SOLVING SIMPLE EQUATIONS WORKSHEET

Question 1 :

Solve for x.

x + 11 = 17

Question 2 :

Solve for p.

3p + 4  =  25

Question 3 :

Solve for y.

5y + 3 = 17 - 2y

Question 4 :

Solve for x.

12 - x = 8

Question 5 :

Solve for s.

ˢ⁄₆ = 7

Question 6 :

Solve for a.

⁽³ᵃ ⁺ ²⁾⁄₄ = 5

Question 7 :

Solve for y.

⁴ʸ⁄₃ - 1 = ¹⁴ʸ⁄₁₅ + ¹⁹⁄₅

Question 8 :

Solve for x.

⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11

Question 9 :

Solve for x.

⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄

Question 10 :

Solve for x.

ˣ⁄₀.₅ - ¹⁄₀.₀₅ + ˣ⁄₀.₀₀₅ - ¹⁄₀.₀₀₀₅ = 0

Question 11 :

How many cookies are in a box, if the box plus three more equals 23 cookies?

Question 12 :

6 less than 7 times of a number is 57. What is the number?

x + 11 = 17

Subtract 11 from both sides.

(x + 11) - 11 = 17 - 11

x + 11 - 11 = 6

x = 6

3p + 4 = 25

Subtract 4 from both sides.

(3p + 4) - 4 = 25 - 4

3p + 4 - 4 = 21

3p = 21

Divide both sides by 3.

³ᵖ⁄₃ = ²¹⁄₃

p = 7

5y + 3 = 17 - 2y

(5y + 3) + 2y = (17 - 2y) + 2y

5y + 3 + 2y = 17 - 2y + 2y

7y + 3 = 17

Subtract 3 from both sides.

(7y + 3) - 3 = 17 - 3

7y + 3 - 3 = 14

7y = 14

Divide both sides by 7.

⁷ʸ⁄₇ = ¹⁴⁄₇

y = 2

12 - x = 8

Subtract 12 from both sides.

(12 - x) - 12 = 8 - 12

12 - x - 12 = -4

-x = -4

Multiply both sides by -1.

-1(-x) = -1(-4)

x = 4

ˢ⁄₆ = 7

Multiply both sides by 6.

6(ˢ⁄₆) = 6(7)

s = 42

⁽³ᵃ ⁺ ²⁾⁄₄ = 5

Multiply both sides by 4.

4(⁽³ᵃ ⁺ ²⁾⁄₄) = 4(5)

3a + 2 = 20

Subtract 2 from both sides.

(3a + 2) - 2 = 20 - 2

3a + 2 - 2 = 18

3a = 18

Divide both sides by 3.

³ᵃ⁄₃ = ¹⁸⁄₃

a = 6

⁴ʸ⁄₃ - 1 = ¹⁴ʸ⁄₁₅ + ¹⁹⁄₅

The least common multiple of the denominators (3, 15, 5) is 15.

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

15(⁴ʸ⁄₃ - 1) = 15(¹⁴ʸ⁄₁₅ + ¹⁹⁄₅)

Using distributive property,

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

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

20y - 15 = 14y + 57

Subtract 14y from both sides.

(20y - 15) - 14y = (14y + 57) - 14y

20y - 15 - 14y = 14y + 57 - 14y

6y - 15 = 57

(6y - 15) + 15 = 57 + 15

6y - 15 + 15 = 72

6y = 72

Divide both sides by 6.

⁶ʸ⁄₆ = ⁷²⁄₆

y = 12

⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11

The least common multiple of the denominators (4, 3) is 12.

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

12(⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃) = 12(11)

Using distributive property,

12(⁽ˣ ⁺ ⁴⁾⁄₄) + 12(⁽ˣ ⁻ ⁵⁾⁄₃) = 132

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

Using distributive property,

3x + 12 + 4x - 20 = 132

7x - 8 = 132

(7x - 8) + 8 = 132 + 8

7x - 8 + 8 = 140

7x = 140

Divide both sides by 7.

⁷ˣ⁄₇ = ¹⁴⁰⁄₇

x = 20

⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄

The least common multiple of the denominators (5, 4) is 20.

Multiply both sides of the equation by 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 5x from both sides.

(4x + 96) - 5x = (80 + 5x) - 5x

4x + 96 - 5x = 80 + 5x - 5x

-x + 96 = 80

Subtract 96 from both sides.

(-x + 96) - 96 = 80 - 96

-x + 96 - 96 = -16

-x = -16

Multiply both sides by -1.

x = 16

ˣ⁄₀.₅ - ¹⁄₀.₀₅ + ˣ⁄₀.₀₀₅ - ¹⁄₀.₀₀₀₅ = 0

ˣ/⁽⁵⁄₁₀₎ - ¹/⁽⁵⁄₁₀₀₎ + ˣ/⁽⁵⁄₁₀₀₀₎ - ¹/⁽⁵⁄₁₀₀₀₀₎ = 0

x(¹⁰⁄₅) - 1(¹⁰⁰⁄₅) + x(¹⁰⁰⁰⁄₅) - 1(¹⁰⁰⁰⁰⁄₅) = 0

¹⁰ˣ⁄₅ - ¹⁰⁰⁄₅ + ¹⁰⁰⁰ˣ⁄₅ - ¹⁰⁰⁰⁰⁄₅ = 0

5(¹⁰ˣ⁄₅ - ¹⁰⁰⁄₅ + ¹⁰⁰⁰ˣ⁄₅ - ¹⁰⁰⁰⁰⁄₅) = 5(0)

5(¹⁰ˣ⁄₅) + 5(-¹⁰⁰⁄₅) + 5(¹⁰⁰⁰ˣ⁄₅) + 5(-¹⁰⁰⁰⁰⁄₅) = 5(0)

10x - 100 + 1000x - 10000 = 0

1010x - 10100 = 0

(1010x - 10100) + 10100 = 0 + 10100

1010x - 10100 + 10100 = 10100

1010x = 10100

Divide both sides by 1010.

¹⁰¹⁰ˣ⁄₁₀₁₀ = ¹⁰¹⁰⁰⁄₁₀₁₀

x = 10

Let x be the number of cookies in the box.

It is given that the box plus three more equals 23 cookies.

x + 3 = 23

Subtract 3 from both sides.

(x + 3) - 3 = 23 - 3

x + 3 - 3 = 20

x = 20

There are 20 cookies in a box.

Let x be the number.

It is given that 6 less than 7 times of a number is 57.

7x - 6 = 57

(7x - 6) + 6 = 57 + 6

7x -6 + 6 = 63

7x = 63

Divide both sides by 7.

⁷ˣ⁄₇ = ⁶³⁄₇

x = 9

The number is 9.

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