7.5 – Rational Equations/Work Problems
Algebra 2H
7.5 – Solving Rational Equations
1. FACTOR, FACTOR, FACTOR denominators
2. Multiply the ENTIRE equation through by the common denominator (this should cancel out the
denominators)
3. Solve
Example #1: Solve
5
2
1


24 3  x 4
5
2
1


24 3  x 4
Brightstorm Video
Factor the denominators
24   2  2  2  3 
 3  x  already
4   2  2 
factored
Common denominator =  2 2 2 3  3  x   24 3  x 
 24  3  x    5   24  3  x    2   24  3  x    1 

    
 
  
  
1
1
1

  24  
 3  x  
 4 
 24  3  x    5   24  3  x 




  24  
1
1



Multiply by the common
denominator
  2   24 6  3  x    1 

 

 4 
  3  x  
1


5  3  x    24  2   6 3  x 
15  5 x  48  18  6 x
5 x  63  18  6 x
x  63  18
x  45
Solve
7.5 – Rational Equations/Work Problems
Algebra 2H
Example #2: Solve
p2  p  1 p2  7
 2
p
p 1
p 1
Brightstorm Video 2
p2  p  1
p2  7
p


p 1
p

1
p

1


 1
Factor denominators
Common denominator =  p  1 p  1
Multiply by common denominator
   p  1 p  1   p 
  p  1 p  1   p2  p  1    p  1 p  1  
p2  7


  

 


 

  

1
1
1

 p 1  
   p  1 p  1  
 1 
  p  1  p  1   p2  p  1    p  1 p  1




  p  1  
1
1




p2  7

   p  1 p  1

   p  1 p  1  p
 

 

 
1
 1 

 p  1  p2  p  1  p2  7   p  1 p  1 p 
p3  p 2  p 2  p  p  1  p 2  7 
 p
2
 
 1  p
Multiply through
p3  p 2  p 2  p  p  1  p 2  7  p3  p
p3  2 p 2  2 p  1  p 3  p 2  p  7
Combine like terms
p3  2 p 2  2 p  1  p3  p 2  p  7
Solve
3 p 2  3 p  6  0
3 p2  3 p  6  0

Factor

3 p2  p  2  0
3  p  2  p  1  0
p20
p2
p  1 0
p  1
-1 is an extraneous solution – doesn’t work as it
would make the denominator of the original
equation 0.
7.5 – Rational Equations/Work Problems
Algebra 2H
Example #3: Working along, Private Eye can dig a hole in 20 minutes. Working alone, Corporal
Punishment can dig a hole in 15 minutes. How long will it take them to dig a hole together?
How much of the job can Private Eye do in 1 minute?
1
20
Corporal Punishment?
1
15
t = time working together
t
Private Eye =
of job
20
t
t

1
20 15
Corporal Punishment =
t
of job
15
Together they do 1 job
 60   t   60   t   60 
 1   20    1   15    1  1
       
3t  4t  60
Multiply by common denominator
7t  60
t
60
 8.57
7
How would the problem change if Corporal Punishment were filling the hole back up as Private Eye was
digging?
You would have to subtract Corporal Punishments amount since it is working against the goal.
What if Private Eye and Corporal Punishment wanted to dig two holes?
You would set the equation equal to 2 instead of 1.
Brightstorm Video