Chapter 11.2, Problem 1P

(a)

To determine

Multiply: $\frac{5}{y}·\frac{3}{y^3}$

Answer to Problem 1P

  $\frac{5}{y}·\frac{3}{y^3}=\frac{15}{y^4}$

Explanation of Solution

Given:

Multiply: $\frac{5}{y}·\frac{3}{y^3}$

Concept Used:

Rational expression is in the form $\frac{\text{P}}{\text{Q}}$ , where $\text{Q}\ne 0$ , denominator cannot be zero in any rational expression.

Excluded values are also called points of discontinuity. These are the values that make the denominator equal to zero and are not part of the domain.

In any rational function, the denominator cannot be zero.

To find the excluded value of the function, make denominator equal to zero and simplify

Calculation:

  $\frac{5}{y}·\frac{3}{y^3}=\frac{5·3}{y^1·y^3}=\frac{15}{y^{1+3}}=\frac{15}{y^4}$; Excluded value: $y\ne 0$

Thus, $\frac{5}{y}·\frac{3}{y^3}=\frac{15}{y^4}$

(b)

To determine

Multiply: $\frac{x}{x-2}·\frac{x+1}{x-3}$

Answer to Problem 1P

  $\frac{x}{x-2}·\frac{x+1}{x-3}=\frac{x(x+1)}{(x-2)(x-3)}$

Explanation of Solution

Given:

Multiply: $\frac{x}{x-2}·\frac{x+1}{x-3}$

Concept Used:

Rational expression is in the form $\frac{\text{P}}{\text{Q}}$ , where $\text{Q}\ne 0$ , denominator cannot be zero in any rational expression.

Excluded values are also called points of discontinuity. These are the values that make the denominator equal to zero and are not part of the domain.

In any rational function, the denominator cannot be zero.

To find the excluded value of the function, make denominator equal to zero and simplify

Calculation:

  $\frac{x}{x-2}·\frac{x+1}{x-3}=\frac{x(x+1)}{(x-2)(x-3)}$

Excluded values: $x-2\ne 0\Rightarrow x\ne 2\text{and}x-3\ne 0\Rightarrow x\ne 3$

Thus, $\frac{x}{x-2}·\frac{x+1}{x-3}=\frac{x(x+1)}{(x-2)(x-3)}$