Chapter 6.7, Problem 18E

To determine

To simplify: the expression $\frac{1-2u}{1+\frac{1}{2u}}÷\frac{1+2u}{1-\frac{1}{2u}}$.

Answer to Problem 18E

  $-{\left(\frac{(2u-1)}{(2u+1)}\right)}^2$.

Explanation of Solution

Given information:

  $\frac{1-2u}{1+\frac{1}{2u}}÷\frac{1+2u}{1-\frac{1}{2u}}$

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 ,

  $\frac{1-2u}{1+\frac{1}{2u}}÷\frac{1+2u}{1-\frac{1}{2u}}$

Re-writing the expression we get,

  $\frac{\frac{1-2u}{1+\frac{1}{2u}}}{\frac{1+2u}{1-\frac{1}{2u}}}$

Therefore,

Solving the first term $\frac{1-2u}{1+\frac{1}{2u}}$

Numerator $=1-2u$

Denominator $=1+\frac{1}{2u}$

LCM $=2u$

  $=\frac{2u+1}{2u}$

Hence, the first term is

  $\frac{1-2u}{\frac{2u+1}{2u}}$

  $=\frac{1-2u}{\frac{2u+1}{2u}}=\frac{1-2u}{1}÷\frac{2u+1}{2u}$

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

Now, solving the second term $\frac{1+2u}{1-\frac{1}{2u}}$

Numerator $=1+2u$

Denominator $=1-\frac{1}{2u}$

LCM $=2u$

  $=\frac{2u-1}{2u}$

Hence, the second term is

  $\frac{1+2u}{\frac{2u-1}{2u}}$

  $=\frac{1+2u}{\frac{2u-1}{2u}}=\frac{1+2u}{1}÷\frac{2u-1}{2u}$

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

With end with the expression

  $\frac{\frac{2u(1-2u)}{2u+1}}{\frac{2u(2u+1)}{2u-1}}=\frac{2u(1-2u)}{2u+1}÷\frac{2u(2u+1)}{2u-1}$

  $\begin{array}{l}=\frac{2u(1-2u)}{2u+1}\times \frac{2u-1}{2u(2u+1)}\\ =\frac{-(2u-1)(2u-1)}{(2u+1)(2u+1)}\\ =-{\left(\frac{(2u-1)}{(2u+1)}\right)}^2\end{array}$

Thus, the solution is $-{\left(\frac{(2u-1)}{(2u+1)}\right)}^2$.