SOLVING LINEAR EQUATIONS IN ONE VARIABLE

A linear equation in one variable is any equation that can be written in the form 

ax + b  =  0

where a and b are real numbers and x is a variable. This form is called standard form of a linear equation in one variable.

But, all the linear equations in one variable can not be expected to be in this form. Always we can not expect x as variable and we may have some other English alphabet as variable. 

In this section, we are going to learn, how to solve linear equations in one variable. That is, how to find the value of the variable. 

Key Concept

To solve linear equations in one variable, we have to isolate the variable x. That is, we have to get rid of all the values around the variable.

Rule 1 :

Simplify each side of the equation, if needed.

Rule 2 :

An equal quantity may be added to each side of the equation.

Rule 3 :

An equal quantity may be subtracted from each side of the equation.

Rule 4 :

An equal, non-zero quantity may multiply each side of the equation.

Rule 5 :

An equal, non-zero quantity may divide each side of the equation.

Example 1 :

Solve for x : 

x - 3  =  2

Solution : 

To solve for x, add 3 to each side of the equation.

x - 3 + 3  =  2 + 3

x  =  5

Example 2 :

Solve for x : 

3x - 2(x - 1)  =  5

Solution : 

Simplify the left side of the equation.

3x - 2(x - 1)  =  5

3x - 2x + 2  =  5

Combine the like terms.

x + 2  =  5

Subtract 2 from each side of the equation.

x + 2 - 2  =  5 - 2

x  =  3

Example 3 :

Solve for x : 

3x - 8  =  x + 10

Solution : 

Subtract x from each side of the equation.

3x - 8 - x  =  x + 10 - x

Combine the like terms.

2x - 8  =  10

Add 8 to each side of the equation.

2x - 8 + 8  =  10 + 8

2x  =  18

Divide each side by 2.

2x / 2  =  18 / 2

x  =  9

Example 4 :

Solve for x : 

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

Solution : 

Simplify each side of the equation. 

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

5x - 15 - 42 + 7x  =  24 - 24 + 3x - 3

Combine the like terms. 

12x - 57  =  3x - 3

Subtract 3x from each side of the equation. 

12x - 57 - 3x  =  3x - 3 - 3x

9x - 57  =  - 3

Add 57 to each side of the equation. 

9x - 57 + 57  =  - 3 + 57

9x  =  54

Divide each side by 9. 

9x / 9  =  54 / 9

x  =  6

Example 5 :

Solve for x : 

4x/5 - 7/4  =  x/5 + x/4

Solution : 

In the given equation, we have the denominators 4 and 5. The least common multiple of 4 and 5 is 20. 

Now, we can proceed as follows. 

4x/5 - 7/4  =  x/5 + x/4

16x/20 - 35/20  =  4x/20 + 5x/20

(16x - 35)/20  =  (4x + 5x)/20

(16x - 35)/20  =  9x/20

Multiply each side by 20.

20 ⋅ [(16x - 35)/20]  =  (9x/20) ⋅ 20

16x - 35  =  9x

Subtract 9x from each side. 

16x - 35 - 9x  =  9x - 9x

7x - 35  =  0

Add 35 to each side. 

7x - 35 + 35  =  0 + 35

7x  =  35

Divide each side by 7. 

7x / 7  =  35 / 7

x  =  5

Example 6 :

Solve for x : 

4x/3 - 1  =  14x/15 + 19/5

Solution : 

Simplify each side of the equation. 

4x/3 - 1  =  14x/15 + 19/5

4x/3 - 3/3  =  14x/15 + 57/15

(4x - 3) / 3  =  (14x + 57) / 15

Least common multiple of 3 and 15 is 15. So, multiply each side of the equation by 15. 

15 ⋅ [(4x - 3) / 3]  =  15 ⋅ [ (14x + 57) / 15 ]

Simplify. 

(4x - 3)  =  14x + 57

20x - 15  =  14x + 57

Subtract 14x from each side of the equation. 

20x - 15 - 14x  =  14x + 57 - 14x

6x - 15  =  57

Add 15 to each side of the equation. 

6x - 15 + 15  =  57 + 15

6x  =  72

Divide each side by 6. 

6x / 6  =  72 / 6

x  =  12

Example 7 :

The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator, the fraction increases by unity. Find the fraction. 

Solution : 

Let x be the numerator of the fraction. 

Then the fraction is

x / (x + 2) -----(1)

Given : If 5 be added to the numerator, the fraction increases by unity. 

(x + 5) / (x + 2)  =  [x / (x + 2)] + 1

Simplify. 

(x + 5) / (x + 2)  =  [x / (x + 2)] + [(x + 2) / (x + 2)]

(x + 5) / (x + 2)  =  (x + x + 2) / (x + 2)

Multiply each side by (x + 2). 

x + 5  =  2x + 2

Subtract x from each side.

x + 5 - x  =  2x + 2 - x

5  =  x + 2

Subtract 2 from each side. 

5 - 2  =  x + 2 - 2

3  =  x

Substitute x  =  3 in (1).

(1)----->  x / (x + 2)  =  3 / (3 + 2)  

x / (x + 2)  =  3 / 5

So, the required fraction is 3/5. 

Example 8 :

Three consecutive integers add up to 51. What are these integers?

Solution :

Let x be the first integer.

Then the remaining two consecutive integers are

(x + 1) and (x + 2)

So, the three consecutive integers are

x, (x + 1), and (x + 2)

Given : Three consecutive integers add up to 51. 

So, we have 

x + (x + 1) + (x + 2)  =  51

x + x + 1 + x + 2  =  51

Combine the like terms.

3x + 3  =  51

Subtract by 3 from each side side.

3x + 3 - 3  =  51 - 3

3x  =  48

Divide each side by 3.

3x / 3  =  48 / 3

x  =  16

So, the three consecutive integers are 16, 17 and 18.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Power Rule of Logarithms

    Oct 04, 22 11:08 PM

    Power Rule of Logarithms - Concept - Solved Problems

    Read More

  2. Product Rule of Logarithms

    Oct 04, 22 11:07 PM

    Product Rule of Logarithms - Concept - Solved Problems

    Read More

  3. Quotient Rule of Logarithms

    Oct 04, 22 11:06 PM

    Quotient Rule of Logarithms - Concept - Solved Problems

    Read More