Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\bf{\text{Solution Outline:}}$ Solve the solution sets of the given statements, \begin{array}{l}\require{cancel} \dfrac{3}{5}a+\dfrac{1}{5}=2 \\\text{and}\\ 3a+1=10 .\end{array} If the solution sets are the same, then they are equivalent. Otherwise, they are not equivalent. $\bf{\text{Solution Details:}}$ Using the properties of equality, the first equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{3}{5}a+\dfrac{1}{5}=2 \\\\ 5\left( \dfrac{3}{5}a+\dfrac{1}{5} \right)=5(2) \\\\ 3a+1=10 .\end{array} Since the equation above is the same as the second given equation, then the given are $\text{ equivalent equations .}$