Chapter 9.6, Problem 33PPS

a.

To determine

To calculate: the equation for x.

Answer to Problem 33PPS

Only one solution is $x=1$ .

Explanation of Solution

Given information: Consider $\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}$ .

Calculation:

Consider $\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}$ .

The LCD for the terms is $x\left(x-3\right)$ .

Multiply both side by LCD

  $\begin{array}{l}2x+x-3=x^2-x\\ x^2-4x+3=0\\ x^2-3x-x+3=0\\ \left(x-3\right)\left(x-1\right)=0\\ x=3,1\end{array}$

The value $x=3$ is extraneous.

b.

To determine

To graph: Graph $y_1=\frac{2}{x-3}+\frac{1}{x}$ and $y_2=\frac{x-1}{x-3}$ .

Explanation of Solution

Given information: Graph $y_1=\frac{2}{x-3}+\frac{1}{x}$ and $y_2=\frac{x-1}{x-3}$ on the same graph for $0<x<5$ .

Graph: Graph $y_1=\frac{2}{x-3}+\frac{1}{x}$ and $y_2=\frac{x-1}{x-3}$ on the same graph for $0<x<5$ .

  

c.

To determine

To find: for what value of x both functions intersect.

Answer to Problem 33PPS

the two functions intersect for $x=1$ .

Explanation of Solution

Given information: Function $y_1=\frac{2}{x-3}+\frac{1}{x}$ and $y_2=\frac{x-1}{x-3}$ .

Calculation:

Function $y_1=\frac{2}{x-3}+\frac{1}{x}$ and $y_2=\frac{x-1}{x-3}$

Graphed both functions on the same graph for $0<x<5$ .

  

From the graph notice that the two functions intersect for $x=1$ .

The graphs do not intersect at $x=3$ because they both have a vertical asymptote in $x=3$ .