POLYNOMIAL FRACTIONS

Objective:  To recall how to convert rational equations to their equivalent polynomial equations.

Recall 1: A rational expression is a ratio of two polynomials.  A   rational equation is an equation involving rational   expressions.  For example,

  is a rational equation.  This equation can be converted to a polynomial equation by multiplying both sides by 2x^2   We   get:

Simplifying,

    x^2 - 1 = 2x(x - 1),

  or

    x^2 - 2x + 1 = 0.

  or

    (x - 1)^2 = 0.

  This--of course--means that x = 1.

Note 1
Once again, be aware that you should check your final answer by substituting it for the variable in the original equation.  It may happen that one of your solutions is not valid.  Consider the next example.

Recall 2: Consider the equation

  Multiply both sides by the least common divisor (x - 2)(4 - x).
  You get:

             3(4 - x) = (x - 2)x.

  This is equivalent to the polynomial

            x^2 + x - 12 = 0.

  This is a quadratic polynomial that can be solved for x.

Recall 3: (Terminology)

  A pole of a rational expression is a value of the variable where the expression is not defined.

For example 0 is a pole of the expression 1/x.

  On rare occasions, you may find that one of the candidates  for a solution to a rational equation turns out to also be a  pole of one of the expressions in the equation.  If this   happens, the candidate for the solution must be rejected.   Example 2 demonstrates this case.

Note 2
It may happen that a rational expression does not have any solution.  Consider Example 2.

Examples

Example 1. Find the values of x that satisfy:

Solution: This is the same equation that is in Recall 1.  Only this time work it out by considering a method that you may have seen once before.  The method is called cross multiplication.  The method uses the fact that if

            a/b =c / d

then
          a*d = c*b.

Applying the method to our equation, we get

(x^2 - 1)x = (x - 1)(2x^2).

This reduces to the polynomial x^2 - 2x + 1 = 0.  which means that x = 1.

Note 3

In the last example, we did not use the idea of dividing both sides by (x - 1).
It may seem like a quick way to simplify the equation, since
x^2 - 1 = (x - 1)(x + 1).  Notice, that the equation is

(x - 1)(x + 1)x = (x - 1)(2x^2).

Dividing both sides by (x - 1) gives us

(x + 1)*x = (2x^2).

But is this legal?  After all, if x = 1 turns out to be a solution, then we cannot divide by (x - 1), since it would mean dividing by 0.

Example 2. Find all solutions to the rational equation

Solution: The least common divisor is (x - 3)(x + 4).

Note that   x^2 + x -12 = (x - 3)(x + 4).

By multiplying both sides of the equation by the least common divisor, we get

which simplifies to

(x + 4) - 2(x - 3) = 7.

We solve this last equation to find that x = 3.  However, x = 3 is a pole.   (The first term of the original equation is not defined at x = 3.)

So we must reject x = 3 as a solution.  Since there are no other solutions, we must conclude that the equation does not have any solutions.  This can happen.

Warning

Example 2 shows that a candidate for a solution may not be a solution.  The algebra seemed to admit an answer, but the answer did not make sense.  This means that you should always check that your answer makes sense.

Example 3  Find all solutions to the rational equation

Solution: The least common divisor is 3(x - 1).

By multiplying both sides of the equation by the least common divisor, we get

which simplifies to
                  x(x - 1) + 6 = 3(x + 1),
or

                   x^2 - 4x + 3 = 0.
We solve this last equation to find that it has two solutions, x = 3 and x = 1.  However, x = 1 is a pole.   (The right side of the original equation is not defined at x = 1.)

So we must reject x = 1 as a solution.  This means that x = 3 is the only solution to our equation.

Note 4

 The next two examples illustrate word problems. Such problems generally scare students into thinking that the problem is difficult.  Word problems are no more difficult than symbolic algebra problems; however, they require the added task of having to rewrite the words in mathematical or symbolic language. 

Example 4

    A group of lawyers plan to open an office that requires startup costs of $150,000.  Since the group could not raise that much money from its own members, it decided to recruit four additional partners in the corporation.  This decreased the cost for each partner by $10,000.  With the additional partners, the group could manage to raise the required startup cost.  How many partners were in the original group?

Solution: Let x stand for the number of partners in the original group.

Then x + 4 = number in the new group.

We may set up a relation:

[cost per partner in original group] - 10,000 = [cost per partner in new group]

 150,000/x - 10,000=150,000/(x + 4)

Putting the left side over a common denominator and dividing through by 10,000, we get:

which reduces to (x + 4)(15 - x) = 15x.  This, in turn, reduces to the quadratic equation x^2 + 4x - 60 = 0.  This equation is equivalent to

(x + 10)(x - 6) = 0.

We therefore have two solutions x = -10 and x = 6.  We may reject the first solution, because we cannot have a negative number of partners.

Example 5      Two businessmen arrange to meet at a Paris airport.  They both leave from New York; but one must take an 11:00 a.m. flight, the other a 12:00 p.m. flight.  If the flights arrive at the same time and the first flight travels at an average speed that is 20 miles per hour slower than the second, what was the average speed of the first flight?  Assume that Paris is 3,600 flight miles from New York.

Solution:
    Let v stand for the speed of the first flight.
    Then v + 20 = speed of the second flight.

We may set up a relation:

[flight time of first flight in hours] - 1 = [flight time of second flight in hours]

3600/v - 1 = 3600/ (v+20)

Putting the left side over a common denominator, we get:

which reduces to (v + 20)(3600 - v) = 3600v.  This, in turn, reduces to the quadratic equation v^2 + 20v - 72000 = 0.  We may use the quadratic formula to find v:

which--ignoring the negative solution--is approximately 278.51 miles/hour.

In exercises 1 through 9, solve each of the rational equations for x.

1.
 

2. 
 

3. (x + 1)/ (x - 1) = 1/ (x - 2)
 

4.  (x + 1)/ (x - 1) + x / (x + 1)=9
 
 
 

5. (2x + 1)/ x = 7 / (x - 1)
 
 

6. (2x^2 + 1)/x + 1/ (x - 1)=3
 
 

7. 1/(x + 1) - 1= -x / (x + 2)
 
 

8 . 1/ (x - 2) + 2/ (x + 3) = 3/ (x62 + x - 6)
 
 

9. 2x/ (x^2 -4) =3x / (x^2 - 3x + 2)
 
 

10.  Several students plan a party that costs $120.00.   Two new students  decide to contribute to the party.  This decreases the cost per student  by $10.00.  How many students were in the original group?
 
 

11 Achilles and Mr. Tortoise have a 10 mile race.  Achilles maintains a  speed that is 5 miles per hour faster than Mr. Tortoise and finishes 1  hour before Mr. Tortoise.  How fast was Achilles traveling?