Linear Equations, Rational Expressions
Goal: Recognize
a. Identities
b. Conditional Equations
c. Inconsistent Equations
Solve Rational Expressions
Def: Equivalent Equations: Two equations
with the same solution set.
Ex: 2 x = 10, 2 x – 10 = 0 both have the solution set {5} We say
that the equation is true when x = 5 and false if x has any other value.
Properties of Equality:
If A = B and C is a real number:
A+C = B+C
A–C = B - C
CA = CB, C ≠ 0
Addition property
Subtraction property
Multiplication property
A B

C C,
Division Property
C≠0
Def: Identity: An equation that is satisfied by
every real number for which both sides are
defined.
Examples: a. 3x – 1 = 3 x – 1 for all x.
b.
x
1
x
, x ≠ 0 {x|x≠0}
Def: Conditional Equation: An equation that
is satisfied by at least one real number, but is
not an identity.
Examples: 3x = 9, x = {3}
x2
=9, x = {-3, 3}
Def: Inconsistent Equation: An equation with
no solution.
Example: x – 5 = x + 3
Def: Solution Set: The set of all solutions to a
problem.
What happens when we solve equations involving
fractions (rational expressions)?
Example 1:
y
3
3
y 3
y 3
Multiplying through by LCD we obtain:
y + 3(y – 3) = 3
Distribute:
y + 3y – 9 = 3
4y = 12
y=3
What happens when we substitute 3 for y in the
original equation?
Notice that we get a 0 in the denominator when we
substitute 3 for y. This is an inconsistent equation,
it has no solution.
The solution set is
{}

, which may also be written as
Example 2:
1
1

1
2 x 1
Multiplying through by the LCD we obtain:
x-1 + 2 = 2(x-1)
Distribute:
x – 1 + 2 = 2x – 2
3=x
Substituting 3 for x in the original equation, we
observe that the equation is consistent.
1 1
 1
2 2
This is a conditional equation, as it is true when x
= 3. The solution set is {3}
Example 3
1
1
2

 2
x 1 x 1 x 1
You should recognize that x - 1 as the difference
of two squares, which may be factored as
(x+1)(x-1)
2
Multiplying through by the LCD we obtain;
x+1 – (x-1) = 2
Distribute:
x+1–x+1=2
2=2
This is an identity equation. Is this equation true
for all real numbers?
Observe that substituting either 1 or -1 in the
original equation results in a zero in the
denominator. The solution set is the set of all real
numbers, excluding x = 1 and x = -1. This is
written {x|x≠-1, x≠1} and read, the set of all real
numbers such that x is not equal to -1 and x is not
equal to 1.
Example 4
2
6
12

 2
x  3 x  2 x  5x  6
You should recognize that (x+3)(x+2) =
x2  5x  6
Multiplying through by the LCD we obtain
2(x+2) + 6(x+3) = 12
Distribute:
2x + 4 + 6x + 18 = 12
8x = -10
x =  108   54
This is a conditional equation, with the solution set
{- 54 }