# SOLVING LINEAR EQUATIONS IN ONE VARIABLE WORKSHEET

## About "Solving Linear Equations in One Variable Worksheet"

Solving Linear Equations in One Variable Worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on how to solve linear equations in one variable.

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## Solving Linear Equations in One Variable Worksheet - Problems

Problem 1 :

Solve for x :

x - 3  =  2

Problem 2 :

Solve for x :

3x - 2(x - 1)  =  5

Problem 3 :

Solve for x :

3x - 8  =  x + 10

Problem 4 :

Solve for x :

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

Problem 5 :

Solve for x :

4x/5 - 7/4  =  x/5 + x/4

Problem 6 :

Solve for x :

4x/3 - 1  =  14x/15 + 19/5

Problem 7 :

The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator, the fraction increases by unity. Find the fraction.

Problem 8 :

Three consecutive integers add up to 51. What are these integers?

Problem 1 :

Solve for x :

x - 3  =  2

Solution :

To solve for x, add 3 to each side of the equation.

x - 3 + 3  =  2 + 3

x  =  5

Problem 2 :

Solve for x :

3x - 2(x - 1)  =  5

Solution :

Simplify the left side of the equation.

3x - 2(x - 1)  =  5

3x - 2x + 2  =  5

Combine the like terms.

x + 2  =  5

Subtract 2 from each side of the equation.

x + 2 - 2  =  5 - 2

x  =  3

Problem 3 :

Solve for x :

3x - 8  =  x + 10

Solution :

Subtract x from each side of the equation.

3x - 8 - x  =  x + 10 - x

Combine the like terms.

2x - 8  =  10

Add 8 to each side of the equation.

2x - 8 + 8  =  10 + 8

2x  =  18

Divide each side by 2.

2x / 2  =  18 / 2

x  =  9

Problem 4 :

Solve for x :

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

Solution :

Simplify each side of the equation.

5(x - 3) - 7(6 - x)  =  24 - 3(8 - x) - 3

5x - 15 - 42 + 7x  =  24 - 24 + 3x - 3

Combine the like terms.

12x - 57  =  3x - 3

Subtract 3x from each side of the equation.

12x - 57 - 3x  =  3x - 3 - 3x

9x - 57  =  - 3

Add 57 to each side of the equation.

9x - 57 + 57  =  - 3 + 57

9x  =  54

Divide each side by 9.

9x / 9  =  54 / 9

x  =  6

Problem 5 :

Solve for x :

4x/5 - 7/4  =  x/5 + x/4

Solution :

In the given equation, we have the denominators 4 and 5. The least common multiple of 4 and 5 is 20.

Now, we can proceed as follows.

4x/5 - 7/4  =  x/5 + x/4

16x/20 - 35/20  =  4x/20 + 5x/20

(16x - 35)/20  =  (4x + 5x)/20

(16x - 35)/20  =  9x/20

Multiply each side by 20.

20 ⋅ [(16x - 35)/20]  =  (9x/20) ⋅ 20

16x - 35  =  9x

Subtract 9x from each side.

16x - 35 - 9x  =  9x - 9x

7x - 35  =  0

Add 35 to each side.

7x - 35 + 35  =  0 + 35

7x  =  35

Divide each side by 7.

7x / 7  =  35 / 7

x  =  5

Problem 6 :

Solve for x :

4x/3 - 1  =  14x/15 + 19/5

Solution :

Simplify each side of the equation.

4x/3 - 1  =  14x/15 + 19/5

4x/3 - 3/3  =  14x/15 + 57/15

(4x - 3) / 3  =  (14x + 57) / 15

Least common multiple of 3 and 15 is 15. So, multiply each side of the equation by 15.

15 ⋅ [(4x - 3) / 3]  =  15 ⋅ [ (14x + 57) / 15 ]

Simplify.

(4x - 3)  =  14x + 57

20x - 15  =  14x + 57

Subtract 14x from each side of the equation.

20x - 15 - 14x  =  14x + 57 - 14x

6x - 15  =  57

Add 15 to each side of the equation.

6x - 15 + 15  =  57 + 15

6x  =  72

Divide each side by 6.

6x / 6  =  72 / 6

x  =  12

Problem 7 :

The denominator of a fraction exceeds the numerator by 2. If 5 be added to the numerator, the fraction increases by unity. Find the fraction.

Solution :

Let x be the numerator of the fraction.

Then the fraction is

x / (x + 2) -----(1)

Given : If 5 be added to the numerator, the fraction increases by unity.

(x + 5) / (x + 2)  =  [x / (x + 2)] + 1

Simplify.

(x + 5) / (x + 2)  =  [x / (x + 2)] + [(x + 2) / (x + 2)]

(x + 5) / (x + 2)  =  (x + x + 2) / (x + 2)

Multiply each side by (x + 2).

x + 5  =  2x + 2

Subtract x from each side.

x + 5 - x  =  2x + 2 - x

5  =  x + 2

Subtract 2 from each side.

5 - 2  =  x + 2 - 2

3  =  x

Plug x  =  3 in (1).

(1)----->  x / (x + 2)  =  3 / (3 + 2)

x / (x + 2)  =  3 / 5

Hence, the required fraction is 3/5.

Problem 8 :

Three consecutive integers add up to 51. What are these integers?

Solution :

Let x be the first integer.

Then the remaining two consecutive integers are

(x + 1) and (x + 2)

So, the three consecutive integers are

x, (x + 1), and (x + 2)

Given : Three consecutive integers add up to 51.

So, we have

x + (x + 1) + (x + 2)  =  51

x + x + 1 + x + 2  =  51

Combine the like terms.

3x + 3  =  51

Subtract by 3 from each side side.

3x + 3 - 3  =  51 - 3

3x  =  48

Divide each side by 3.

3x / 3  =  48 / 3

x  =  16

Hence, the three consecutive integers are 16, 17 and 18.

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