Bell Work: Graph on a number line I x I > 0

Answer: All reals

In Lesson 78 we discussed the solution of rational equations in which all denominators are integers. In these problems we found that the recommended first step is to multiply the numerator of every term in the equation by the least common multiple of the denominators. Since every denominator is a factor of the least common multiple, this procedure permits us to cancel every denominator, as we see in the following example.

Example: Solve y + 1 = y 2 4 6

Answer: y = -3/4

The denominators from the previous example are all real numbers. In this lesson we will discuss equations whose denominators contain variables. If there are variables in the denominator, the replacement values of the variables are restricted.

For example, if the given equation is t – 2 = t 3t 3 The number zero would not be a permissible value for t because of division by zero.

If our equation is n = 3 n We cannot accept -2 as a value for n because of division by zero.

Thus, as our first step in the solution of rational equations whose terms have variables in one or more denominators, we will list the unacceptable values of the variable, which, of course, are those values of the variable that would cause any denominator to equal zero.

Example: Solve: t – 2 = t 3t 3

Answer: 3t – 6 = 14 – t t = 5

Example: Solve: n - 3 = 0 n + 2 5

Answer: 5n – 3(n + 2) = 0 n = 3

Example: Solve: = 0 3n n + 4

Answer: 2(n + 4) – 6n = 0 n = 2

Example: Solve: = 0 x x – 2

Answer: 4x – 8 – 7x = 0 x = -8/3

Example: Solve: = 0 p p - 4

Answer: 4p – 16 – 3p = 0 p = 16

HW: Lesson 103 #1-30