Transcript 8-6 Solve Rational Equations

Quiz 8-5
1.
Simplify
7
x

2
9x
3x
2.
12
3

2
x  5 x  24 x  3
3.
( 2 x  7 ) ( x  4)
 2
2
x 2
x 2
8-6
Solve Rational Equations
Vocabulary:
What does solve a single variable equation mean?
3x + 2 = 11
What is a factor?
What is a least common multiple?
Solving Rational Equations
Method #1: eliminate the denominators one at a time.
Method #2: Obtain common denominators for each term.
Method #3: Find the least common multiple of all the
denominators then multiply (left/right) by the LCM.
Rational Equations
x
4
5
How do you get the ‘5’ out of the denominator?
Multiply both sides by ‘5’
x 5
5* 4  *
5 1
x = 20
Rational Equations
10
2
x
How do you get the ‘x’ out of the denominator?
Multiply both sides by ‘x’
10 x
x*2  *
x 1
2x = 10
x=?
Rational Equations
12
4
x2
How do you get the ‘x-2’ out of the denominator?
Multiply both sides by ‘x-2’
12
( x  2)
4( x  2) 
*
( x  2)
1
4(x – 2) = 12
÷4
÷4
x–2 =3
+2
+2
x=5
Your turn:
21
1. Solve 7 
x
10
 12
2. Solve 18 
x
20
3. Solve 5 
( x  1)
4. Solve
30
4
6
( x  2)
Variable on both sides of the equation
3
2

x x 1
x 3
2
x
* 
*
1 x x 1 1
2x
3
x 1
How do you get the ‘x’ out of the
left-side denominator?
Multiply both sides by ‘x’
Variable on both sides of the equation
3
2

x x 1
2x
3
x 1
How do you get the (x+1) out of the
right-side denominator?
2 x ( x  1)
( x  1) * 3 
*
Multiply both sides by (x+1)
x 1
1
3x  3  2x
3 3
3x  2x  3
 2x  2x
x  3
Another example:
3
9

x  1 4x  5
The book teaches you to “cross-multiply” (yuck!)
There is no property called “cross multiply”.
There are the properties of equality.
Cross multiplication is the result of multiplying
both sides by the left denominator, then multiplying
the both sides by the right denominator.
DON’T THINK Cross multiply!! Forget it completely!!
Eliminate one Denomiator at a time.
3
9
* (4x + 5)

(4x + 5) *
x  1 4x  5
(x + 1) *
3(4 x  5)
 9 * (x + 1)
x 1
3(4 x  5)  9( x  1)
Distributive property
12x 15  9x  9
-9x both sides
3x  15  9
-15 both sides
3x  6
÷3 both sides
x  2
Or: obtain a common denominator
3
9
* (x + 1)
(4x + 5) *

(4x + 5) * x  1 4 x  5 * (x + 1)
Multiply both sides by the common denominator
(4 x  5)( x  1) 3(4 x  5)
9( x  1) (4 x  5)( x  1)

(4 x  5)( x  1) (4 x  5)( x  1)
3(4 x  5)  9( x  1)
Your Turn:
6. Solve
9
4

3x ( x  2)
7. Solve
x
1

2
x 2 x
8. Solve
x
3
3
2
4x
What about this wrinkle?
“multiply to eliminate the denominators one at a time”
4
x5
x
4 x  x  5x
2
4

x *   x  5 * x
x

Now what?
It’s a quadratic,
solve the quadratic!
put into standard form !!!
x  5x  4  0
2
Your turn:
9. solve x 2  5 x  4  0
Rational equations with 2 solutions.
8
9
1

x5
x
Eliminate denominators one at at time.
( x  5) 
8   9  ( x  5) Multiply left/right by (x – 5)
1 
   
1  x  5   x  1 Carefull: distributive property
8   9  ( x  5)
 ( x  5) 1   ( x  5)
    
*   
*

1  1
( x  5)   x  1
 1
( x  5)  8   9( x  5)
  
1
x
1
Rational equations with 2 solutions.
8
9
1

x5
x
Eliminate denominators one at at time.
( x  5) 
8   9  ( x  5) Multiply left/right by (x – 5)
1 
   
1  x  5   x  1 Carefull: distributive property
8   9  ( x  5)
 ( x  5) 1   ( x  5)
    
*   
*

1  1
( x  5)   x  1
 1
( x  5)  8   9( x  5)
  
1
x
1
Rational equations with 2 solutions.
8
9
1

x5
x
Eliminate denominators one at at time.
 9( x  5)
Combine like terms.
( x  5)  8 
x
 9( x  5)
Multiply left/right x
x3
x
 9( x  5) x
x( x  3) 
*
x
1
x 2  3x  9( x  5)
Rational equations with 2 solutions.
8
9
1

x5
x
Eliminate denominators one at at time.
x 2  3x  9( x  5) Distributive property.
x  3x  9 x  45
2
x 2  12 x  45  0
( x  15)( x  3)  0
x   15, 3
What kind of equation is this?
Quadratic: get into standard form then
solve
Your turn:
10. Solve
x3
x
x5


x 3 x 5 x 5
Put all factors inside parentheses to
avoid silly distributive property errors
later on.
Vocabulary
Extranious Solution: a solution obtained algebraically
but which doesn’t work in the original equation.
2
1 x 1
 
x 3 x x 3
2
x  4x  3  0
( x  3)( x  1)  0
x  1, 3
2
1 3 1
 
33 3 33
2 1 3 1
 
0 3
0
2
1 1 1
 
1 3 1 1 3
2 1 0
 
2 1 2
1  1  0
Check for Extraneous solutions.
Your Turn:
11. Solve.
3x  6 x  1

2
x 4 x2
12. Check to see if either one is extraneous.
Multiply for the LCD
Multiply both sides by the LCD!!
What is the least common denominator?
2 1 4
 
3x 6 3x
Factor the
denominator
2
1
4


3x 2 * 3 3x
1. Look for what is common to each denominator. 3
2. What does the 1st denominator need to be common
with the other two denominators? Needs a ‘2’
3. What does the 2nd denominator need to be common
with the other two denominators? Needs an ‘x’
4. What does the 3rd denominator need to be common
with the other two denominators?
Needs an ‘2’
Find the least common denominator
2 1 4
 
3x 6 3x
2 * 3x
2*3 * x
2 *3x
The least common denominator is:
Multiply both sides by the LCD.
2 1 4
 
6x
3x 6 3x
6x
6x
Find the least common denominator
2 1 4
 
3x 6 3x
The LCD is: 6x
Multiply both sides by the LCD.
2 1 4
 
6x
3x 6 3x
6x
Notice how I factored 6x
so that it is easier to simplify.
(3 x * 2) * 2 6 * x *1 4 * (2 * 3 x)


3x
6
3x
4 x 8
Wow !
x=4