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Students who are looking for the practice material on linear equations can stay tuned to this page. Here, we are giving several problems on linear equations. Interested students can solve them by using addition, subtraction, multiplication, and division operation. You can also find the step by step solution guide for each and every question. While practicing this Linear Equations Word Problems just have a look at the important notes mentioned below.

How to Solve Linear Equations?

Go through the below steps on solving linear equations. Follow them and arrive at the solutions easily. They are listed as under

• A linear equation in one variable is one that contains only one variable and its highest power is 1.
• One can add or subtract the same number to both sides of the equation.
• One can divide or multiply by both sides of the equation by the same non-zero integer.
• The process in which any term in the equation can be moved to the other side of the equal symbol by changing its sign from (+ to -), (- to +), (x to ÷), and (÷ to x) is called the transposition.
• Cross multiplication means the process of multiplying the numerator of L.H.S with the denominator of the R.H.S and multiplying the denominator of L.H.S with the numerator of R.H.S.

Linear Equations Questions and Answers

Example 1.

Solve the following linear equations.

(a) (3 – 7x)/(15 + 2x) = 1

(b) 8x + 9 – 3x = 8 + 4x + 1

(c) 3x – 12 = 0

Solution:

(a) (3 – 7x)/(15 + 2x) = 1

Multiply both sides by (15 + 2x).

(3 – 7x) = 1(15 + 2x)

3 – 7x = 15 + 2x

Transfer – 7x to R.H.S becomes 7x

3 = 15 + 2x + 7x

3 = 15 + 9x

Transfer 15 from R.H.S to L.H.s becomes -15.

3 – 15 = 9x

-12 = 9x

Divide both sides by 9

9x / 9 = -12/9

x = -4/3

Therefore required solution is x = -4/3.

(b) 8x + 9 – 3x = 8 + 4x + 1

5x + 9 = 9 + 4x

Transfer 4x from R.H.S to L.H.S becomes -4x

5x – 4x + 9 = 0

x + 9 = 0

Subtract both sides from -9

x + 9 – 9 = 0 – 9

x = -9

Therefore, required solution set is x = -9.

(c) 3x – 12 = 0

3x = 12

Divide both sides of the equation by 3.

3x/3 = 12/3

x = 4

Therefore, required solution set is x = 4.

Example 2.

Solve the following equations and represent them on a graph.

(a) (3y – 2)/3 + (2y + 3)/3 = (y + 7)/6

(b) 5x – 11 = 3x + 9

(c) (0.5y – 9)/0.25 = 4y – 3

Solution:

(a) (3y – 2)/3 + (2y + 3)/3 = (y + 7)/6

[(3y – 2) + (2y + 3)] / 3 = (y + 7) / 6

(5y + 1) = (y + 7) / 2

Multiply both sides by 2.

2(5y + 1) = (y+7) / 2 x 2

10y + 2 = y + 7

Transferring y from R.H.S to L.H.S becomes -y, 2 from L.H.S to R.H.S becomes -2.

10y – y = 7 – 2

9y = 5

y = 5/9

Therefore, the required solution set is y = 5/9.

(b) 5x – 11 = 3x + 9

Transferring 3x from R.H.S to L.H.S becomes -3x, -11 from L.H.S to R.H.S becomes 11.

5x – 3x = 9 + 11

2x = 20

Divide both sides by 2.

2x/2 = 20/2

x = 10.

Therefore, the required solution set is x = 10

(c) (0.5y – 9)/0.25 = 4y – 3

Multiply both sides by 0.25.

(0.5y – 9)/0.25 x 0.25 = (4y – 3)0.25

0.5y – 9 = y – 0.75

Transferring 0.5y from L.H.S to R.H.S becomes -0.5y, -0.75 from R.h.s to L.H.S becomes 0.75.

-9 + 0.75 = y – 0.5y

-8.25 = 0.5y

Divide both sides by 0.5.

0.5y/0.5 = -8.25/0.5

y = -16.5

Therefore, the required solution set is y = -16.5

Example 3.

Solve the equations and verify them.

(a) (x – 3)/4 + (x – 1)/5 – (x – 2)/3 = 1

(b) y/2 – 1/2 = y/3 + 1/4

Solution:

(a) (x – 3)/4 + (x – 1)/5 – (x – 2)/3 = 1

L.C.M of 4, 5, 3 is 60.

[15(x – 3) + 12(x – 1) – 20(x – 2)] / 60 = 1

Multiply both sides of the equation by 60.

[15(x – 3) + 12(x – 1) – 20(x – 2)] / 60 x 60 = 1 x 60

[15(x – 3) + 12(x – 1) – 20(x – 2)] = 60

15x – 45 + 12x – 12 – 20x + 40 = 60

27x – 20x – 57 + 40 = 60

7x – 17 = 60

7x = 60 + 17

7x = 77

x = 77/7

x = 11

Therefore, the required solution set is x = 11.

Verification:

L.H.S = (x – 3)/4 + (x – 1)/5 – (x – 2)/3

Substitute x = 11

L.H.S = (11 – 3)/4 + (11 – 1)/5 – (11 – 2)/3

= 8/4 + 10/5 – 9/3

= 2 + 2 – 3

= 4 – 3 = 1

Hence, L.H.S = R.H.S

(b) y/2 – 1/2 = y/3 + 1/4

(y – 1)/2 = (4y + 3)/12

Cross multiply the fractions.

12(y – 1) = 2(4y + 3)

12y – 12 = 8y + 6

12y – 8y = 6 + 12

4y = 18

y = 18/4

y = 4.5

Verification:

L.H.S = y/2 – 1/2

Substitute y = 4.5

L.H.S = 4.5/2 – 1/2

= (4.5 – 1)/2

= 3.5/2 = 1.75

R.H.S = y/3 + 1/4

Substitute y = 4.5

R.H.S = 4.5/3 + 1/4

= (4.5 x 4 + 3) / 12

= (18 + 3) / 12

= 21/12 = 1.75

Hence, L.H.S = R.H.S

Example 4.

Solve the below-mentioned linear equations.

(a) 8a – (4a + 32) = 16

(b) 4(x + 5) = 3(x – 2) – 2(x + 2)

Solution:

(a) 8a – (4a + 32) = 16

8a – 4a – 32 = 16

4a – 32 = 16

Transferring -32 from L.H.S to R.H.S becomes +32.

4a = 16 + 32

4a = 48

Divide both sides of the equation by 4.

4a/4 = 48/4

a = 12

Therefore, the required solution set a = 12.

(b) 4(x + 5) = 3(x – 2) – 2(x + 2)

4x + 20 = 3x – 6 – 2x – 4

4x + 20 = x – 10

Transferring x from R.H.S to L.H.S becomes -x, 20 from L.H.S to R.H.S becomes -20.

4x – x = -10 – 20

3x = -30

Divide both sides by 3.

3x/3 = -30/3

x = -10

Therefore, the required solution set x = -10.

Example 5.

Solve the following equations and verify them.

(a) ⅓ (21 – 3x) = ½ (8 – 4x)

(b) (0.4y – 3)/(1.5y + 9) = -7/5

Solution:

(a) ⅓ (21 – 3x) = ½ (8 – 4x)

Cross-multiply the fractions.

2(21 – 3x) = 3(8 – 4x)

42 – 6x = 24 – 12x

Transferring -12x from R.H.S to L.H.S becomes 12x, 42 from L.H.S to R.H.S becomes -42.

-6x + 12x = 24 – 42

6x = -18

Divide both sides by 6.

6x/6 = -18/6

x = -3

Verification:

L.H.S = ⅓ (21 – 3x)

Put x = -3

L.H.S = ⅓ (21 – 3(-3))

= ⅓ (21 + 9)

= ⅓ (30)

= 10

R.H.S = ½ (8 – 4x)

Put x = -3

R.H.S = ½ (8 – 4(-3))

= ½ (8 + 12)

= ½ (20)

= 10

L.H.S = R.H.S

Hence proved.

(b) (0.4y – 3)/(1.5y + 9) = -7/5

Cross multiply the fractions.

5(0.4y – 3) = -7(1.5y + 9)

2y – 15 = -10.5y – 63

Transferring -10.5y from R.H.S to L.H.S becomes 10.5y, -15 from L.H.S to R.H.S becomes 15.

2y + 10.5y = 15 – 63

12.5y = -48

Divide both sides by 12.5

12.5y/12.5 = -48/12.5

y = -3.84

Verification:

L.H.S = (0.4y – 3)/(1.5y + 9)

Substitute y = -3.84

L.H.S = (0.4(-3.84) – 3) / (1.5(-3.84) + 9)

= (-1.536 – 3) / (-5.76 + 9)

= -4.536 / 3.24

= -1.4

R.H.S = -7/5

= -1.4

L.H.S = R.H.S.

Hence proved.

We have provided several problems that involve relations among known and unknown numbers and can be put in the form of linear equations. Those equations can be stated in words and it is the main reason we prefer these Word Problems on Linear Equations. You can practice as many types of questions as you want to get an expert in this concept. For better understanding, we even listed linear equations examples with solutions.

Steps to Solve Word Problems on Linear Equations

Below are the simple steps to solve the linear equations word problems. Follow these instructions and solve the questions carefully.

• Read the problem carefully and make a note of what is given in the question and what is required.
• Denote the unknown things as the variables like x, y, z, a, b, . . .
• Translate the given word problem into mathematical statements.
• Form the linear equations in one variable by using the conditions provided in the question.
• Solve the unknown parameters from the equation.
• Verify the condition with the obtained answer ad cross check whether it is correct or not.

Linear Equations Examples with Answers

Example 1.

A motorboat goes downstream in the river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water?

Solution:

Let the speed of the boat in still water = x km/hr

Speed of the boat down stream = (x + 3) km/hr

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) x 5

Speed of the boat upstream = (x – 3) km/hr

Time taken to cover the distance = 6 hrs

Therefore, distance covered in 6 hrs – (x – 3) x 6

Therefore, the distance between the two coastal towns is fixed, i.e., the same.

As per the question

5(x + 3) = 6(x – 3)

5x + 15 = 6x – 18

15 + 18 = 6x – 5x

33 = x

x = 33

Required speed of the boat is 33 km/hr

Example 2.

The perimeter of a rectangular swimming pool is 144 m. Its length is 2 m more than twice its width. What are the length and width of the pool?

Solution:

Let l be the length of the swimming pool, w be the width of the swimming pool.

According to the question,

length l = 2w + 2

The perimeter of swimming pool = 144 m

2l + 2w = 144

Substitute l = 2w + 2

2(2w + 2) + 2w = 144

4w + 4 + 2w = 144

6w = 144 – 4

6w = 140

w = 140 / 6

w = 23.3

Then, the length is

l = 2(23.3) + 2

= 46.6 + 2

= 48.6

Hence, the length and width of the rectangular swimming pool is 48.6 m, 23.3 m.

Example 3.

The sum of three consecutive even numbers is 126. What are the numbers?

Solution:

Let the first even number be x, the second number be (x + 2), the third number be (x + 4).

According to the question, the sum of consecutive even numbers is 126.

First Number + Second Number + Third Number = 126

x + (x + 2) + (x + 4) = 126

3x + 6 = 126

Subtract 6 from both sides.

3x + 6 – 6 = 126 – 6

3x = 120

Divide both sides by 3.

3x / 3 = 120 / 3

x = 40

The first number is 40, the second number is (x + 2) = 40 + 2 = 42, third number is (x + 4) = 40 + 4 = 44.

Hence, the three consecutive even numbers are 40, 42, 44.

Example 4.

When five is added to three more than a certain number, the result is 19. What is the number?

Solution:

Let us take the number as x.

According to the question,

Add 5 to the three more than a certain number.

5 + x + 3 = 19

x + 8 = 19

Subtract 8 from both sides of the equation.

x + 8 – 8 = 19 – 8

x = 11.

So, the number is 11.

Example 5.

Eleven less than seven times a number is five more than six times the number. Find the number?

Solution:

Let the number be x.

According to the question,

11 less than the seven times a number is five more than six times the number.

7x – 11 = 6x + 5

7x – 6x = 5 + 11

x = 16

Hence the required number is 16.

Example 6.

Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find the measure of the angles.

Solution:

Let the triangle be ∆ABC.

So, ∠A = ∠B and ∠C = ∠A – 12 degrees

The sum of three angles of a triangle = 180 degrees

∠A + ∠B + ∠C = 180

∠A + ∠A + ∠A – 12 = 180

3∠A – 12 = 180

Add 12 to both sides of the equation.

3∠A – 12 + 12 = 180 + 12

3∠A = 192

Divide both sides by 3.

3∠A / 3 = 192/3

∠A = 64

Hence, the first and second angles of the triangle are 64 degrees, 64 degrees and the third angle is 64 – 12 = 52 degrees.

Example 7.

The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions.

Solution:

Let the rectangle width is w.

Length of rectangle l = w + 15 cm

Given that, the perimeter of a rectangle is 150 cm

2l + 2w = 150

Substitute l = w + 15 cm in above equation.

2(w + 15) + 2w = 150

2w + 30 + 2w = 150

4w + 30 = 150

Subtract 30 from both sides.

4w + 30 – 30 = 150 – 30

4w = 120

Divide both sides by 4.

4w/4 = 120/4

w = 30

Hence, the rectangle width is 30 cm, the length is (30 + 15) = 45 cm.

Example 8.

If Mr. David and his son together had 220 dollars, and Mr. David had 10 times as much as his son, how much money had each?

Solution:

Let Mr. David’s son has x dollars.

The amount at Mr. David = 10x dollars

Given that, Mr. David and his son together had 220 dollars

x + 10x = 220

11x = 220

Divide both sides by 11.

x = 220 / 11

x = 20

Hence, Mr. David has 20 x 10 = 200 dollars and his son has 20 dollars.

Solved questions on linear equations in one variable are provided below with a detailed explanation. You have to eliminate brackets and expand the given linear equation. Find the variable and translate the problem to the mathematical statement. Use the conditions and solve the equation. After that verify whether the answer satisfies the condition or not.

How To Solve Linear Equation in One Variable?

Go through the step by step procedure listed for Solving Linear Equations in One Variable. They are mentioned below

• Read the given question twice or many times until you observe the parameters like what is given and what you need to find.
• To make a strategy, represent the unknown values as variables.
• Convert the word problem into mathematics.
• Portray the problem as a linear equation in one variable using the conditions provided.
• Solve the equation for unknown.
• Verify to be sure whether the answer satisfies the conditions of the problem.

Linear Equations in One Variable Word Problems with Solutions

Example 1.

The sum of two numbers is 44. If one exceeds the other by 6, find the numbers?

Solution:

Let x be one of the two numbers.

Then, another number is (x + 6).

The Sum of two numbers is 44.

x + x + 6 = 44

2x + 6 = 44

2x = 44 – 6

2x = 38

x = 38/2

x = 19.

The second number is x + 6 = 19 + 6 = 25

So, the two numbers are 19 and 25.

Example 2.

The sum of four consecutive multiples of 5 is 650. Find these multiples?

Solution:

If x is a multiple of 5, the next multiple is x + 5, third number is x + 10, and fourth multiple is x + 15.

The sum of four consecutive multiples of 5 is 650

x + x + 5 + x + 10 + x + 15 = 650

4x + 30 = 650

Subtract 30 from both sides

4x + 30 – 30 = 650 – 30

4x = 620

Divide both sides by 4.

4x/4 = 620/4

x = 155

The first multiple is 155, the second multiple is 155 + 5 = 160, the third multiple is 155 + 10 = 165, fourth multiple is 155 + 15 = 170.

So, the consecutive multiple of 5 is 155, 160, 165, and 170.

Example 3.

Two numbers are in the ratio 8 : 1. If they differ by 126, what are those numbers?

Solution:

Given ratio is 8:1

From the ratio two numbers can be assumed as 8x, x.

The difference between numbers is 126

8x – x = 126

7x = 126

Divide both sides by 7.

7x/7 = 126/7

x = 18

8x = 8 x 18 = 144.

Hence, two numbers are 18, 144.

Example 4.

The ratio of the three numbers is 5: 6: 7. If the sum of those three numbers is 54, find those numbers?

Solution:

The given ratio of three numbers are 5: 6: 7

From the ratio, the numbers can be assumed as 5x, 6x, and 7x.

The sum of three numbers is 54

5x + 6x + 7x = 54

18x = 54

Divide both sides by 18.

18x / 18 = 54 / 18

x = 3

So, the numbers be

5x = 5 x 3 = 15

6x = 6 x 3 = 18

7x = 7 x 3 = 21.

Hence, the three numbers are 15, 18, and 21.

Example 5.

If you subtract 2/3 from a number and multiply the result by 2/3, you will obtain 1/6. What is the number?

Solution:

Let us say x is the required number.

From the given information, we can write

(x – 2/3) * 2/3 = 1/6

2x/3 – 4/9 = 1/6

2x/3 = 1/6 + 4/9

2x/3 = (3 + 8)/18

2x/3 = 11/18

Cross multiply the fractions.

2x * 18 = 11 * 3

36x = 33

x = 33/36

So, the number is 33/36.

Example 6.

A total of $1500 is distributed among 100 persons as compensation for the work. The given compensation is either of $50 or $100. Find the number of compensations of each type?

Solution:

Total number of compensations = 100

Let the number of compensations of $50 is x

Then the number of compensations of $100 is ($100 – x)

Amount spend on x compensations of $50 = $50x

Amount spend on (150 – x) compensations of $100 = $100(100 – x)

The total amount spent for compensation = $1500

According to the question,

50x + 100(100 – x) = 1500

50x + 10000 – 100x = 1500

-50x + 10000 = 1500

-50x = 1500 – 10000

-50x = -8500

50x = 8500

x = 8500 / 50

x = 170

100 – x = 100 – 170 = 70

Therefore, compensation of $50 are 170, and compensations of $100 are 70.

Example 7.

Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Solution:

Let one part be x.

Then other art = 28 – x

It is given 6/5 of one part = 2/3 of the other.

6x/5 = 2/3(28 – x)

6x/5 = (28 x 2)/3 – 2x/3

6x/5 + 2x/3 = 56/3

(18x + 10x) / 15 = 56/3

28x/15 = 56/3

Cross multiply the fractions.

28x * 3 = 56 * 15

84x = 840

x = 840 / 84

x = 10

Then the two parts are 10 and 28 – 10 = 18.

Example 8.

The numerator of a rational number is less than its denominator by 3. If the denominator is increased by 7 and the numerator is decreased by 1, the new number becomes 3/2. Find the original number?

Solution:

Let the denominator of a rational number = x

Then the numerator of the rational number = x – 3

When denominator is increased by 7, then new denominator = x + 7

When the numerator is decreased by 1, then new numerator = x – 3 – 1 = x – 4

The new number formed = 3/2

According to the question,

(x – 4) / (x + 7) = 3/2

Cross multiply.

2(x – 4) = 3(x + 7)

2x – 8 = 3x + 21

-8 – 21 = 3x – 2x

-29 = x

x = -29

The original number i.e., (x – 3) / x = (-29 – 3) / -29 = (-32)/-29 = 32/29.

Do you want to learn how to solve linear equations? Then stay on this page, here we are giving the detailed steps of solving the linear equations. Go through the simple steps and some worked example problems to get a clear idea on the topic. You can easily solve the linear equations by using addition, subtraction, multiplication, and division operations.

Steps to Solve a Linear Equation

Have a look at the detailed steps and solve the linear equations effortlessly. Follow the guidelines and arrive at the solutions easily. They are as follows

• Transfer variables to one side and constants to the other side of the equation.
• Use arithmetic operations to transfer those.
• Find the value of a variable.
• The variable value becomes the required solution.

Linear Equations Questions with Solutions

Example 1.

Solve the equation (x + 4) / 2 = 8 + (x – 6) / 5 and verify your answer?

Solution:

Given linear equation is (x + 4) / 2 = 8 + (x – 6) / 5

(x + 4) / 2 = [40 + (x – 6)] / 5

Cross multiply the fraction.

5(x + 4) = 2(40 + x – 6])

5x + 20 = 2(x + 34)

5x + 20 = 2x + 68

Transferring 2x to L.H.S. changes to negative 2x and 20 to R.H.S. changes to -20.

5x – 2x = 68 – 20

3x = 48

x = 48/3

x = 16

Therefere, x = 16 is the solution for the equation.

Verification:

L.H.S = (x + 4) / 2

Put x = 16

L.H.S = (16 + 4) / 2

= 20/2 = 20

R.H.S = 8 + (x – 6) / 5

Put x = 16

R.H.S = 8 + (16 – 6) / 5

= 8 + 10/5

= (40 + 10) / 5

= 50/5 = 10

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

Example 2.

Solve the equation 12 – x/6 = 7?

Solution:

Given linear equation is 12 – x/6 = 7

(72 – x) / 6 = 7/1

Cross multiply.

72 – x = 7 x 6

72 – x = 42

72 – 42 = x

30 = x

x = 30

Therefore, x = 30 is the solution for the equation.

Example 3.

Solve the equation 3(x – 3) = 4(2x + 1) and represent the solution graphically?

Solution:

Given linear equation is 3(x – 3) = 4(2x + 1)

3x – 9 = 8x + 4

-9 – 4 = 8x – 3x

-15 = 5x

x = -15/5

x = -3

Therefore, x = -3 is the solution for the equation.

Example 4.

Solve (x + 5) / (x – 5) = 1/2?

Solution:

Given linear equation is (x + 5) / (x – 5) = 1/2

Cross multiply the fraction

2(x + 5) = x – 5

2x + 10 = x – 5

2x – x = -5 – 10

x = -15

Therefore, x = -15 is the solution for the equation.

Find the steps of solving first-order linear equations with an example question. By following these steps, one can solve the linear equations in one variable, two variables, and three variables easily. Also, learn how to represent the linear equations on a graph from this page. So, read this complete page to get useful information. For the sake of your convenience, we even listed Linear Equations Examples with Solutions explained in detail.

Steps on How to Solve Linear Equations?

Find the step-by-step process for solving the linear equations by addition, subtraction, multiplication, and division in the following sections.

• Transfer the variable to the one side and constants to the other side of the equality sign by performing the required arithmetic operations.
• Now, make the variable equal to constant.
• Take a number line graph.
• Represent the solution on a graph by marking *.

What is Cross Multiplication?

If an equation has a fraction on both sides, then we will use cross multiplication. The process of multiplying the numerator of the left-hand side fraction with the denominator of the right-hand side fraction and multiplying the denominator of the left-hand side fraction with the numerator of the right-hand side fraction is called cross-multiplication. And equate both products to get a linear equation.

Solving Linear Equations Examples

Example 1.

Solve 0.5 – 1.5x = 0.8 – 0.28x.

Solution:

Given that,

0.5 – 1.5x = 0.8 – 0.28x

Transfer – 1.5x from the left side to right side. And transfer 0.8 from right side to the left side.

0.5 – 0.8 = 1.5x – 0.28x

-0.3 = 1.22x

x = -0.3 / 1.22

x = -0.245

Therefore, -0.245 is the required solution.

Verification:

L.H.S = 0.5 – 1.5x

Put x = -0.245

L.H.S = 0.5 – 1.5 x (-0.245)

= 0.5 + 1.5 x 0.245

= 0.5 + 0.3675

= 0.8675

R.H.S = 0.8 – 0.28x

Put x = -0.245

R.H.S = 0.8 – 0.28 x (-0.245)

= 0.8 + 0.28 x 0.245

= 0.8 + 0.0686

= 0.8686

Since, L.H.S = R.H.S. Hence proved.

Example 2.

Solve the equation 5 – 2(x – 1) = 4(3 – x) – 2x and represent the solution graphically.

Solution:

Given linear equation is 5 – 2(x – 1) = 4(3 – x) – 2x

Remove the braces and then simplify

5 – 2x + 2 = 12 – 3x – 2x

7 – 2x = 12 – 6x

Transfer -6x to L.H.S. changes to positive 6x and 7 to R.H.S. changes to negative 7.

6x – 2x = 12 – 7

4x = 5

x = 5/4

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

Example 3.

Solve the equation 6(2x + 3) + 5(4x – 6) – 8x = 5(3x + 1) + 6 (7x – 6) and verify your answer.

Solution:

Given linear equation is 6(2x + 3) + 5(4x – 6) – 8x = 5(3x + 1) + 6 (7x – 6)

12x + 18 + 20x – 30 – 8x = 15x + 5 + 42x – 36

24x – 12 = 57x – 31

31 – 12 = 57x – 24x

19 = 33x

19/33 = x

x = 19/33

Therefore, x = 19/33 is the solution for the equation.

Now we will verify both sides of the equation,

L.H.S = 6(2x + 3) + 5(4x – 6) – 8x

Sunstitute x = 19/33

L.H.S = 6[2 x 19/33 + 3] + 5[4 x 19/33 – 6] – 8 x 19/33

= 6[38/33 + 3] + 5[76/33 – 6] – 152/33

= 228/33 + 18 + 380/33 – 30 – 152/33

= 456/33 – 12

= (456 – 396)/33 = 60/33

R.H.S = 5(3x + 1) + 6 (7x – 6)

Sunstitute x = 19/33

R.H.S = 5[3 x 19/33 + 1] + 6[7 x 19/33 – 6]

= 5[57/33 + 1] + 6[133/33 – 6]

= 285/33 + 5 + 798/33 – 36

= 1083/33 – 31

= (1083 – 1023)/33 = 60/33

Since, L.h.S = R.H.S. Hence proved.

Example 4.

Solve 12/(x – 8) = 3/(x + 4).

Solution:

Given linear equation is 12/(x – 8) = 3/(x + 4)

Cross multiply and then remove the braces.

12(x + 4) = 3(x – 8)

12x + 48 = 3x – 24

12x – 3x = -24 – 48

9x = -72

x = -72/9 = -8

Therefore, x = -8 is the solution for the equation.

Verification:

L.H.S = 12/(x – 8)

Put x = -8

L.H.S = 12/(-8 – 8)

= 12/-16

= -3/4

R.H.S = 3/(x + 4)

Put x = -8

R.H.S = 3/(-8 + 4)

= 3/-4

= -3/4

Since L.H.S = R.H.S. Hence verified.

Linear equations are the first-order equations. The highest power of variables is 1 in the linear equation. These equations are defined as the straight lines in the coordinate geometry. The representation of straight-line equation is y = mx + c. Learn about the linear equations in one variable, two variables, and three variables and solving linear equations from the following sections.

Forms of Linear Equation

Some common forms of the linear equations are mentioned below:

General Form of Linear Equation:

Generally, linear equations are a combination of variables and constants. The standard form in one variable is ax + b = 0. Where a ≠ 0 and x is the variable. The standard form of a linear equation in two variables is ax + by + c = 0, Where a ≠ 0, b ≠ 0, x, and y are variables. The standard form of a linear equation in three variables is ax + by + cz + d = 0. Where a ≠ 0, b ≠ 0, c ≠ 0, x, y, z are variables.

Slope Intercept Form:

It is the most common form of the linear equation, which is represented as y = mx + c.
where x, y are the variables, m is the slope of the line, c is the intercept.

Point Slope Form:

Here the straight-line equation is formed by considering the points in the xy plane. such that (y – y₁) = m(x – x₁)

Where (x₁, y₁) are the coordinates of the line.

We can also express the point-slope form as y = mx + y₁ – mx₁.

Intercept Form:

A line that is neither parallel to axes of the coordinate plane nor passes through the origin but intersects the axes in two different points represents the intercept form. The intercept values x₀, y₀ are non zero and form the equation as x / x₀ + y / y₀ = 1

Two-Point Form:

If a line passes through the two points (x₁, y₁) and (x₂, y₂), then that equation of the line is given as y – y₁ = [(y₂ – y₁) / (x₂ – x₁)](x – x₁)

Here (y₂ – y₁) / (x₂ – x₁) is the slope of the line and x₁ ≠ x₂.

How to Solve Linear Equations?

We have given the step by step detailed explanation of how to solve the linear equation in one variable. Have a look at the following points and solve them easily. The first rule is both sides of the equation should be balanced. The equality sign between the expressions represents that both are equal. To balance that equation, we need to perform certain mathematical operations on both sides that should not change the actual equation.

Solution of Linear Equations in Two Variables

You can follow any of these methods to solve the linear equations in two variables. They are the method of substitution, method of elimination, cross multiplication method, and determinant methods

Solution of Linear Equations in Three Variables

To solve the linear equations with 3 variables, you need to follow the matrix method.

What is transposition?

Any term of an equation can be shifted to the other side with a change in its sign without affecting the equality. This process is called transposition.

While transposing a term from one side to the other side of equality you need to follow the below instructions

• We can simply change its sign and carry it to the other side.
• ‘+’ sign changes to ‘-‘ sign and vice-versa.
• ‘x’ sign of a term changes to ‘÷’ sign and vice-versa.
• Simplify the left-hand side expression such that each side contains one term.
• Finally, simplify the equation to get the variable value.

Linear Equations Questions with Solutions

Example 1.

Solve the linear equations in one variable.

(a) x + 8 = 56

(b) x – 5 = 20

Solution:

(a) x + 8 = 56

Subtract 8 from both sides.

x + 8 – 8 = 56 – 8

x = 48

(b) x – 5 = 20

Add 5 to both sides.

x – 5 + 5 = 20 + 5

x = 25

Example 2.

(a) x + ⅓ = 25

(b) 3x – 6 = 9

Solution:

(a) x + ⅓ = 25

Subtract ⅓ from both sides.

x + ⅓ – ⅓ = 25 – ⅓

x = (25 x 3 – 1) / 3

x = (75 – 1) / 3

x = 74 / 3.

(b) 3x – 6 = 9

Add 6 to both sides.

3x – 6 + 6 = 9 + 6

3x = 15

Divide 3 by both sides.

3x / 3 = 15 / 3

x = 5.

Example 3.

Solve the following linear equations.

(a) 0.5x = 0.3

(b) 2x = 12

Solution:

(a) 0.5x = 0.3

Divide both sides by 0.5

0.5x / 0.5 = 0.3 / 0.5

x = 0.6

(b) 2x = 12

Divide both sides by 2

2x / 2 = 12 / 2

x = 6.

The equation is a mathematical expression that is enclosed between two statements. It contains equals (=) symbol. It says that two things are equal. ‘=’ is an operator that is used only in the equations. Read further to know more about the mathematical sentence, open sentence, equations, and type of equations. You can also find some examples of the equations in the below sections along with solutions for a better understanding of the concepts.

Equation Definition

A statement of equality of two algebraic expressions that involve one or more variables is called the equation. Every equation has an “=” symbol. The statement or expression at the left side of the “=” symbol is called the left-hand side expression and the expression which is present at the right side of the symbol is called the right-hand side expression.

Example:

4 – x = 10

The set of values of x which satisfy the equation is called the solution set.

Open Sentence & Mathematical Sentence

The mathematical sentence is a statement whether it is true or false but not both. Some examples are 2 + 3 = 5 which is true, 8 – 5 > 1 which is also true. 8 x 2 <15 a false statement.

Let us take one sentence 3 + x = 18 that may be true or false based on the value of x. So, these type of mathematical sentences having variables become either true or false depending on the literal value is called an open sentence.

Types of Algebraic Equations

We have different types of equations and they are given along the lines

1. Linear Equation: The terms of the linear equations are either a constant or a single variable or a product of both. The general form of the linear equation with two variables is the slope-intercept form. It is given by y = mx + c

Here m is not zero.

m is the slope

c is the y-intercept

The examples of Linear equations with one variable is x + 1 = 8, with two variables is x + y = 1.

2. Quadratic Equation: It is a second-order linear equation with two variables. In these equations, any one of the variable exponents must be 2. The general form of quadratic equation is ax² + bx + c = 0, a ≠ 0.

An example of quadratic equation is 5x² + 2y = 25

3. Radical Equation: In radical equations, the variable highest exponent is ½ or you can say that the variable is lying inside the square root.

An example of a radical equation is √x – 6 = 30

4. Exponential Equation: In this type of equation, the variables are there in place of exponents. By using the exponential equation property, it can be solved.

An example of an exponential equation is 2^x = 56.

5. Rational Equation: Rational math equations contain rational expressions.

An example is 8 / x = (x + 7) / 25

Equation Examples with Solutions

Example 1.

Find the solution set for the following open sentences.

(a) x + 5 = 13

(b) x/2 > 5

Solution:

(a) x + 5 = 13

If x = 6, then 6 + 5 ≠ 13

If x = 7, then 7 + 5 ≠ 13

If x = 8, then 8 + 5 = 13

Therefore, the solution set for the open sentence x + 5 = 13 is 8.

(b) x/2 > 5

If x = 8, then 8/2 ≯  5

If x = 9, then 9/2 ≯  5

If x = 10, then 10/2 ≯  5

If x = 11, then 11/2 > 5

If x = 12, then 12/2 > 5

Therefore, the solution set for the open sentence x/2 > 5 is 11, 12, 13, . . .

Example 2.

Find the solution set for the equations.

(a) 2x + 8 = 18    (b) x + 7 = 14

Solution:

(a) 2x + 8 = 18

If x = 3, then 2(3) + 8 = 6 + 8 ≠ 18

If x = 4, then 2(4) + 8 = 8 + 8 ≠ 18

If x = 5, then 2(5) + 8 = 10 + 8 = 18

Therefore, the solution set for the equation 2x + 8 = 18 is 5.

(b) x + 7 = 14

If x = 5, then 5 + 7 ≠ 14

If x = 6, then 6 + 7 ≠ 14

If x = 7, then 7 + 7 = 14

Therefore, the solution set for the equation x + 7 = 14 is 7.

Example 3.

Find the values of variables for the following.

(a) 5^x = 125

(b) x/2 = (x + 2) / 4

Solution:

(a) 5^x = 125

125 can be written as 5 x 5 x 5 = 5³

5^x = 5³

Here the bases are equal so equate the powers.

x = 3.

(b) x/2 = (x + 2) / 4

Cross multiply the numerators and denominators

4x = 2(x + 2)

4x = 2x + 4

4x – 2x = 4

2x = 4

If x = 1, then 2(1) ≠ 4

If x = 2, then 2(2) = 4.

Therefore, x = 2.

Before you start practicing the Difference of Two Squares Concept know the step-by-step process to solve the Factoring Difference of Two Squares. Therefore, students can easily learn all types of factoring problems here. Without any late, begin your practice and finish solving every problem included. Refer to Solved Examples on Difference of Two Squares with Answers Provided.

Difference of Two Squares Questions

1. m⁴ – (n + r)⁴

Solution:
Given expression is m⁴ – (n + r)⁴
Rewrite the given expression in the form of a² – b².
(m²)² – ( (n + r)²)²
Now, apply the formula of a² – b² = (a + b) (a – b), where a = m² and b = (n + r)²
[m² + (n + r)²] [m² – (n + r)²]
[m² + n² + r² + 2nr] [(m)² – (n + r)²]
From the above equation, [(m)² – (n + r)²] is in the form of a² – b².
[(m)² – (n + r)²]
Now, apply the formula of a² – b² = (a + b) (a – b), where a = m and b = (n + r)
[m + (n + r)] [m – (n + r)]
Now, [m² + n² + r² + 2nr] [(m)² – (n + r)²]
[m² + n² + r² + 2nr] [m + (n + r)] [m – (n + r)]
[m² + n² + r² + 2nr] [m + n + r] [m – n – r]

The final answer is [m² + n² + r² + 2nr] [m + n + r] [m – n – r]

2. 4a² – b² + 6b – 9.

Solution:
Given expression is 4a² – b² + 6b – 9.
Rewrite the given expression.
4a² – (b² – 6b + 9)
b² – 6b + 9 is in the form of a² – b² + 2ab where a = b, b = 3
We know that a² – b² + 2ab = (a – b)²
Therefore, b² – 6b + 9 = (b – 3)²
So, 4a² – (b – 3)²
The above equation 4a² – (b – 3)² is in the form of a² – b².
[(2a)² – (b – 3)²]
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 2a and b = (b – 3)
(2a + b – 3) {2a – (b – 3)},
(2a + b – 3) (2a – b – 3)

The final answer is (2a + b – 3) (2a – b – 3)

3. 25x² – (4m² – 12mn + 9n²)

Solution:
Given expression is 25x² – (4m² – 12mn + 9n²)
(4m² – 12mn + 9n²) is in the form of a² – b² + 2ab where a = 2m, b = 3n
We know that a² – b² + 2ab = (a – b)²
Therefore, (4m² – 12mn + 9n²) = (2m – 3n)²
So, 25x² – (2m – 3n)²
The above equation 25x² – (2m – 3n)² is in the form of a² – b².
[(5x)² – (2m – 3n)²]
Now, apply the formula of a² – b² = (a + b) (a – b), where a = 5x and b = (2m – 3n)
[5x + (2m – 3n)] [5x – (2m – 3n)]
(5x + 2m – 3n) (5x – 2m + 3n)

The final answer is (5x + 2m – 3n) (5x – 2m + 3n)

Know the process to find Factorization by using identities. Based on the identities, we can simply factorize an algebraic equation. That means, depending on the identities or identity values, we can easily reduce the number of expressions into n number of terms. Generally, for the simplest factorization process, we have to follow some basic expressions. They are

(i) (x + y)^2 = (x)^2 + 2xy + (y)^2.
(ii) (x – y)^2 = (x)^2 – 2xy + (y)^2.
(iii) x^2 – y^2 = (x + y) (x – y).

Factoring Polynomials Identities Steps

Go through the below-listed guidelines on how to factor polynomials using algebraic identities. They are in the following fashion

• Note down the given expression and compare the expression with the basic expressions.
• If there are three terms and all are identified with the positive sign, then that is related to (x + y)^2.
• If the first and last terms are with the positive sign and the middle term is identified with the negative sign, then that is related to (x – y)^2.
• If there are only two terms with one positive sign and one negative sign, then that is related to x^2 – y^2.
• Now, compare the co-efficient values with the basic expressions.
• Find the values of coefficients.
• Based on the values, reduce the given expression into simple terms.

Factorization Using Identities Examples with Answers

1. Factorize using the formula of a square of the sum of two terms

(i) a² + 6a + 9

Solution: Given expression is a² + 6a + 9
There are three terms are identified with the positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a² + 6a + 9 with the expression (x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 6 then y = 3.
So, (x + y)^2 = (a + 3)^2.

Finally, our expression a^2 + 6a + 9 is reduced to (a + 3)^2.

(ii) a^2 + 20a + 100

Solution:
Given expression is a^2 + 20a + 100.
There are three terms that are identified with a positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.
Now, compare the co-efficient values of a^2 + 20a + 100 with the expression(x)^2 + 2xy + (y)^2.
Here, x = a, 2y = 20 then y = 10.
So, (x + y)^2 = (a + 10)^2.

Finally, our expression a^2 + 20a + 100 is reduced to (a + 10)^2.

2. Factorize using the formula of the square of the difference of two terms

(i) 4p^2 – 12pq + 9q^2

Solution:
Given expression is 4p^2 – 12pq + 9q^2.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of 4p^2 – 12pq + 9q^2with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = 4p^2 then x =2p, y^2 = 9q^2 then y = 3q.
So, (x – y)^2 = (2p – 3q)^2.

Finally, our expression 4p^2 – 12pq + 9q^2is reduced to (2p – 3q)^2.

(ii) a^2 – 10a + 25

Solution:
Given expression is a^2 – 10a + 25.
There are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.
Then it is related to the expression (x)^2 – 2xy + (y)^2 = (x – y)^2.
Now, compare the co-efficient values of a^2 – 10a + 25 with the expression (x)^2 – 2xy + (y)^2.
Here, x^2 = a^2 then x =a, y^2 = 25 then y = 5.
So, (x – y)^2 = (a – 5)^2.

Finally, our expression a^2 – 10a + 25 is reduced to (a – 5)^2.

3. Factorize using the formula of a difference of two squares:

(i) 49a^2 – 64

Solution:
Given expression is 49a^2 – 64
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 49a^2 – 64 with the expression (x)^2 – (y)^2.
Here, x^2 = 49a^2 = (7a)^2 then x =7a, y^2 = 64 = (8)^2 then y = 8.
So, (x)^2 – (y)^2 = (7a)^2 – (8)^2.
(x + y) (x – y) = (7a + 8) (7a – 8)

Finally, our expression 49a^2 – 64is reduced to (7a + 8) (7a – 8).

(ii) 16a^2 – 36b^2

Solution:
Given expression is 16a^2 – 36b^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of 16a^2 – 36b^2 with the (x)^2 – (y)^2.
Here, x^2 = 16a^2 = (4a)^2 then x =4a, y^2 = 36b^2 = (6b)^2 then y = 6b.
So, (x)^2 – (y)^2 = (4a)^2 – (6b)^2.
(x + y) (x – y) = (4a + 6b) (4a – 6b)

Finally, our expression 16a^2 – 36b^2 is reduced to (4a + 6b) (4a – 6b).

(iii) 1 – [5(2p – 5q)]^2

Solution:
Given expression is 1 – [5(2p – 5q)]^2
We can write it as (1)^2 – [5(2p – 5q)]^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (1)^2 – [5(2p – 5q)]^2 with the (x)^2 – (y)^2.
Here, x^2 = (1)^2 then x =1,
y^2 = [5(2p – 5q)]^2then y = 5(2p – 5q).
So, (x)^2 – (y)^2 = (1)^2 – [5(2p – 5q)]^2.
(x + y) (x – y) = [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].

Finally, our expression 1 – 25(2p – 5q)^2 is reduced to [1 + 5(2p – 5q)] [1 – 5(2p – 5q)].

4. Factor completely using the formula of a difference of two squares

(i) p^4 –q^4

Solution:
Given expression is p^4 –q^4
We can write it as (p^2)^2 – (q^2)^2
There are two terms identified with the positive sign of the first term and the negative sign of the last term.
Now, compare the co-efficient values of (p^2)^2 – (q^2)^2with the (x)^2 – (y)^2.
Here, x^2 = (p^2)^2 then x =p^2,
y^2 = (q^2)^2 then y = q^2.
So, (x)^2 – (y)^2 = (p^2)^2 – (q^2)^2.
(x + y) (x – y) = (p^2 + q^2) (p^2 –q^2).

Finally, our expression p^4 – q^4 is reduced to (p^2 + q^2) (p^2 – q^2).

Factorization of Quadratic Trinomials is the process of finding factors of given Quadratic Trinomials. If ax^2 + bx + c is an expression where a, b, c are constants, then the expression is called a quadratic trinomial in x. The expression ax^2 + bx + c has an x^2 term, x term, and an independent term. Find Factoring Quadratics Problems with Solutions in this article.

Factorization of Quadratic Trinomials Forms

The Factorization of Quadratic Trinomials is in two forms.

(i) First form: x^2 + px + q
(ii) Second form: ax^2 + bx + c

How to find Factorization of Trinomial of the Form x^2 + px + q?

If x^2 + px + q is an Quadratic Trinomial, then x² + (m + n) × + mn = (x + m)(x + n) is the identity.

Solved Examples on Factorization of Quadratic Trinomial of the Form x^2 + px + q 

1. Factorize the algebraic expression of the form x^2 + px + q

(i) a² – 7a + 12

Solution:
The Given expression is a² – 7a + 12.
By comparing the given expression a² – 7a + 12 with the basic expression x^2 + px + q.
Here, a = 1, b = – 7, and c = 12.
The sum of two numbers is m + n = b = – 7 = – 4 – 3.
The product of two number is m * n = a * c = -4 * (- 3) = 12
From the above two instructions, we can write the values of two numbers m and n as – 4 and -3.
Then, a² – 7a + 12 = a² – 4a -3a + 12.
= a (a – 4) – 3(a – 4).
Factor out the common terms.

Then, a² – 7a + 12 = (a – 4) (a – 3).

(ii) a² + 2a – 15

Solution:
The Given expression is a² + 2a – 15.
By comparing the given expression a² + 2a – 15 with the basic expression x^2 + px + q.
Here, a = 1, b = 2, and c = -15.
The sum of two numbers is m + n = b = 2 = 5 – 3.
The product of two number is m * n = a * c = 5 * (- 3) = -15
From the above two instructions, we can write the values of two numbers m and n as 5 and -3.
Then, a² + 2a – 15 = a² + 5a – 3a – 15.
= a (a + 5) – 3(a + 5).
Factor out the common terms.

Then, a² + 2a – 15 = (a + 5) (a – 3).

How to find Factorization of trinomial of the form ax^2 + bx + c?

To factorize the expression ax^2 + bx + c we have to find the two numbers p and q, such that p + q = b and p × q = ac

Solved Examples on Factorization of trinomial of the form ax^2 + bx + c 

2. Factorize the algebraic expression of the form ax2 + bx + c

(i) 15b² – 26b + 8

Solution:
The Given expression is 15b² – 26b + 8.
By comparing the given expression 15b² – 26b + 8 with the basic expression ax² + bx + c.
Here, a = 15, b = -26, and c = 8.
The sum of two numbers is p + q = b = -26 = 5 – 3.
The product of two number is p * q = a * c = 15 * (8) = 120
From the above two instructions, we can write the values of two numbers p and q as -20 and -6.
Then, 15b² – 26b + 8 = 15b² – 20 – 6b + 8.
= 5b (3b – 4) – 2(3b – 4).
Factor out the common terms.

Then, 15b² – 26b + 8 = (3b – 4) (5b – 2).

(ii) 3a² – a – 4

Solution:
The Given expression is 3a² – a – 4.
By comparing the given expression 3a² – a – 4 with the basic expression ax² + bx + c.
Here, a = 3, b = -1, and c = -4.
The sum of two numbers is p + q = b = -1 = 5 – 3.
The product of two number is p * q = a * c = 3 * (-4) = -12
From the above two instructions, we can write the values of two numbers p and q as -4 and 3.
Then, 3a² – a – 4 = 3a² – 4a -3a – 4.
= a (3a – 4) – 1(3a – 4).
Factor out the common terms.

Then, 3a² – a – 4 = (3a – 4) (a – 1).