Solutions to Algebraic Equations

When we solve an algebraic equation, instead of plugging in a given number for the variable, we find a number that, when plugged in for the variable, would make the equation true. Such a number is called a solution to an equation. 58 is a solution to the equation h + 2 = 60 , because 58 + 2 = 60 . 46 is not a solution to h + 2 = 60 , because 46 + 2 does not equal 60.

Some equations have more than one solution. For example, 4 and -4 are both solutions to r ² = 16 . Most of the equations we will deal with, however, have only one solution.

Fundamentals of Equations

The goal in solving an equation is to get the variable by itself on one side of the equation and a number on the other side of the equation.

Generally, the variable will start on one side with operations being performed on it. We must reverse these operations by performing the inverse of each operation. However, we cannot just perform the inverse operation on on e side, because that would change the equation. However, if you perform the same operation on both sides of an equation the equation will not change.

Performing an operation on one side of an equation will change the equation and make it false.

Given, 5×6 = 30
5×6 = 30×3 ; 5×6 = 30 while 30×3 = 90
5×6 = 30 + 18 ; 5×6 = 30 while 30 + 18 = 48
5×6 = 30/10 ; 5×6 = 30 while 30/10 = 3

Performing the same operation on each side of an equation won't change the equation:

Given, 7 + 4 = 11
(7 + 4)×12 = 11×12 ; both sides equal 132
(7 + 4) + 3 = 11 + 3 ; both sides equal 14
- (7 + 4) = - 11 ; both sides equal -11

Herein lies a vital role of solving algebraic equations: whatever operation is carried out on one side of the equal sign in an equation must be carried out on the other side as well.

Solving Algebraic Equations

To solve an algebraic equation, reverse all the operations on the variable side of the equation by performing their inverse operations on both sides of the equation.

Here are some examples of how to solve an algebraic equation:

Example 1: y + 11 = 18

(y + 11) - 11 = 18 - 11
y = 7

Example 2: 4×q = 12

(4×q)/4 = 12/4
q = 3

Example 3: m/5 = 6

(m/5)×5 = 6×5
m = 30

Example 4: h - 2 = 7

(h - 2) + 2 = 7 + 2
h = 9

When there is more than one operation, reverse the "outside" operation first. Follow the reverse of the order of operations to reverse the operations in an algebraic equation:

Example 5: 7g + 5 = 33

(7g + 5) - 5 = 33 - 5
7g = 28
(7g)/7 = 28/7
g = 4

Example 6: (4 + w)/6 = 8

((4 + w)/6)×6 = 8×6
(4 + w) = 48
(4 + w) - 4 = 48 - 4
w = 44

Example 7: 7×(2y + 6) = 56

(7×(2y + 6))/7 = 56/7
(2y + 6) = 8
(2y + 6) - 6 = 8 - 6
2y = 2
(2y)/2 = 2/2
y = 1

Example 8: 22 - x = 25

(22 - x) - 22 = 25 - 22
- x = 3
x = - 3

Remember to check your answers--plugging the answer in for the variable should make the equation true.