# Word Problems on Linear Equations | How to Solve Linear Equations Word Problems?

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.

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