US Common Core K-12 Curricum Algebra : Solving Simple Equations

Problem 1-13 : Solve each equation.

Problem 1 :

x - 3 = -5

Solution :

x - 3 = -5

Add 3 to both sides.

x = -2

Problem 2 :

0.9 = y + 2.8

Solution :

0.9 = y + 2.8

Subtract 2.8 from both sides.

-1.9 = y

Problem 3 :

g - 1/3 = -2/3

Solution :

g - 1/3 = -2/3

Add 1/3 to both sides.

g = -2/3 + 1/3

g = (-2 + 1)/3

g = -1/3

Problem 4 :

-n/5 = -3

Solution :

-n/5 = -3

Multiply both sides by -5.

(-5)(-n/5) = (-5)(-3)

5n/5 = 15

n = 15

Problem 5 :

πx = -2π

Solution :

πx = -2π

Divide both sides by π.

x = -2

Problem 6 :

1.3z = 5.2

Solution :

1.3z = 5.2

Divide both sides by 1.3.

z = 4

Problem 7 :

8x + 17(x - 3) = -1

Solution :

8x + 17(x - 3) = -1

Use the Distributive Property.

8x + 17x - 51 = -1

Combine the like terms.

25x - 51 = -1 

Add 51 to both sides.

25x = 50

Divide both sides by 25.

x = 2

Problem 8 :

5x - 3 = 3x + 5

Solution :

5x - 3 = 3x + 5

Subtract 3x from both sides.

2x - 3 = 5

Add 3 to both sides of the equation.

2x = 8

Divide both sides by 2.

x = 4

Problem 9 :

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

Solution :

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

Use the Distributive Property.

5x - 15 - 42 + 7x = 24 - 24 + 3x - 3

Simplify.

12x - 57 = 3x - 3

Subtract 3x from both sides.

9x - 57 = -3

Add 57 to both sides.

9x = 54

Divide both sides by 9.

x = 6

Problem 10 :

Solution :

In the given equation, we find fractions with denominators 5 and 4. The least common multiple of 5 and 4 is 20. Multiply both sides of the equation by 20 to get rid of the denominators.

4(4x) + 5(-7) = 4x + 5x

16x - 35 = 9x

Subtract 9x from both sides.

7x - 35 = 0

Add 35 to both sides.

7x = 35

Divide both sides by 7.

x = 5

Problem 11 :

Solution :

Since we find a fraction on each side of the given equation, we can do cross multiplication. That is, multiply numerator on the left side by denominator on the right side and numerator on the right side by denominator on the left side.

5(x - 4) = 7(x - 10)

5x - 20 = 7x - 70

Subtract 5x from both sides.

-20 = 2x - 70

Add 70 to both sides of the.

50 = 2x

Divide both sides by 2.

25 = x

Problem 12 :

Solution :

In the given equation, we find fractions with denominators 3, 15 and 5. The least common multiple of 3, 5 and 15 is 15. Multiply both sides of the equation by 15 to get rid of the denominators.

5(4x) - 15 = 14x + 3(19)

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

Problem 13 :

5(x + 2) + 4 = 7(2 - x) + 12x

Solution :

5(x + 2) + 4 = 7(2 - x) + 12x

Use the Distributive Property.

5x + 10 + 4 = 14 - 7x + 12x

5x + 14 = 14 + 5x

Subtract 5x from both sides.

14 = 14

The variable x vanishes in the last step and 14 = 14 is a true statement. Therefore, the given equation has infinitely many solutions.

Problem 14 :

5 - ax = ½(10 - 4x)

If the equation above has infinitely many solutions, determine the value of a ?

Solution :

5 - ax = ½(10 - 4x)

Use the Distributive Property.

5 - ax = 5 - 2x

If a = 2,

5 - 2x = 5 - 2x

Add 2x to both sides of the equation.

5 = 5

When a = 2, the variable x vanishes in the last step and 5 = 5 is a true statement. 

If the given equation has infinitely many solutions, then 

a = 2

Problem 15 :

7(x - 2) + bx = 4x + 13

If the equation above has no solution, what is the value of b?

Solution :

7(x - 2) + bx = 4x + 13

Use the Distributive Property.

7x - 14 + bx = 4x + 13

7x + bx - 14 = 4x + 13

(7 + b)x - 14 = 4x + 13

If b + 7 = 4,

4x - 14 = 4x + 13

Subtract 4x from both sides.

-14 = 13

The variable x vanishes in the last step and -14 = 13 is a false statement. 

b + 7 = 4

b = -3

If the given equation has no solution, then

b = -3

Related Pages

US Common Core K-12 Curriculum (Solving System of Equations)

US Common Core K-12 Curriculum (Word Problems on System of Linear Equations)

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