Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

Step 3


Distributive property
Multiply.
Distributive property
Combine like terms.
Subtract 15.
Proposed solution Divide by

Step 4 Since the proposed solution, 3, is not in the domain, it cannot be a solution


of the equation. Substituting 3 into the original equation shows why.


CHECK Original equation


Let

Division by 0 is undefined.

The equation has no solution. The solution set is 0. NOW TRY


2


0


-


3


6


^12


0


x=3.


2


3 - 3


-


3


3 + 3


^12


32 - 9


2


x- 3


-


3


x+ 3


=


12


x^2 - 9


x= 3 - 1.


- x=- 3


- x+ 15 = 12


2 x+ 6 - 3 x+ 9 = 12


21 x+ 32 - 31 x- 32 = 12


= 1 x+ 321 x- 32 c


12


1 x+ 321 x- 32


d


1 x+ 321 x- 32 a


2


x- 3


b - 1 x+ 321 x- 32 a


3


x+ 3


b


388 CHAPTER 7 Rational Expressions and Functions


NOW TRY
EXERCISE 3
Solve.


1
t- 7

+


2


t+ 7

=


14


t^2 - 49

NOW TRY
EXERCISE 4
Solve.


2


t^2 - 16

2


t^2 - 2 t- 8

-


4


t^2 + 6 t+ 8

=


Solving a Rational Equation

Solve


Factor each denominator to find the domain and the LCD.


Factor the denominators.

The domain excludes 1, , and. Multiply each side of the equation by the LCD,


Distributive property
Distributive property
Combine like terms.
Subtract 4p. Add 7.
Proposed solution Divide by 3.

Note that 3 is in the domain. Substitute 3 for pin the original equation to check that


the solution set is 536. NOW TRY


3 =p


9 = 3 p


4 p+ 2 = 7 p- 7


6 p+ 6 - 2 p- 4 = 7 p- 7 2 # 3 =6;


2 # 31 p+ 12 - 21 p+ 22 = 71 p- 12


= 21 p- 121 p+ 221 p+ 12 c


7


21 p+ 221 p+ 12


d


21 p- 121 p+ 221 p+ 12 c


3


1 p- 121 p+ 22


-


1


1 p+ 121 p- 12


d


21 p- 121 p+ 221 p+ 12.


- 2 - 1


=


7


21 p+ 221 p+ 12


3


1 p- 121 p+ 22


-


1


1 p+ 121 p- 12


3


p^2 + p- 2


-


1


p^2 - 1


=


7


21 p^2 + 3 p+ 22


.


EXAMPLE 4


NOW TRY ANSWERS



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