Solving Problems Involving Rational Equations
TOPIC: PROBLEM SOLVING INVOLVING RATIONAL EQUATION
FEATURING: WORK PROBLEM
At the end of the lesson the students are expected to:
1. Gain skill in solving work and motion problems.
2. Understand the techniques in solving work and motion problems.
3. Gain skill in manipulating rational equations.
4. Gain ability in analyzing and solving Algebra Word Problems.
INTRODUCTION: Why a lesson in solving rational equation?
Rational Equations is one of the most important concept in Algebra. This subject has sought many great applications in many fields of Mathematics.
In Trigonometry, applications involving solving oblique triangle require the use of rational equation. The Law of Lines, the Law of Corines and Law of Tangents are usually expressed as a rational equations.
In Physics, majority of the application problems involve rational equations. Most of the problem in Mechanics involve skill in manipulation of rational equations.
May it be a problem involving motion or force, it requires skill in handling rational equations.
In Mathematics of Investment, applying the concept of rational equations remains inevitable. Problems in interest and discount require skills in handling rational equations.
In fact, all fields of Mathematics requires handling of rational equations. Next to linear equation, I believe firmly that The concept of “Rational Equations” is the most important concept in Algebra.
Problem Involving Work
Problem Number One:
Leonor mows a lawn in 4 hours Alex can mow the same lawn in 5 hours. How long would it take both of them, working together, to mow the lawn.
In solving this problem, one consider first how much of the job is done in one hour, two hours and so on.
It takes Leo 4 hours to mow the entire lawn . Thus after one hour, he has done ¼ of the lawn. It takes Alex 5 hours to mow the entire lawn. Thus after one hour he has done 1/5 of the entire lawn.
Let t = the number of hours it will take both Leo and Alex
to finish the lawn working together
t (1/4 + 1/5) = 1
(t/4 + t/5 = 1) 20
5t + 4t = 20
9t/9 = 20/9
t = 20/9 or 2 and 2/9 hours
Answer : 2-2/9
Sample Problem Two:
It takes Ramil 9 hours longer to construct a fence than it takes Ency. If they work together, they can construct the fence in 20 hours. How long would it takes each working along to construct the fence?
Let t = the amount of time it would take Ency working alone.
t + 9 = the amount of time it would take Ramil working alone
Using the same reasoning in the previous problem, we can assume then that Ency has finished 1/t of the fence after one hour and Ramil has finished 1/t+9 of the fence after one hour. Since Ramil and Ency can complete the entire fence in 20 hours, we have:
20 (1/t + 1/t+9) = 1
as our working equation:
20/t + 20/t+9 = 1 LCM: (t) (t + 9)
(t) (t + 9) (20/t + 20/t+9) = 1 (t) (t + 9)
(t + 9)20 + t . 20 = (t) (t + 9)
20 + 180 + 20t = t2 + 9t
40T + 180 = t2 + 9t
0 = t2 – 31t – 180
(t – 36) (t + 5) = 0
t – 36 = 0 t + 5 = 0
t = 36 t = -5
It takes 36 hours for Ency to do the fence alone and it takes 45 hours for Ramil to do the work alone.
Check: 20/36 + 20/45 = 1
5/9 + 4/9 = 1
9/9 = 1
Part Two: MOTION PROBLEM
Biker A bikes 15 km.hr faster than Biker B. By the time it takes Biker A to reach 80 km. Biker B has gone 50 km. Find the speed of each biker.
Let r = The rate in km/hr. of biker B
r + 15 = the rate in km/hr of Biker A
Distance Speed Time
A 80 r + 15 t
B 50 r t
Since time = distance/rate
Distance Speed Time
A 80 r + 15 80/r+15
B 50 r 50/r
From the problem, it is given that the time is the same, we can now have our working equation as:
50/t = 80/r+15
r (r+15) = 50/r = 80/r+15 = (r) (r+15)
50 (r+15) = 80R
50r + 750 = 80r
750 = 80r – 50r
750 = 80r – 50r
750/30 = 30r/30
r = 25
Biker B is going at 25 km/hr. and Biker A’s speed is 40 km./hr.
Mastering Intermediate Algebra
by: Simon L Chua
Benson S. Tan
Roberto G. Degolacion
Ma. Salome B. Aguinaldo
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