# 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. 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

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 = 1

⁽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

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

6x = 72

Divide both sides by 6.

x = 12

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

x = -8

⁽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

7x = 140

Divide both sides by 7.

x = 20

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

⁽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

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

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

2x2 - 2x - 18 = 2x2 - 5x

Subtract 2xfrom both sides.

-2x - 18 = -5x

3x - 18 = 0

3x = 18

Divide both sides by 3.

x = 6

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

1010x = 10100

Divide both sides by 1010.

x = 10

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.

From rthe given information, we have

⁵ʸ⁄₈ = 15

Multiply both sides by 8.

5y = 120

Divide both sides by 5.

y = 24

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.

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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