Chapter 1.7, Problem 30WE

To determine

Whether the given numbers are the solution or not.

Answer to Problem 30WE

y = 0 is not the solution and y = 2 is the solution.

Explanation of Solution

Given:

The given expression is:

  $\frac{2y}{2y-1}=\frac{y+2}{y+1}$

Calculation:

Put $y=0$ in the given expression.

  $\begin{array}{l}\frac{2y}{2y-1}=\frac{y+2}{y+1}\\ \frac{2\left(0\right)}{2\left(0\right)-1}=\frac{\left(0\right)+2}{\left(0\right)+1}\\ \frac{0}{-1}=\frac{2}{1}\\ 0=2\\ \text{not possible}\end{array}$

Put $y=2$ in the given expression.

  $\begin{array}{l}\frac{2y}{2y-1}=\frac{y+2}{y+1}\\ \frac{2\left(2\right)}{2\left(2\right)-1}=\frac{2+2}{2+1}\\ \frac{4}{4-1}=\frac{4}{3}\\ \frac{4}{3}=\frac{4}{3}\end{array}$

Hence, y = 0 is not the solution and y = 2 is the solution.