6. What is the number?

Solution

Let the unknown number be '$a$'

Three times a certain number added to $14=14+3xa=14+3a$

Two times the same number subtracted from $6=6-2xa=6-2a$

$14+3a=6-2a$

collect like terms

$3a+2a=6-14$

$5a=-8$

divide both sides by the coefficient 5

$\frac{\cancel{5}a}{\cancel{5}}=\frac{\cancel{-8}}{\cancel{5}}$

$a=\frac{-8}{5}$

$a=-1\frac{3}{5}$

verify your answer:

$14+3a=6-2a$

$a=-1\frac{3}{5}$

substitute $-1\frac{3}{5}$ for $a$ to get:

$14+3\left(-1\frac{3}{5}\right)=6-2\left(-1\frac{3}{5}\right)$

$14-4\frac{4}{5}=6+3\frac{1}{5}$

$9\frac{1}{5}=9\frac{1}{5}$

Example 6

A father is three times as old as the son. Nine years ago, the father was four times as old as the son. Find their present ages.

Solution

Let the son's age be $y$

The father's age will then be $3xy=3y$

Nine years' ago, their ages were:

son $=y-9$

father $=3y-9$

but $3y-9$ (father's age) was four times of $y-9$ (son's age)

∴ $3y-9=4\left(y-9\right)$

clear brackets and collect like terms

$3y-9=4y-36$

$3y-4y=-36+9$

$-y=-27$

divide both sides by the coefficient −1

$\frac{\cancel{-}y}{\cancel{-1}}=\frac{\cancel{-27}}{\cancel{-1}}$

$y=27$

Their ages now are:

son $=y=27$ years

father $=3y=3=2781$

Verification can be made for this solution by checking whether the father was four times as old as the son in 9 years' ago.

9 years' ago, their ages were:

Note that $y=27$

son $=y-9=27-9=18$

father $=3y-9=3\left(27\right)-9=81-9=72$

From our verification above, it can be seen that the father was four times as old as the son $\left(18x4=72\right)$ 9 years' ago.

Example 7

The sum of three consecutive numbers is 78. What are the numbers?

Solution

Any time you are given a question similar to the one above, do the the following:

• Let the first number be any alphabet
• Let the second number be the alphabet chosen plus (+) one
• Let the third number be the alphabet chosen plus (+) two

In the case of the question we want to solve now, do the following:

• Let the first number be $x$
• Let the second number be $x+1$
• Let the third number be $x+2$

( N.B. If you are given 4 consecutive numbers to solve, you start at $x$ and stop at $x+3$. By now you should be able to solve for 5 or even 6 consecutive numbers. )

Add the 3 consecutive numbers together and make them equal to 78 like so:

$x+x+1+x+2=78$

Re-arrange to get:

$3x+3=78$

collect like terms

$3x=78-3$

$3x=75$

divide both sides by the coefficient 3:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{75}}{\cancel{3}}$

$x=25$

Verify your answer:

$x+x+1+x+2=78$

$x=25$

$25+25+1+25+2=78$

$78=78$

Example 8

The sum of three consecutive odd numbers is 60. What are the numbers?

Now that you know what the definition and examples of odd numbers are, let us solve the question above.

Solution

This is similar to solving for sum of three consecutive numbers but you must choose the first three odd numbers in this case.

The first three odd numbers are 1, 3 and 5; therefore, our solution will look like so:

• 1st odd number $=x+1$
• 2nd odd number $=x+3$
• 3rd odd number $=x+5$

Add them together and make them equal to 60

$x+1+x+3+x+5=60$

re-arrange to get:

$3x+9=60$

collect like terms

$3x=60-9$

$3x=51$

divide both sides by the coefficient 3:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{51}}{\cancel{3}}$

$x=17$

Verify your answer:

$x+1+x+3+x+5=60$

$x=17$

$17+1+17+3+17+5=60$

$60=60$

N.B.: For 4 consecutive odd numbers, add 1, 3, 5 and 7 to your chosen alphabet. By now you should be familiar with what to do if it is 5 or 6 consecutive odd numbers.

Example 9

The sum of three consecutive even numbers is 51. Find the numbers.

Now that you have the knowledge of the definition and examples of even numbers, let us solve the question above.

Solution

The first three even numbers must be chosen in order to solve the above question.

The first three even numbers are 2, 4 and 6; therefore, our solution will look like so:

• Let the 1st even number be $=x+2$
• Let the 2nd even number be $=x+4$
• Let the 3rd even number be $=x+6$

Add the numbers together and make them equal to 51

$x+2+x+4+x+6=51$

re-arrange to get:

$3x+12=51$

collect like terms

$3x=51-12$

$3x=39$

divide both sides by the coefficient 3:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{39}}{\cancel{3}}$

$x=13$

Verify your answer:

$x+2+x+4+x+6=51$

$x=13$

$13+2+13+4+13+6=51$

$51=51$

N.B.: I believe you know what to do when given 4, 5 or 6 consecutive even numbers to solve. Add 2, 4, 6 and 8 to your chosen alphabet if given 4 consecutive even numbers to solve. .

Example 10

The sum of three consecutive prime numbers is 34. What are the prime numbers?

Having known what prime numbers are and its examples, let us solve the question above.

Solution

The first three prime numbers must be chosen in order to solve the above question.

The first three prime numbers are 2, 3 and 5; therefore, our solution will look like so:

• Let the 1st prime number be $=x+2$
• Let the 2nd prime number be $=x+3$
• Let the 3rd prime number be $=x+5$

Add the numbers together and make them equal to 34

$x+2+x+3+x+5=34$

re-arrange to get:

$3x+10=34$

collect like terms

$3x=34-10$

$3x=24$

divide both sides by the coefficient 3:

$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{24}}{\cancel{3}}$

$x=8$

Verify your answer:

$x+2+x+3+x+5=34$

$x=8$

$8+2+8+3+8+5=34$

$34=34$

N.B.: Add 2, 3, 5 and 7 to your chosen alphabet if given 4 consecutive prime numbers to solve.

It is also possible for examiners at any level to give you 'sum of three consecutive numbers with different description' of the numbers and make it equal to a given number.

Let us consider the question below to drive home our lectures on 'Word Problems in Linear Equation'.

Example 11

The sum of three numbers is 52. The second number is four more than the first while the third is twice the first. What are the numbers?

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