## Introductory Algebra for College Students (7th Edition)

$y=\pm\dfrac{5}{4}$
Using the properties of equality, the given equation, $16y^2=25 ,$ is equivalent to \begin{array}{l}\require{cancel} y^2=\dfrac{25}{16} .\end{array} Taking the square root of both sides, then the solutions of the given equation are \begin{array}{l}\require{cancel} y=\pm\sqrt{\dfrac{25}{16}} \\\\ y=\pm\sqrt{\left(\dfrac{5}{4}\right)^2} \\\\ y=\pm\dfrac{5}{4} .\end{array}