Transcript Lesson 7: Solving Rational Equations

Math 20-1 Chapter 6 Rational Expressions and Equations
6.4 Solve Rational Equations
Teacher Notes
Expression
x
1
x
1 x2

x x4
Equation
x5
1
5
x
1 x2

5
x x4
What is one main difference?
Which one can you Simplify ?
Solve
When do you need to use a LCD?
abc
bc
2 x3
x3 8
x2  5x  6 x  3

2
x 9
x 3
ab 1

b bc
2
x3

x3
8
What do you do
with the LCD?
6.4 Solve Rational Equations
A rational equation is an equation containing at least one rational
expressions.
1
2

x
x3
x2  3
x 1
 3
x 1
x  2x
5
 3 are rational equations.
x 1
To solve a rational equation:
1. Determine the LCD of the denominators. (factor)
2. Reduce denominators to 1 by multiplying
both sides of the equation by the LCD.
Multiply all terms by LCD on both sides.
3. Solve the resulting polynomial equation.
4. Check the solutions.
6.4.1
Solve Rational Equations
a)
7
6

x  2 x 5
Domain x | x  2, x  5, x  R
1
7
6
(x  2)( x  5) x  2  x  5 (x  2)( x  5)
1
1
1
7( x  5)  6( x  2)
7x  35  6x 12
x  35  12
x  47
Do you need to use a
LCD?
What do you do with the
LCD?
Multiply each term by the
LCD (x + 2)(x - 5)
Divide out the common
factors
Verify by substitution
7
6

47  2 47  5
7
6

49 42
1 1

7 7
6.4.2
Solve Rational Equations
Solve :
4
1
2


x 2  x  6 x 2  4 x 2  5x  6
Domain
4
1
2


( x  3)( x  2) ( x  2)( x  2) ( x  3)( x  2)
x | x  2, 3, 2, x  R
Multiply each term by the
LCD (x + 2)(x + 3)(x – 2)






4
1
2
( x  3)( x  2)( x  2) 

(
x

3)(
x

2)(
x

2)

(
x

3)
(
x

2)(
x

2)





 ( x  3)( x  2) 
 ( x  2)( x  2) 
 ( x  3)( x  2) 
4( x  2)  1( x  3)  2( x  2)
4x  8  x  3  2 x  4
3x  5  2x  4
Verify by substitution
x  9
4
1
2


(9) 2  (9)  6 (9) 2  4 (9)2  5(9)  6
4
1
2


66 77 42
1 1
=
21 21
Do you need to
use a LCD?
What do you do
with the LCD?
6.4.3
Solve Rational Equations
2
1
 x 2  3x
Solve :

 2
x  3 x 1 x  2x  3
Domain
2
1
 x ( x  3)


( x  3) ( x  1) ( x  3)( x  1)
x | x  3,1, x  R
Multiply each term by the
LCD (x + 3)(x – 1)
 2 
 1 
  x ( x  3) 
( x  3)( x  1) 

(
x

3)(
x

1)

(
x

3)(
x

1
)





 ( x  3) 
 ( x  1) 
 ( x  3)( x  1) 
2( x  1)  1( x  3)   x ( x  3)
2 x  2  x  3   x 2  3x
x 2  4x  5  0
( x  5)( x  1)  0
x  5 or x  1
since x  1 Solution: x  –5
6.4.4
Your Turn
4
x 1
a)

x  1 12
x  7
4t  3 4  2t
b)

1
5
3
t 2
10
2m  5 2m  5
c)


2
m 1 m 1
m 1
http://www.regentsprep.org/Regents/math/al
gtrig/ATE11/RationalEqPract.htm
5
m
3
6.4.5
Find the value of two integers. One positive integer is 5
more than the other. When the reciprocal of the larger
number is subtracted from the reciprocal of the smaller the
result is 5 .
14
smaller
larger
x
x5
1
1
5


x x  5 14
Application of Solving Rational Equations
1. A traveling salesman drives from home to a client’s store 150 miles away. On the
return trip he drives 10 miles per hour slower and adds one-half hour in driving time.
At what speed was the salesperson driving on the way to the client’s store?
Let r be the rate of travel (speed) in miles per hour.
distance
rate
Trip to client
150
r
Trip home
150
r – 10
d  st
time
150
r
150
r  10
Longer Time (slower speed) - Shorter Time (faster speed) = Time Difference
150 150 1


r  10
r
2
LCD = 2r (r – 10).
6.4.6
150 150 1


r  10 r
2
2r (r  10)
LCD = 2r (r – 10).
150
150
1
 2r (r  10)
 2r (r  10)
r  10
r
2
300r – 300(r – 10) = r(r – 10)
300r – 300r + 3000 = r2 – 10r
0 = r2 – 10r – 3000
0 = (r – 60)(r + 50)
r = 60 or – 50
Why is -50 not an
acceptable answer?
The salesman drove from home to the client’s store at
60 miles per hour.
6.4.7
Check:
Distance
Time=
Rate
At 60 mph the time taken to drive the 150 miles
from the salesman’s home to the clients store is
= 2.5 h.
At 50 mph (ten miles per hour slower) the
time taken to make the return trip of 150 miles is
= 3 h.
150
60
150
50
The return trip took one-half hour longer.
6.4.8
Application of Solving Rational Equations
2. If a painter can paint a room in 4 hours and her assistant can paint the
room in 6 hours, how many hours will it take them to paint the room working
together?
Let t be the time it takes them to paint the room together.
rate of work
time worked
Portion of work
completed
1
t
t
4
4
1
t
t
assistant
6
6
Write an equation, in terms of t, to represent completing the job
working together.
painter
(
)+(
Portion of job
completed by the
painter
) =( )
Portion of job
completed by the
assistant
t t
 1
4 6
1
whole
job
6.4.9
Application of Solving Rational Equations
t t
  1 LCD = 12
4 6
t
t
(12)  (12)  1(12)
4
6
3t  2t  12
5t  12
12
t
5
Working together they will
paint the room in 2.4 hours.
6.4.10
Your Turn
Andrea can wallpaper a bathroom in 3 hr. Erin can
wallpaper the same bathroom in 5 hr. How long
would it take them if they worked together?
Let t be the time it takes them to paint the room together.
Andrea
Erin
rate of work
Part of work
time worked completed
per hour
1
t
t
3
3
1
t
t
5
5
t t
 1
3 5
15
t
8
Working together they will
paper the room in 1.875 hours.
6.4.11
Suggested Questions:
Part A:
Page 348:
1a, 2, 3a,c, 4, 5, 7, 8
Part B:
Page 348:
11, 12, 14, 15, 18, 27
6.4.12