Transcript Rational Equations: Applications and Problem Solving

Chapter 6 Section 7
Rational Equations:
Applications and
Problem Solving
Setup and Solve Applications
Containing Rational Equations:
Steps.
1. Read
2. Reread – to make sure that you understand the problem.
Look for key words.
3. Write the equation
4. Solve
5. Check
6. Answer the question originally asked.
Rational Equations:
Applications and Problem Solving
We need to check our solutions to make sure.
1. We did not make a mistake in our calculations.
2. The solution is valid for the original equation.
3. The solution is correct for the content.
Example: a negative number would not be
correct for the length or width.
Types of Application Problems
 Geometry
1. Area of Rectangle = length ● width
A = LW
2. Area of Triangle = ½ base ● height
A = ½bh
3. Area of a Square = side2
A = s2
Types of Application Problems
 Motion
1. distance = rate ● time
d = rt
2. time = distance / rate
t = d/r
3. rate = time / distance
r= t/d
Types of Application Problems
 Work
1. Amount = rate ● time
A = rt
 Number
1. Two numbers are Reciprocals of each other when
their product is 1.
example:
the reciprocal of 3 is 1/3
the reciprocal of 3/5 is 5/3
0 has no reciprocal
Geometry Problem - The area of a rectangle is 160 square
meters. Determine the length and width if the width is 6
meters less than the length.
A = LW
Let x = length
X = L = 16
W = (x – 6) = 16 – 6 = 10
Then (x - 6)(x) = 160
x2 – 6x = 160
x2 – 6x – 160 = 0
(x – 16 )(x + 10) = 0
x - 16 = 0
x + 10 = 0
x = 16
x ≠ -10
-10 not an answer
Number Problem - One number is 5 times another. The
difference of their reciprocals is 2/5. Determine the
numbers.
Let x = first number
1
reciprocal =
x
5x = second number
reciprocal =
1 1 2


x 5x 5
5 1  2x
1st number = x = 2
LCD  5 x
2
1 1 
(5 x)     (5 x)
5
 x 5x 

4  2x
1
5x
2nd number = 5x = (5)(2) = 10
1
 1 
2
(5 x )    ( 5x ) 
  ( 5 x)  
 x
 5x 
5


2x
Motion Problem - The current in the river is 3 mph. If it
takes the same amount of time to travel 14 miles
downstream as 2 miles upstream, find the speed of the boat
in still water.
r = still water
Direction
r + 3 = downstream (with current)
Downstream
r – 3 = upstream (against current)
Upstream
d
t
r
Distance Rate Time
14
r+3
14/(r+3)
2
r-3
2/(r-3)
Because the time it takes to travel upstream is the
same as downstream set times equal to each other
and cross multiply to solve.
14
2

r 3 r 3

14r  42  2r  6
(14)(r  3)  (2)(r  3)

14r  2r  6  42


(14r  42)  (2r  6)
12r  48
The speed of the boat in still water is 4 mph

r4
Motion Problem - A person rides a bike on a trail part of
the time at 6 mph and part of the time at 10 mph. If a
total distance of 28 miles is traveled in 4 hours, how long
did the person ride at each speed?
d = at 10 mph
t
28 - d = at 6 mph
d
r
Direction
Distance
Rate
Time
A
d = 10
10
d/10=10/10
1 hour at 10 mph
B
28-d = 18
6
(28-d)/6 = (28-10)/6 = 18/6
3 hours at 6 mph
d 28  d

4
10 6

6d  10d  240  280
 6   d   28  d   10 
 60     
  60   (4)(60)
   10   6   

 4d  40

d  10

6d  (280  10d )  240
Motion Problem - Tom runs 8 mph and Jake runs at 6 mph
If they start at the same time and Tom finishes the course
1.5 hours ahead of Jake, Determine the length of the course.
d = length of course
d
t
r
Direction
Distance
Rate
Time
Tom
d
8
d/8 = 36/8 = 4.5
Jake
d
6
d/6 = 36/6 = 6
d d
 -1.5
8 6
6d  72  8d


 6  d   d  8 
 48        48   1.5(48) 
  8   6  
72  8d  6d

72  2d
Distance traveled is 36 miles.
Check: 6 - 4.5 = 1.5 hours ahead of Jake.


6d  8d  72
36  d
Work Problem – One person can do a task in 6 hours. A
second person can do a task in 4 hours. How long will it take
them working together to do the task?
a = rt
t = time working together
Rate of Work
Time Worked
Part of Task
1st person
6 hrs
t
t/6
2nd person
4 hrs
t
t/4
t t
 =1
6 4
4t  6t  24
 4  t   t  6 
 24        24   (1)(24)

 6   4 



10t  24

t  24 /10

t  2.4
It takes 2.4 hours to complete the task working together.
Work Problem – One pipe can fill a tank in 4 hours. Another
pipe can empty a tank in 6 hours. If both pips are open how
long will it take to fill the empty tank.
t = time to fill the tank both valves open
Rate of Work
Time
Part of Task
1st tank
1/6 hrs
t
t/6
2nd tank
1/4 hrs
t
t/4
t t
 1
6 4
4t  6t  24


 4  t   t  6 
 24        24   (1)(24)

 6   4 

-2t  24

t  24 /  2

It takes 12 hours to fill the tank with both valves open.
t  12
Work Problem – Tom and Frank can paint an apartment in
12 hours. Tom can paint the apartment by himself in 18
hours. How long does it take Frank to paint the apartment
by himself?
t = time Frank to paint apartment by himself
Rate of Work
Time
Part of Task
Tom
1/18
12
12/18
Frank
1/t
12
12/t
12 12
 1
18 t
12t  216  18t
 12
18 t 
 18
 


  12 
    18 t  (1)(18t )
  t 
216  18t  12t



216  6t

36  t
It takes Frank 36 hours to paint the apartment by himself.
Work Problem – Bill by himself can sweep up sand in a
parking lot in 20 hours. Sally by herself can sweep up the sand
in 25 hours. After Sally sweeps sand for 10 hours, she stops
and Bill takes over. How long will it take Bill to finish the job?
t = time Frank to paint apartment by himself
Rate of Work
Time
Part of Task
Bill
1/20
t
t/20
Sally
1/25
10
10/25 = 2/5
2 t

1
5 20
8  t  20


 4  2   t 
 20     
 20  (1)(20)

  5   20 
 
t  20  8

t  12
It takes Bill 12 hours to finish the job by himself.
Remember
 It is important that you understanding the problem,
rather than memorizing the procedures.
 Work Problems: times for working together will be less
than times for working alone.
 Rectangle: area = (length)(width)
 distance = (rate)(time)
 It is important that you check the answer in the equation
as well as making sure your answer is reasonable
 If the equation is setup wrong the equation could be
right but the solution could still be incorrect
HOMEWORK 6.7
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