# Problems on Linear Equations in One Variable | Linear Equations in One Variable Practice Questions

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.