Download x + 2

October 31st
copyright2009merrydavidson
Simplifying Rational Expressions
What is the
difference between
a factor and a
term?
An expression has
NO equal sign.
( x  5)(2 x  3)
(3x  5)( x  2)
TERMS are separated by +, - signs.
FACTORS are separated by
multiplication sign.
The REDUCTION PROPERTY
OF FRACTIONS
You are allowed to divide out
(cancel) common factors (not
common terms) that appear in both
the numerator and the denominator.
You may NOT cancel
4x  5
the “4x”’s. They are
terms not factors!
4x  3
Example 2:
2
2x p
4 pr
There are NO
add or subtract
Can you cancel
signs so these
anything out?
are all factors.
Why or Why not???
Reduce the 2/4 to ½ and cancel out the p’s.
2
2
x
List domain
x

; r0
restrictions.
2r
2r
domain restrictions are called
critical numbers.
CRITICAL NUMBERS:
• Found in the
numerator by
solving for x.
• Found in the
denominator by
solving for x.
2x  3
(4 x  3)( x  5)
• C.N. = 3/2
• C.N. = ¾ ,-5
Example 3:
( x  5)(2 x  3)
(4 x  3)( x  5)
DO domain restrictions and critical numbers
before canceling.
3
List domain restrictions.
4
List critical numbers. 3
2
SIMPLIFY:
2x  3

4x  3
2x  3
3
; x
4x  3
4
Example 4:
x  3x  10
2
x 4
2
Can you cancel anything out? Why or Why not?
These are all terms so no. Factor to see if
anything will match to cancel.
( x  5)( x  2)
( x  2)( x  2)
x5

x2
SIMPLIFY:
List domain restrictions. -5,2,-2
List critical numbers. -5,2,-2
x5
; x  -2
x2
Example 5:
For this
example only
simplify and
list answer
with domain
restrictions.
x3
x7
x 9
( x  3)( x  7)
2
( x  3)( x  3)
( x  3)( x  7)
Remember to do
domain restrictions
BEFORE canceling!!!
x3
; x  3, -7
x7
Example 6:
( x  5)(2 x  3)
(4 x  3)(5  x)
(x – 5) and (5 – x) are opposites…..
In order to cancel you must factor
out a (-1) from one of those terms.
1( x  5)(2 x  3)
(4 x  3)(5  x)
1(5  x)(2 x  3)
(4 x  3)(5  x)
1(2 x  3) 2 x  3
3
; x  ,5
4x  3
4x  3
4
7. Solve the following rational equation.
Step 1: Find the LCD. 3
6x
LCD = 6
Step 2: Multiply
each term by the
LCD over 1.
Step 3: Cancel
2
26 6
 7
1 2 31 1
3x + 4 = 42
x = 38/3
Step 4: Re-write the equation.
Step 5: Solve the equation.
8. Solve the following rational
equation.
LCD is: 6
2
3
6 x  1 26
6
 x
1 2
31 1
3(x + 1) + 4 = 6x
3x + 3 + 4 = 6x
7 = 3x
7/3 = x
Extraneous Solutions:
You are not allowed to have a zero in the
denominator of a fraction. Therefore, If you
get x = 5 and 5 would make the denominator
= 0, 5 would be an extraneous solution. In
other words, if algebraically you get a
solution, but that makes the denominator
zero it is called an extraneous solution. For
any fraction with a variable in the
denominator you must list domain
restrictions first (that is what x can not be).
9. Solve the following rational equation.
There
is
a
variable
in
3
8
13
5x
5x
5x
 
the denominator.
1 x 51 x 1
What can x not be?
x =0
LCD is 5x
By using the LCD
15 + 8x = 65
and canceling you
eliminate the
8x = 50
denominator.
x = 25/4 Since the answer is not a
domain restriction we are done.
10. Solve the following rational
List domain equation.
restrictions
5
2
x

5
( x  5)
(
x

5)
(
x

5)
FIRST. x  5
1

1
Solve
Check for
extraneous
solutions.
x  51
x 5
1
x – 5 + 5 = 2x - 5
x = 2x - 5
5=x
But x cannot be 5. the answer is:
no real solution and 5 is extraneous.
11. Solve the following rational
equation.
x 1
x4
3


2
2
2
x  2 x x  x x  3x  2
Factor everything possible.
x 1
x4
3


x( x  2) x( x  1) ( x  2)( x  1)
domain restrictions: x = 0, -2, -1
domain restrictions: x = 0, -2, -1
LCD:
x( x  2)( x  1)
1
x 1
x4
3 x( x  2)( x  1)


1
x( x  2) x( x  1) ( x  2)( x  1)
x( x  2)( x  1)
1
(x + 1)(x + 1) – (x + 2)(x + 4) = -3x
x2 + 2x + 1 – x2 – 6x - 8 = -3x
x = -7
12. Solve the following rational
x  3, 1, 4 equation.
3x  1
3
2x  5
 2
 2
2
x  4 x  3 x  x  12 x  3x  4
( x  3)( x  1)( x  4)
13x  1
( x  3)( x  1)
( x  3)( x  1)( x  4)
13

( x  4)( x  3)

( x  3)( x  1)( x  4)
2 x  51
( x  4)( x  1)
Be sure that everything in the
denominator gets canceled out
12.
x  3, 1, 4
(x – 4)(3x – 1) – 3(x + 1) = (x + 3)(2x – 5)
( x  3)( x  1)( x  4)
13x  1
( x  3)( x  1)
( x  3)( x  1)( x  4)
13

( x  4)( x  3)

( x  3)( x  1)( x  4)
2 x  51
( x  4)( x  1)
3x2 – 13x + 4 – 3x – 3 = 2x2 + x - 15
x2 – 17x + 16 = 0
x = 1, 16
HW: WS 4-4