LINEAR EQUATIONS IN ONE VARIABLE

Linear equation in one variable is an equation which contains only one variable with the largest exponent 1. A linear equation in one variable can be expressed in the form

ax + b = 0

Here, a and b are constants. The above equation has only one solution.

Solving an equation is to find the value of the varianble. To solve a linear equation in one variable, you have to get rid of all the values around the variable. That is, you have to isolate the variable.

Solved Problems

Solve each of the following linear equations in one variable.

Problem 1 :

2x + 1 = 9

Solution :

2x + 1 = 9

Subtract 1 from both sides.

2x = 8

Divide both sides by 2.

x = 4

Problem 2 :

6y – 5 = 7

Solution :

6y – 5 = 7

Add 5 to both sides.

6y = 12

Divide both sides by 6.

y = 2

Problem 3 :

8 = 3a – 4

Solution :

8 = 3a – 4

Add 4 to both sides.

12 = 3a

Divide both sides by 3.

4 = a

Problem 4 :

ᵐ⁄₅ + 9 = 11

Solution :

ᵐ⁄₅ + 9 = 11

Subtract 9 from both sides.

ᵐ⁄₅ = 2

Multiply both sides by 5.

m = 10

Problem 5 :

13 + 7x = 27

Solution :

13 + 7x = 27

Subtract 13 from both sides.

7x = 14

Divide both sides by 7.

x = 2

Problem 6 :

17 – q = 6

Solution :

17 – q = 6

Subtract 17 from both sides.

-q = -11

Multiply both sides by -1.

q = 11

Problem 7 :

Solution :

Multiply both sides by 2.

x – 31 = 8

Add 31 to both sides.

x = 39

Problem 8 :

1 + 2r = 35

Solution :

1 + 2r = 35

Subtract 1 from both sides.

2r = 34

Divide both sides by 2.

r = 17

Problem 9:

42 + 5t = 8t

Solution :

42 + 5t = 8t

Subtract 5t from both sides.

42 = 3t

Divide both sides by 3.

14 = t

Problem 10 :

4p – 3 = 17 + 2p

Solution :

4p – 3 = 17 + 2p

Subtract 2p from both sides.

2p – 3 = 17

Add 3 to both sides.

2p = 20

Divide both sides by 2.

p = 10

Problem 11 :

5 more than 3 times of a number is equal to 35. Find the number.

Solution :

Let x be the number.

From the given information,

3x + 5 = 35

Subtract 5 from bopth sides.

3x = 30

Divide both sides by 3.

x = 10

Therefore, the number is 10.

Problem 12 :

If Claire paid $255 dollars for a computer that was only 20 dollars more than half the original price, what was the original price, in dollars?

Answer :

Let x be the original price of the computer.

From the given information,

ˣ⁄₂ + 20 = 255

Subtract 20 from both sides.

ˣ⁄₂ = 235

Multiply bothn sides by 2.

x = 470

Therefore, the original price of the computer was $470.

Problem 13 :

If thrice of A’s age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A’s present age.

Answer :

Let a be the present age of A.

From the given information,

2a - 3(a - 6) = a

Use the distributive property.

2a - 3a + 18 = a

-a + 18 = a

Add a to both sides.

18 = 2a

Divide both sides by 2.

9 = a

Therefore, A's present age is 9 years.

Problem 14 :

Divide 56 into two parts such that three times the first part exceeds one third of the second by 48. Find the two parts of 56. 

Answer :

Let x be the first part. Then the second part is (56 - x).

From the given information,

3x = ()(56 - x) + 48

Multiply both sides by 3.

3(3x) = 3[()(56 - x) + 48]

9x = 3()(56 - x) + 3(48)

9x = 56 - x + 144

9x = 200 - x

Add x to both sides.

10x = 200

Divide both sides by 10.

x = 20

56 - x = 36

Therefore, the two parts of 56 are 20 and 36.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Trigonometric Identities with Solutions

    Mar 03, 24 08:27 PM

    Problems on Trigonometric Identities with Solutions

    Read More

  2. Solved Problems on Binomial Expansion

    Mar 03, 24 10:46 AM

    Solved Problems on Binomial Expansion

    Read More

  3. P-Series Test for Convergence and Divergence

    Mar 03, 24 06:28 AM

    P-Series Test for Convergence and Divergence

    Read More