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. 

After having gone through the stuff given above, we hope that the students would have understood, "Solving Linear Equations in One Variable Worksheet". 

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