Beginning Algebra, 11th Edition

(Marvins-Underground-K-12) #1

86 CHAPTER 2 Linear Equations and Inequalities in One Variable


OBJECTIVES An equationis a statement asserting that two algebraic expressions are equal.


The Addition Property of Equality


2.1


1 Identify linear
equations.
2 Use the addition
property of
equality.
3 Simplify, and then
use the addition
property of
equality.

CAUTION Remember that an equation includes an equals symbol.


Equation (to solve) Expression (to simplify or evaluate)

Left side Right side

x- 5  2 x- 5


OBJECTIVE 1 Identify linear equations.The simplest type of equation is a


linear equation.


Linear Equation in One Variable

A linear equation in one variablecan be written in the form


where A, B, and Care real numbers, and AZ 0.


AxBC,


Linear equations

and Nonlinear equations


A solutionof an equation is a number that makes the equation true when it


replaces the variable. An equation is solved by finding its solution set,the set of all


solutions. Equations with exactly the same solution sets are equivalent equations.


A linear equation in is solved by using a series of steps to produce a simpler


equivalent equation of the form


xa number or a numberx.


x


x^2 + 2 x=5, | 2 x+ 6 | = 0


1


x


=6,


4 x+ 9 =0, 2 x- 3 = 5, and x= 7


OBJECTIVE 2 Use the addition property of equality.In the linear equation


both and 2 represent the same number because that is the mean-


ing of the equals symbol. To solve the equation, we change the left side from


to just x, as follows.


Given equation
Add 5 to eachside to keep them equal.
Additive inverse property
Additive identity property

The solution is 7. We check by replacing xwith 7 in the original equation.


CHECK Original equation


Let.

✓ True


Since the final equation is true, 7 checks as the solution and 576 is the solution set.


2 = 2


7 - 5  2 x= 7


x- 5 = 2


x= 7


x+ 0 = 7


x- 5 + 5 = 2 + 5


x- 5 = 2


x- 5


x- 5 =2, x- 5


The left side equals
the right side.

Add 5. It is the opposite
(additive inverse) of ,
and .- 5 + 5 = 0


  • 5


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