How to Solve Linear Equations?

How to solve linear equations?

Step-by-step instructions are given in the examples of solving linear equations. We will learn how to solve one variable linear equations using addition, subtraction, multiplication and division.


Examples on solving linear equations: 

1. Solve the equation 2x - 1 = 14 - x and represent the solution graphically. 

Solution: 

2x - 1 = 14 - x 

⇒ 2x + x = 14 + 1

(Transfer -x from right hand side to the left hand side, then negative x changes to positive x. Similarly again transfer -1 from left hand side to the right hand side, then negative 1 change to positive 1.


Therefore, we arranged the variables in one side and the numbers in the other side.)

⇒ 3x = 15

⇒ 3x/3 = 15/3 (Divide both sides by 3)

⇒ x = 5

Therefore, x = 5 is the solution of the given equation.

The solution may be represented graphically on the number line by graphing linear equations.

graphing linear equations



2. Solve the equation 10x = 5x + 1/2 and represent the solution graphically.

Solution:

10x = 5x + 1/2

⇒ 10x – 5x = 1/2

(Transfer 5x from right hand side to the left hand side, then positive 5x changes to negative 5x).

⇒ 5x = 1/2

⇒ 5x/5 = 1/2 ÷ 5 (Divide both sides by 5)

⇒ x = 1/2 × 1/5

⇒ x = 1/10

Therefore, x = 1/10 is the solution of the given equation.


The solution may be represented graphically on the number line.

solution graphically



3. Solve the equation 6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3) and verify your answer

Solution:

6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3)

⇒ 18x + 12 + 35x - 30 - 12x = 30x - 5 + 6x - 18

⇒ 18x + 35x - 12x + 12 - 30 = 30x + 6x - 5 - 18

⇒ 41x - 18 = 36x - 23

⇒ 41x - 36x = - 23 + 18

⇒ 5x = -5

⇒ x = -5/5

⇒ x = -1

Therefore, x = -1 is the solution of the given equation.

Now we will verify both the sides of the equation,

6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3) are equal to each other;

Verification:

L.H.S. = 6(3x + 2) + 5(7x - 6) - 12x

Plug the value of x = -1 we get;

= 6[3 × (-1) + 2] + 5 [7 × (-1) - 6] - 12 × (-1)

= 6[-3 + 2] + 5[-7 - 6] + 12

= 6 × (-1) + 5 (-13) + 12

= - 6 - 65 + 12

= -71 + 12

= -59

Verification:

R.H.S. = 5(6x - 1) + 6(x - 3)

Plug the value of x = - 1, we get

= 5[6 × (-1) - 1] + 6[(-1) - 3]

= 5(-6 - 1) + 6(-1 -3)

= 5 × (-7) + 6 × (-4)

= - 35 - 24

= - 59

Since, L.H.S. = R.H.S. hence verified.

What is cross multiplication?

The process of multiplying the numerator on the left hand side with the denominator on the right hand side and multiplying the denominator on left hand side with the numerator on right hand side is called cross multiplication.

And then equating both the products we get the linear equation.

On solving it we get the value of variable for which L.H.S. = R.H.S. Then, it is an equation of the form.

(mx + n)/(ox + p) = q/r where m, n, o, p, q, r are numbers and ox + p ≠ 0
⇒ r(mx + n) = q(ox + p)

It’s an equation in one variable x but it is not a linear equation as L.H.S. is not a linear polynomial.

We convert this into linear equation by the method of cross multiplication and further solve it step-by-step.


Examples on cross multiplication while solving linear equations:

1. (3x + 4)/5 = (2x - 3)/3

Solution:

(3x + 4)/5 = (2x - 3)/3

On cross multiplication, we get;

⇒ 3(3x + 4) = 5(2x - 3)

⇒ 9x + 12 = 10x - 15

⇒ 9x - 10x = -15 - 12

⇒ -x = -27

⇒ x = 27

Verification:

L.H.S. = (3x + 4)/5

Plug x = 27, we get;

(3 × 27 + 4)/5

= 81 + 4/5

= 85/5

= 17

Verification:

R.H.S. = (2x - 3)/3

Plug x = 27, we get;

(2 × 27 - 3)/3

= 54 - 3/3

= 51/3

= 17

Since, L.H.S. = R.H.S. hence verified.


2. Solve 0.8 - 0.28x = 1.16 - 0.6x

Solution:

0.8 - 0.28x = 1.16 - 0.6x

⇒ 0.6x - 0.28x = 1.16 - 0.8

⇒ 0.32x = 0.36

⇒ x = 0.36/0.32

⇒ x = 36/32

⇒ x = 9/8

Therefore, 9/8 is the required solution.

Verification:

L.H.S. = 0.8 - 0.28x

Plug x = 9/8, we get;

= 0.8 - 0.28 × 9/8

= 8/10 - 2̶8̶/100 × 9/8̶

= 8/10 - 63/200

= (160 - 63)/200

= 97/200

Verification:

R.H.S. = 1.16 - 0.6x

= 1.16 - 0.6 × 9/8

= 116/100 - 6̶/10 × 9/8̶

= 116/100 - 27/40

= (232 - 135)/200

= 97/200

Since, L.H.S. = R.H.S. hence verified.


 Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations


 Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation









7th Grade Math Problems

8th Grade Math Practice 

From How to Solve Linear Equations? to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Adding 1-Digit Number | Understand the Concept one Digit Number

    Apr 26, 24 01:55 PM

    Add by Counting Forward
    Understand the concept of adding 1-digit number with the help of objects as well as numbers.

    Read More

  2. Subtracting 2-Digit Numbers | How to Subtract Two Digit Numbers?

    Apr 26, 24 12:36 PM

    Subtracting 2-Digit Numbers
    In subtracting 2-digit numbers we will subtract or minus a two-digit number from another two-digit number. To find the difference between the two numbers we need to ‘ones from ones’ and ‘tens from

    Read More

  3. 1st Grade Word Problems on Subtraction | Subtracting 2-Digit Numbers

    Apr 26, 24 12:06 PM

    1st Grade Word Problems on Subtraction
    In 1st grade word problems on subtraction students can practice the questions on word problems based on subtraction. This exercise sheet on subtraction can be practiced by the students to get more ide…

    Read More

  4. Subtracting 1-Digit Number | Subtract or Minus Two One-Digit Number

    Apr 26, 24 11:21 AM

    Cross Out 6 Objects
    In subtracting 1-digit number we will subtract or minus one-digit number from one-digit number or one-digit number from 2-digit number and find the difference between them. We know that subtraction me…

    Read More

  5. Perimeter of a Square | How to Find the Perimeter of Square? |Examples

    Apr 25, 24 05:34 PM

    Perimeter of a Square
    We will discuss here how to find the perimeter of a square. Perimeter of a square is the total length (distance) of the boundary of a square. We know that all the sides of a square are equal. Perimete…

    Read More