SOLVING LINEAR EQUATIONS IN ONE VARIABLE WORKSHEET

Problems 1-10 : Solve for x in each case.

Problem 1 :

3x + 5 = 19

Problem 2 :

ˣ⁄₃₀ = ²⁄₄₅

Problem 3 :

⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄

Problem 4 :

⁴ˣ⁄₃ - 1 = ¹⁴ˣ⁄₁₅ + ¹⁹⁄₅

Problem 5 :

²⁽ˣ ⁺ ¹⁾⁄₃ = ³⁽ˣ ⁻ ²⁾⁄₅

Problem 6 :

⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11

Problem 7 :

²⁄₍ₓ ₊ ₁₎ = 4 - ˣ⁄₍ₓ ₊ ₁₎

Problem 8 :

⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉ = ⁽ˣ ⁺ ⁷⁾⁄₄

Problem 9 :

(x + 2)(x - 3) + (x + 3)(x - 4) = x(2x - 5)

Problem 10 :

ˣ⁄₀.₁ - ¹⁄₀.₀₁ + ˣ⁄₀.₀₀₁ - ¹⁄₀.₀₀₀₁ = 0

Problem 11 :

Check whether ¼ is a solution to the following linear equation :

3(x + 1) = 3(5 – x) – 2(5 + x)

Problem 12 :

When 5 times a number y is divided by 8, the result is 15. What is the value of y?

Problem 13 :

3 more than 5 times of a number is equal to -12. Find the number.

Problem 14 :

Three times the present age of a man is equal to 10 more than twice his age 5 years hence. Find the present age of the man.

Answers

1. Answer :

To find the value of x, we have to isolate it. To isolate x, we have to get rid of all the values around x using the properties of equality.

3x + 5 = 19

Subtract 5 from both sides.

3x = 15

Divide both sides by 3.

x = 5

2. Answer :

ˣ⁄₃₀ = ²⁄₄₅

To solve the above equation, we have to get rid of the denominators 30 and 45.

Least common multiple of (30, 45) = 90.

Multiply both sides of the equation by the least common multiple 15 to get rid of the denominators.

90(ˣ⁄₃₀) = 90(²⁄₄₅)

3x = 2(2)

3x = 4

Divide both sides by 3.

x = ⁴⁄₃

x = 1⅓

3. Answer :

⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄

Least common multiple of the denominators (5, 4) = 20.

Multiply both sides of the equation by the least common multiple 20 to get rid of the denominators.

20[⁽ˣ ⁺ ²⁴⁾⁄₅] = 20(4 + ˣ⁄₄)

4(x + 24) = 20(4) + 20(ˣ⁄₄)

4x + 96 = 80 + 5x

Subtract 4x from both sides.

96 = 80 + x

Subtract 80 from both sides.

16 = x

4. Answer :

⁴ˣ⁄₃ - 1 = ¹⁴ˣ⁄₁₅ + ¹⁹⁄₅

Least common multiple of the denominators (3, 15, 5) = 15.

Multiply both sides of the equation by the least common multiple 15 to get rid of the denominators.

15(⁴ˣ⁄₃ - 1) = 15(¹⁴ˣ⁄₁₅ + ¹⁹⁄₅)

15(⁴ˣ⁄₃) + 15(-1) = 15(¹⁴ˣ⁄₁₅) + 15(¹⁹⁄₅)

20x - 15 = 14x + 57

Subtract 14x from both sides.

6x - 15 = 57

Add 15 to both sides.

6x = 72

Divide both sides by 6.

x = 12

5. Answer :

²⁽ˣ ⁺ ¹⁾⁄₃ = ³⁽ˣ ⁻ ²⁾⁄₅

Do cross multiplication.

5[2(x + 1)] = 3[3(x - 2)]

10(x + 1) = 9(x - 2)

10x - 10 = 9x - 18

Subtract 9x from both sides.

x - 10 = -18

Add 10 to both sides.

x = -8

6. Answer :

⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11

Least common multiple of the denominators (4, 3) = 12.

Multiply both sides of the equation by the least common multiple 12 to get rid of the denominators.

12[⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃] = 12(11)

12[⁽ˣ ⁺ ⁴⁾⁄₄] + 12[⁽ˣ ⁻ ⁵⁾⁄₃] = 132

3(x + 4) + 4(x - 5) = 132

3x + 12 + 4x - 20 = 132

7x - 8 = 132

Add 8 to both sides.

7x = 140

Divide both sides by 7.

x = 20

7. Answer :

²⁄₍ₓ ₊ ₁₎ = 4 - ˣ⁄₍ₓ ₊ ₁₎

Multiply both sides by (x + 1) to get rid of the denomintors.

2 = 4(x + 1) - x

Use distributive property.

2 = 4x + 4 - x

2 = 3x + 4

Subtract 4 from both sides.

2 - 4 = 3x

-2 = 3x

Divide both sides by 3.

⁻²⁄₃ = x

8. Answer :

⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉ = ⁽ˣ ⁺ ⁷⁾⁄₄

Least common multiple of the denominators (6, 9, 4) = 36.

Multiply both sides of the equation by the least common multiple 36 to get rid of the denominators.

36[⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉] = 36[⁽ˣ ⁺ ⁷⁾⁄₄]

36[⁽ˣ ⁺ ¹¹⁾⁄₆] - 36[⁽ˣ ⁺ ¹⁾⁄₉] = 36[⁽ˣ ⁺ ⁷⁾⁄₄]

6(x + 11) - 4(x + 1) = 9(x + 7)

6x + 66 - 4x - 4 = 9x + 63

2x + 62 = 9x + 63

Subtract 2x from both sides.

62 = 7x + 63

Subtract 63 from both sides.

-1 = 7x

Divide both sides by 7.

⁻¹⁄₇ = x

9. Answer :

(x + 2)(x - 3) + (x + 3)(x - 4) = x(2x - 5)

x² - 3x + 2x - 6 + x² - 4x + 3x - 12 = 2x² - 5x

2x² - 2x - 18 = 2x² - 5x

Subtract 2x² from both sides.

-2x - 18 = -5x

Add 5x to both sides.

3x - 18 = 0

Add 18 to both sides.

3x = 18

Divide both sides by 3.

x = 6

10. Answer :

ˣ⁄₀.₁ - ¹⁄₀.₀₁ + ˣ⁄₀.₀₀₁ - ¹⁄₀.₀₀₀₁ = 0

x(¹⁰⁄₁) - 1(¹⁰⁰⁄₁) + x(¹⁰⁰⁰⁄₁) - 1(¹⁰⁰⁰⁰⁄₁) = 0

10x - 100 + 1000x - 10000 = 0

1010x - 10100 = 0

Add 10100 to both sides.

1010x = 10100

Divide both sides by 1010.

x = 10

11. Answer :

3(x + 1) = 3(5 – x) – 2(5 + x)

To check 1/4 is the solution to the given linear equation, substitute x = 1/4 in the equation.

3(¼ + 1) = 3(5 – ¼) – 2(5 + ¼)

3(⁵⁄₄) = 3(¹⁹⁄₄) - 2(²¹⁄₄)

¹⁵⁄₄ = ⁵⁷⁄₄ - ⁴²⁄₄

¹⁵⁄₄ = ⁽⁵⁷ ⁻ ⁴²⁾⁄₄

¹⁵⁄₄ = ¹⁵⁄₄ 

Since ¼ makes the given equation true, it is the solution.

12. Answer :

From rthe given information, we have

⁵ʸ⁄₈ = 15

Multiply both sides by 8.

5y = 120

Divide both sides by 5.

y = 24

13. Answer :

Let x be the required number.

It is given that 3 more than 5 times of a number is equal to -12.

5x + 3 = -12

Subtract 3 from both sides.

5x = -15

Divide both sides by 5.

x = -3

The number is -3.

14. Answer :

Let x be the present age of the man.

Age of the man 5 years hence = x + 5

It is given that three times the present age of the man is equal to 10 more than twice his age 5 years hence.

3x = 2(x + 5) + 10

3x = 2x + 10 + 10

3x = 2x + 20

Subtract 2x from both sides.

x = 20

The present age of the man is 20 years. 

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