## Intermediate Algebra (6th Edition)

Squaring both sides of the given equation, $\sqrt{y+5}=2-\sqrt{y-4} ,$ results to \begin{array}{l}\require{cancel} (\sqrt{y+5})^2=(2-\sqrt{y-4})^2 \\ y+5=(2)^2-2(2)(\sqrt{y-4})+(\sqrt{y-4})^2 \\ y+5=4-4\sqrt{y-4}+y-4 \\ y+5=y-4\sqrt{y-4} \\ y-y+5=-4\sqrt{y-4} \\ 5=-4\sqrt{y-4} \\ \dfrac{5}{-4}=\sqrt{y-4} \\ \sqrt{y-4}=-\dfrac{5}{4} .\end{array} The radical expression at the left side of the equation above reults to a nonnegative number. This will never be equal to the negative expression at the right. Hence, there is $\text{ no solution }.$