# SOLVING LINEAR EQUATIONS

To solve an equation means to find all values of the variable that make the equation a true statement. One way to do this is to isolate the variable that has a coefficient of 1 onto one side of the equation. You can do this using the rules of algebra called properties of equality.

## Properties of Equality

If a = b, then

a + c = b + c

Example :

If x - 3 = 5, then

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

Subtraction Property :

If a = b, then

a - c = b - c

Example :

If x + 3 = 5, then

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

Multiplication Property :

If a = b, then

⋅ a = c ⋅ b

Example :

If x/2 = 7, then

⋅ (x/2) = 2 ⋅ 7

Division Property :

If a = b and c ≠ 0, then

a/c = b/c

Example :

If 3x = 15, then

3x/5 = 15/5

## Solved Problems

Problem 1 :

Solve for a :

a - (-10) = -12

Solution :

a - (-10) = -12

a + 10 = -12

Subtract 10 from both sides.

a = -22

Problem 2 :

Solve for x :

-24 = 8x

Solution :

-24 = 8x

Divide both sides by 8.

-3 = x

Problem 3 :

-10 + x = 12

Given the above equation, what is the value of 2 - (5 - x)?

Solution :

-10 + x = 12

Add 10 to both sides.

x = 22

Substitute x = 22 in 2 - (5 - x).

= 2 - (5 - 22)

= 2 - (-17)

= 2 + 17

= 19

Problem 4 :

If 33 - y = y + 27 - 5y, what is the value of 33 + 3y?

Solution :

33 - y = y + 27 - 5y

Simplify.

33 - y = 27 - 4y

Add 4y to both sides.

33 + 3y  = 27

Problem 5 :

If x/2 + 3 = 3/4 - x, what is the value of x?

Solution :

x/2 + 3 = 3/4 - x

In the equation above, we find two fractions with two different denominators 2 and 4.

The least common multiple of (2, 4) = 4.

Multiply both sides of the above equation by 4 to get rid of the fractions.

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

4x/2 + 12 = 12/4 - 4x

2x + 12 = 3 - 4x

Add 4x to both sides.

6x + 12 = 3

Subtract 12 from both sides.

6x = -9

Divide both sides by 6.

x = -9/6

x = -3/2

Problem 6 :

If y - (3 - 2y) + (4 - 5y) = -7, what is the value of y?

Solution :

y - (3 - 2y) + (4 - 5y) = -7

Simplify the right side of the equation.

y - 3 + 2y + 4 - 5y = -7

1 - 2y = -7

Subtract 1 from both sides.

-2y = -6

Divide both sides by -2.

y = 3

Problem 7 :

Solve for y :

(4/5)(y - 5) - (1/5)(y - 10) = 22

Solution :

(4/5)(y - 5) - (1/5)(y - 10) = 22

Using Distributive property,

4y/5 - 4 - y/5 + 2 = 22

(4y - y)/5 - 2 = 22

3y/5 - 2 = 22

Add 2 to both sides.

3y/5 = 24

Multiply both sides by 5.

3y = 120

Divide both sides by 3.

y = 40

Problem 8 :

If three quarters of a number decreased by twenty is equal to eighty two, what is that number?

Solution :

Let n be the number.

(3/4)n - 20 = 82

3n/4 - 20 = 82

Add 20 to both sides.

3n/4 = 102

Multiply both sides by 4.

3n = 408

Divide both sides by 3.

n = 136

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