Chapter 7 Rational Functions - College of the Redwoods

Section 7.8 Applications of Rational Functions 733
1
x + 1
2x + 1 = 7 10
Multiply both sides of this equation by the common denominator 10x(2x + 1).
[ 1
10x(2x + 1)
x + 1 ] [ 7
= 10x(2x + 1)
2x + 1 10]
10(2x + 1) + 10x = 7x(2x + 1)
Expand and simplify each side of this result.
20x + 10 + 10x = 14x 2 + 7x
30x + 10 = 14x 2 + 7x
Again, this equation is nonlinear. We will move everything to the right-hand side of
this equation. Subtract 30x and 10 from both sides of the equation to obtain
0 = 14x 2 + 7x − 30x − 10
0 = 14x 2 − 23x − 10.
Note that the right-hand side of this equation is quadratic with ac = (14)(−10) = −140.
The integer pair {5, −28} has product −140 and sum −23. Break up the middle term
using this pair and factor by grouping.
0 = 14x 2 + 5x − 28x − 10
0 = x(14x + 5) − 2(14x + 5)
0 = (x − 2)(14x + 5)
Using the zero product property, either
x − 2 = 0 or 14x + 5 = 0.
These linear equations are easily solved for x, providing
x = 2 or x = − 5
14 .
We still need to answer the question, which was to find two numbers such that the
sum of their reciprocals is 7/10. Recall that the second number was 1 more than twice
the first number and the fact that we let x represent the first number.
Consequently, if the first number is x = 2, then the second number is 2x + 1, or
2(2) + 1. That is, the second number is 5. Let’s check to see if the pair {2, 5} is a
solution by computing the sum of the reciprocals of 2 and 5.
Thus, the pair {2, 5} is a solution.
1
2 + 1 5 = 5 10 + 2 10 = 7 10
Version: Fall 2007