rational.html

RATIONAL EQUATIONS

Objectives: To recall the definition of a rational equation and the legal rules for manipulating such equations.

Note

In this Workout you will study fractional equations that do not necessarily reduce to linear equations in x.
The equations that you will encounter may reduce to higher degree polynomials in x. The solutions to such equations may involve techniques that you have seen before.

Recall 1: (Definition of a rational equation)

A fractional equation is an equation that contains a fraction with a variable in the denominator. The usual method for attacking such equations is to

1. Find the Least Common Divisor of the fractions that occur in the equation.
2. Use the L.C.D. to rewrite the equation as an equation    involving a simple fraction on one side.
3. Use the distributive law or any other arithmetic laws    that you may know about to cancel as many terms as you    can from the fraction.
4. Solve the resulting equation by techniques that you may    have learned in the previous Workouts.

Examples

Example 1. Reduce the rational equation

to a polynomial equation.

Solution: The two fractions on the right hand side have the same denominator, so we may combine those fractions to get the equation

The L.C.D. of these two fractions is (3x + 5)(x + 5). Multiply both sides of the equation by this L.C.D. to get

Now cancel the 3x + 5 from the left hand side and the x + 5 from the right hand side. This will give

(x + 5) (2(x - 5)) = (3x + 5)(x - 2)

Now use the distributive law and multiply the terms together to get

2x^2 - 50 = 3x^2 - x - 10,
or
x^2 - x + 40=0

Example 2. Solve the following equation for x.

Solution: The L.C.D. is 6(x + 1). Multiply both sides by this L.C.D. to get

This reduces to 6(x - 1) = (x + 1)(x^2 - x). Use the distributive law and multiply out the terms on the right side to get

x^3 - 7x + 6 = 0.

Now, this is a cubic polynomial. Therefore, there are three solutions to this equation. There are two ways to solve the equation.

1. Try the factors of 6 as possible solutions. They are ±1, ±2, ±3, and ±6. Try x = 1 to find that it is a solution to the cubic equation above. This means that (x - 1) is a factor of x^3 - 7x + 6. Use long division to find the other factors.

            x2 +   x - 6
x - 1) x^3 + 0x^2- 7x + 6
           x^3 -   x^2
                     x^2 - 7x
                     x^2 - x
                            -6x + 6
                            -6x + 6
                                     0

The quadratic x^2 + x - 6 factors as (x - 2)(x + 3), so that the cubic equation factors as (x - 1)(x - 2)(x + 3) = 0. This means that the three solutions to the original rational equation are x = 1, x = 2 and x = -3.

2. The second way is actually simpler. But watch carefully. We can factor out a copy of x from the term on the right side of the equation
6(x - 1) = (x + 1)(x^2 - x)

to get

6(x - 1) = x(x + 1)(x - 1)

At this point it might be nice to be able to cancel (x - 1) from both sides. Can this be done legally? To do so would involve dividing both sides by (x - 1), which is legal as long as x - 1 is different from 0; in other words, it is legal as long as x is not 1. One solution to the equation is x = 1. So eliminate the one solution x = 1 and look for the other solutions. In other words stay away from x = 1 and it is legal to cancel the (x - 1) from both sides of the equation to get

6 = x(x + 1)

Use the distributive law and collect terms to one side to rewrite the equation as the quadratic
x^2 + x - 6 = 0.

This equation factors as
(x - 2)(x + 3) = 0,
and therefore we have two more solutions, x = 2 and x = -3. These solutions, together with the solution x = 1 that was carefully put aside gives the three solutions to the original rational equation.

Note
Remember to check for solutions that you may have accidentally eliminated. Such was the case with cancelling the (x - 1) from both sides of the equation
6(x - 1) = x(x + 1)(x - 1)

You may cancel these factors; but take care to check if x = 1 is a solution. If it is, remember to list it as a solution, together with the other solutions.

Note
It may happen that the solution that you find is also a pole of the equation. A pole is a value at which the denominator of one of the fractions of the equation becomes zero. If a solution turns out to also be a pole, that solution must be rejected. Example 4 shows a case where one of the solutions turns out to also be a pole.

Example 3. Solve the equation below for x

Solution: Notice that x^2 + x - 6 = (x - 2)(x + 3). Therefore the L.C.D. for the fractions on both sides of the equation is (x - 2)(x + 3).

If you multiply both sides of the equation by this L.C.D. and simplify, you should get
x^2 - 3x + 1 = 0.

This quadratic has solutions that may be found through the use of the quadratic formula. They are

Example.4. Find all solutions to the rational equation

Solution: Notice that x^2 + x - 6 = (x - 2)(x + 3). Therefore the L.C.D. for the fractions on both sides of the equation is (x - 2)(x + 3).

If you multiply both sides of the equation by this L.C.D. and simplify, you should get

(x + 3)(x - 3) = (x - 2)(2x - 6) + (5x - 15),

which simplifies to

x^2 - 5x + 6 = 0.
This factors as

(x - 2)(x - 3) = 0.

It would seem that the solutions are x = 2 and x = 3. However, notice that x = 2 is a pole. If you test x = 2 as a possible solution, you find that the rational equation does not have finite values when x = 2, and therefore does not make mathematical sense. We must therefore discard the solution x = 2 and say that the only solution to the original rational equation is x = 3.

Note
It is possible for the solution to a rational equation to include almost all real numbers. For example, consider the rational equation

The only value of x that is not satisfied by this equation is x = -1. (See Exercises 7 and 10 for examples of this situation.) Indeed, there are many equations whose solutions are all real numbers. If you take x = x as an honest equation in x, then the solution will include any real number that is equal to itself; but, every real number is equal to itself!

Warning
Always check your solutions, not only because you should check if you made some arithmetic mistake, but also because you may have a solution that should be discarded because it is also a pole.

EXERCISES

1. Solve for x in the following rational equation