Defining Algebra
Solving Linear Equation
Dividing Both Sides by the Coefficient
Cross Multiplying Linear Equation

Algebra

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 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 Equation

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

• dividing both sides by a coefficient
• cross multiplication
• collecting like terms
• clearing brackets

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

Let 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 operate in practice.

Dividing Both Sides by the Coefficient

Example 1

Find x in the following equations:

(1a) $4x=40$

(1b) $7a=21$

(1c)$10t=120$

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Solution to Example 1a

$4x=40$
divide both sides by the coefficient 4 to get:
$\frac{4x}{4}=\frac{40}{4}$
$x=10$

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:
$\frac{7a}{7}=\frac{21}{7}$
$a=3$

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:
$\frac{10t}{10}=\frac{120}{10}$
$t=12$

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 a given variable (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:
$\boxed{xx4=\overline{)4x}/8x1=\overline{)8}}$
$4x=8$
divide both sides by the coefficient 4 to get:
$\frac{\cancel{4x}}{\cancel{4}}=\frac{\cancel{8}}{\cancel{4}}$
$x=2$

Cross check your answer

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

---

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:
$\boxed{3xx1=\overline{)3x}/6x3=\overline{)18}}$
$3x=18$
divide both sides by the coefficient 3 to get:
$\frac{\cancel{3x}}{\cancel{3}}=\frac{\cancel{18}}{\cancel{3}}$
$x=6$

Cross check your answer

$\frac{3x}{6}=3$
$x=6$
$\frac{3x6}{6}=3$
$\frac{18}{6}=3$
$3=3$

---

Solution to Example 2c

$\frac{2}{5a}=\frac{1}{10}$
cross multiply to get:
$\boxed{5ax1=\overline{)5a}/2x10=\overline{)20}}$
$5a=20$
divide both sides by the coefficient 5 to get:
$\frac{\cancel{5a}}{\cancel{5}}=\frac{\cancel{20}}{\cancel{5}}$
$a=4$

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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Definition of Algebra || Solving Algebraic Expression

Dividing by the Coefficient || Cross Multiplying Algebraic Expression

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) $10t=120$

---

Solution to Example 1a

$4x=40$
divide both sides by the coefficient 4 to get:
$\frac{\cancel{4}x}{\cancel{4}}=\frac{\cancel{40}}{\cancel{4}}$
$x=10$

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

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

---

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:

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

---

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:
$4x=8$
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:
$3x=18$
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:
$5a=20$
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
$2\left(6\right)-2=10$
$12-2=10$
$10=10$

(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
$4\left(4\right)+3=19$
$16+3=19$
$19=19$

(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
$\frac{3x8}{4}=6$
$\frac{24}{4}=6$
$6=6$

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

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
$2x\frac{\left(2t-1\right)}{3}-12x\frac{\left(3t-1\right)}{4}12x1$
$4\left(2t-1\right)-3\left(3t,-,1\right)=12$
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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Please endeavour to view this page with the latest Moxilla Firefox browser, latest google chrome browser or latest Safari browser for Maximum Benefit. Thank you.
You can also drop comments to let me know how this blog can be improved.