Chapter 5.9, Problem 10WE

To determine

To solve: the given equation for $\mathrm{y}$ . Check the result.

Answer to Problem 10WE

The value of $\mathrm{y}$ in the given equation is $-7$ .

Explanation of Solution

Given information:

Given equation is

  $\frac{{\left(\mathrm{y}+1\right)}^2}{{\left(\mathrm{y}-3\right)}^2}=1$

Calculation:

Consider the given equation,

  $\frac{{\left(\mathrm{y}+1\right)}^2}{{\left(\mathrm{y}-3\right)}^2}=1$

Now, solve the equation as shown,

  $\begin{array}{c}\frac{{\left(\mathrm{y}+1\right)}^2}{{\left(\mathrm{y}-3\right)}^2}=1\\ {\left(\mathrm{y}+1\right)}^2={\left(\mathrm{y}-3\right)}^2\\ {\mathrm{y}}^2+2\mathrm{y}+1={\mathrm{y}}^2-6\mathrm{y}+9\\ {\mathrm{y}}^2-{\mathrm{y}}^2+2\mathrm{y}+6\mathrm{y}=9-1\\ 8\mathrm{y}=8\\ \mathrm{y}=1\end{array}$

Now, check the result by back substitution.

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

Which is true.

Hence, the value of $\mathrm{y}$ in the given equation is $1$ .