# Solving Linear Equations in One Variable

**Abstract**

Elementary Algebra: An introduction to solving linear equations in one variable.

1 Module Overview

Learning how to solve various algebraic equations is one of our main goals in
algebra. This module introduces

the basic techniques for solving linear equations in one variable.
(Prerequisites: Working knowledge of real

numbers and their operations.)

**Objectives
**•

**Define Linear Equations in One Variable**

•

**Solutions to Linear Equations**

•

**Solving Linear Equations**

•

**Combining Like Terms and Simplifying**

•

**Literal Equations**

**2 Define Linear Equations in One Variable**

We begin by establishing some definitions.

**Definition 1: Equation**

An equation is a statement indicating that two algebraic expressions are equal.

**Definition 2: Linear Equation in One Variable**

A linear equation in one variable x is an equation that can be written in the form ax +b = 0 where

a and b are real numbers and a ≠ 0.

Following are some examples of linear equations in one variable, all of which will be solved in the course

of this module.

**3 Solutions to Linear Equations in One Variable
**

The variable in the linear equation 2x + 3 = 13 is x. Values that can replace the variable to make a true

statement compose the solution set. Linear equations have at most one solution. After some thought, you

might deduce that x = 5 is a solution to 2x + 3 = 13. To verify this we substitute the value 5 in for x and

see that we get a true statement, 2 (5) + 3 = 10 + 3 = 13.

**Example 1**

Is x = 3 a solution to -2x - 3 = -9?

Yes, because -2 (3) - 3 = -6 - 3 = -9

**Example 2**

Is a solution to -10a + 5 = 25?

No, because

When evaluating expressions, it is a good practice to replace all variables with parenthesis first , then sub-

stitute in the appropriate values. By making use of parenthesis we could avoid some common errors using

the order of operations .

**Example 3**

Is y = -3 a solution to 2y - 5 = -y - 14 ?

Replace variables with parenthesis. | |

Substitute the appropriate value. | |

Simplify. | |

True. |

Yes because y = -3 produces a true mathematical statement .

**4 Solving Linear Equations in One Variable
**

When the coefficients of linear equations are numbers other than nice easy integers, guessing at solutions

becomes an unreasonable prospect. We begin to develop an algebraic technique for solving by first looking

at the properties of equality .

**4.1 Properties of Equality**

Given algebraic expressions A and B where c is a real number:

**Property 1:**Addition Property of Equality

If A = B then A + c = B + c

**Property 2:**Subtraction Property of Equality

If A = B then A - c = B - c

**Property 3:** Multiplication Property of Equality|

If A = B and c ≠ 0 then cA = cB

**Property 4:** Division Property of Equality

If A = B and c ≠ 0 then

Note: Multiplying or dividing both sides of an equation by zero is carefully
avoided. Dividing

by zero is undefined and multiplying both sides by zero will result in an
equation 0=0.

To summarize , the equality is retained if we add, subtract, multiply and divide
both sides of an equation

by any nonzero real number. The central technique for solving linear equations
involves applying these

properties in order to isolate the variable on one side of the equation.

**Example 4**

Use the properties of equality to solve: x + 3 = -5

x + 3 = -5 | |

x + 3 - 3 = -5 - 3 | Subtract 3 on both sides. |

x = -8 | Simplify |

The solution set is

**Example 5**

Use the properties of equality to solve: -5x = -35

-5x = -35 | |

Divide both sides by -5. | |

x = 7 | Simplify |

The solution set is .

Two other important properties are:

**Property 5:** Symmetric Property

If A = B then B = A.

**Property 6: **Transitive Property

If A = B and B = C then A = C.

When solving, we often see 2 = x but that is equivalent to x = 2.

**4.2 Isolating the Variable
**

The idea behind solving in algebra is to isolate the variable. If given a linear equation of the form ax+ b = c

then we can solve it in two steps. First use the equality property of addition or subtraction to isolate the

variable term. Next isolate the variable by using the equality property of multiplication or division. The

property choice depends on the given operation, we choose to apply the opposite property of the given

operation. For example, if given a term plus three we would first choose to subtract three on both sides of

the equation. If given two times the variable then we would choose to divide both sides by two.

**Example 6**

Solve 2x + 3 = 13 .

2x + 3 = 13 | |

2x + 3 - 3 = 13 - 3 | Subtract 3 on both sides. |

2x = 10 | |

Divide both sides by 2. | |

x = 5 |

The solution set is .

**Example 7**

Solve -3x - 2 = 9 .

-3x - 2 = 9 | |

-3x - 2 + 2 = 9 + 2 | Add 2 to both sides. |

-3x = 11 | |

Divide both sides by -3. | |

The solution set is .

**Example 8**

Solve .

Subtract 1/2 on both sides. | |

Multiply both sides by 3. | |

The solution set is

In order to retain the equality, we must perform the same operation on both
sides of the equation. To isolate

the variable we want to remember to choose the opposite operation not the
opposite number. For example,

if we have -5x = 20 then we choose to divide both sides by -5, not 5.

**Figure 1:** Video Example 01

**4.2.1 Multiplying by the Reciprocal
**Recall that when multiplying reciprocals the result is 1, for example,
. We can use this

fact when the coefficient of the variable is a fraction.

**Example 9**

Solve

Add 5 on both sides. | |

Multiply both sides by - 5/4 . | |

1x = -5 · 5 | Simplify. |

x = -25 |

The solution set is

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