LINEAR EQUATIONS IN ONE 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 = (⅓)(56 - x) + 48

Multiply both sides by 3.

3(3x) = 3[(⅓)(56 - x) + 48]

9x = 3(⅓)(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.

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