Chapter 6.7, Problem 21E

To determine

To express: x in terms of z.

Answer to Problem 21E

  $x=\frac{z}{2-z}$.

Explanation of Solution

Given information:

  $x=\frac{y-1}{y+1}\text{ and }y=\frac{1}{1-z}$

Formula used:

Simplify the numerator and denominator . Then express the fraction as quotient using the ÷ sign. Multiply by the reciprocal of the divisor.

Calculation:

Consider ,

  $x=\frac{y-1}{y+1}\text{ and }y=\frac{1}{1-z}$

Therefore, we have

  $x=\frac{\frac{1}{1-z}-1}{\frac{1}{1-z}+1}$

Numerator $=\frac{1}{1-z}-1$

LCM $=1-z$

  $\begin{array}{l}=\frac{1-1(1-z)}{1-z}\\ =\frac{1-1+z}{1-z}\\ =\frac{z}{1-z}\end{array}$

Denominator $=\frac{1}{1-z}+1$

LCM $=1-z$

  $\begin{array}{l}=\frac{1+1(1-z)}{1-z}\\ =\frac{1+1-z}{1-z}\\ =\frac{2-z}{1-z}\end{array}$

The new expression is

  $\frac{\frac{z}{1-z}}{\frac{2-z}{1-z}}$

  $=\frac{\frac{z}{1-z}}{\frac{2-z}{1-z}}=\frac{z}{1-z}÷\frac{2-z}{1-z}$

  $\begin{array}{l}=\frac{z}{1-z}\times \frac{1-z}{2-z}\\ =\frac{z}{2-z}\end{array}$

Hence, $x\text{ in terms }z\text{ is }\frac{z}{2-z}.$.