Chapter 5.1, Problem 6WE

To determine

To find: Simplify the given expression.

Answer to Problem 6WE

  $\frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3x}{2y}$

Explanation of Solution

Given information:

  $\frac{48x^5{y}^5}{32x^4{y}^6}$

Concept:

Using law of exponents

  $\frac{a^m}{a^n}=a^{m-n}$

Where $a$ be any value.

  $\frac{48x^5{y}^5}{32x^4{y}^6}$

  $\Rightarrow \frac{48x^5{y}^5}{32x^4{y}^6}=\frac{16\times 3x^5{y}^5}{16\times 2x^4{y}^6}$

Cancelling $-5$ in numerator and denominator

We get

  $\begin{array}{l}\Rightarrow \frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3x^5{y}^5}{2x^4{y}^6}\\ \Rightarrow \frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3}{2}{x}^{5-4}{y}^{5-6}\\ \Rightarrow \frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3}{2}x{y}^{-1}\\ \Rightarrow \frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3x}{2y}\end{array}$

Therefore,

  $\frac{48x^5{y}^5}{32x^4{y}^6}=\frac{3x}{2y}$