Solving One-variable Equations Using Different Operations (Definition, Types and Examples) - BYJUS

# Solving One-variable Equations Using Different Operations

A variable is an unknown numerical value in an algebraic equation. We can use math operations like addition, subtraction, multiplication, and division to solve algebraic equations. Here we will learn the steps involved in solving algebraic equations that have one variable using basic math operations. ...Read MoreRead Less ## Equations

An equation is a statement that connects two expressions with the equals sign (=) to express their equality.

## Solving One-variable Equations Using Different Operations

For some values, equations may be true, while for others, they may be false. A value that makes an equation true is called a solution.

The table below shows the formation of the correct solution:

 Value of a Equation           a + 2 = 8 Are they equal or not? 4 4 + 2 = 6 No 5 5 + 2 = 7 No 6 6 + 2 = 8 Yes

Hence, when the value of a = 6, it will form a unique solution for the equation a + 2 = 8.

Opposite operations are also referred to as inverse operations. They are operations that counteract the effects of other operations. Addition, for example, is the inverse of subtraction, and multiplication is the inverse of division.

## Properties

The addition property of equality

The two sides of an equation remain equal when the same number is added to each side.

• Algebraic application:

a – 2 =  4

+ 2   +2

a = 6

• Numerical application:

6 = 6

+4  +4

10 = 10

The subtraction property of equality

The two sides of an equation remain equal when the same number is subtracted from each side.

• Algebraic application:

a + 2 =  4

– 2    – 2

a = 2

• Numerical application:

6 =   6

– 4     – 4

2 = 2

The multiplication property of equality

When both sides of an equation are multiplied by the same non-zero number, the two sides remain equal.

• Algebraic application:

$$\frac{a}{2}=4$$

$$\frac{a}{2}.2=4.2$$

a = 8

• Numerical application:

$$\frac{6}{2}=3$$

$$\frac{6}{2}.2=3.2$$

6 = 6

The multiplicative inverse property

The non-zero number a and its reciprocal, 1/a, have a product of one.

• Algebraic application:

$$n.\frac{1}{n}=\frac{1}{n}.n=1$$, n ≠ 0

• Numerical application:

$$6.\frac{1}{6}=1$$

The division property of equality

When both sides of an equation are divided by the same non-zero number, the two sides remain equal.

• Algebraic application:

2x =  4

$$\frac{2X}{2}=\frac{4}{2}$$

x = 2

• Numerical application:

$$6.2=12$$

$$6.\frac{2}{2}=\frac{12}{2}$$

6 = 6

## Solved Examples

Example 1: Solve the following equation to check whether the equation is forming a solution or not.

(a) 3a=42,a=14

(b) x-7=22,x=30

Solution:

Part (a)

We have: 3a = 42, a = 14

3(14) = 42              Substitute a with 14

42 = 42

It will form a unique solution. Part (b)

We have: x – 7 = 22, x = 30

30 – 7 = 22            Substitute x with 30

23 = 22

It will not form a unique solution. Example 2: Solve the following equations using different operations.

(a) 15 = x – 3

(b) a + 3 = 8

(c) $$\frac{a}{2}$$ = 6

(d) 35 = 5x

Solution:

Part (a)

We have: 15 = x – 3

15 = x – 3

+3     + 3                 Using the addition property of equality

18 = x

The solution is x = 18.

Part (b)

We have: a + 3 = 8

a + 3 = 8

-3  – 3                       Using the subtraction property of equality

a = 5

The solution is a = 5.

Part (c)

We have: $$\frac{a}{2}=6$$

$$\frac{a}{2}=6$$

.2 .2       Using the multiplication property of equality

a = 12

The solution is a = 12.

Part (d)

We have: 35 = 5x

35/5 = 5x/5

Using the division property of equality

7 = x

The solution is x = 7.

Example 3: Your uncle gives you $60 to help you purchase the mobile phone in the picture. You have$25.50 left after purchasing the phone. How much money did you have before your uncle gave you $50? Solution: Write an equation using the word sentence of the given question: Amount left ($) = Initial Amount ($) + Uncle’s Money ($) – Price of the phone ($) Let the initial amount be i and form an equation: = 25.50 = i + 60 – 275.95 = 25.50 + 275.95 = i + 60 – 275.95 + 275.95 Using the addition property of equality = 301.45 = i + 60 = 301.45 – 60 = i + 60 -60 Using the subtraction property of equality = 241.45 = i The total amount of money you have is$241.45.

Example 4: In a park, a rectangular LED “screen” covers 7500 square metres. The screen has a 60-metre width. What is the length of the screen?

Solution:

In the formula for the area of a rectangle, A = lw, substitute values for the area A and the width. Calculate the length l.

The area of a rectangle = A = lw

Substitute the value for A = 7500, and w = 60

7500 = $$60l$$

7500 = $$60l$$    Using the division property of equality

$$\div 60 \div 60$$

$$l=125$$

Therefore, the screen in the park is 125 meters long.

Here are the steps for solving linear equations with one variable:

Step 1: Using LCM, remove any fractions that may exist.

Step 2: Make both sides of the equation as simple as possible.

Step 3: Identify and isolate the variable.