Find the solution of each equation if the replacement sets are $x:\left\{1,3,5,7,9\right\}$and $y:\left\{6,8,10,12,14\right\}$

$4y=32$

We have an equation$4y=32$ and a set$y:\left\{6,8,10,12,14\right\}$

To find the solution of the given equation at every value of $y$.

We substitute the values of $y$in the given equation one by one and check which value of $y$makes both the sides of the equation equal.

Which value of $y$makes both the sides of the equation that is called the solution of the given equation.

Substitute $y=6$in the equation $4y=32$

$$\begin{gathered} 4·6=32 \\ \Rightarrow 24\ne 32 \end{gathered}$$

$y=6$is not a solution

Substitute $y=8$in the equation $4y=32$

$$\begin{gathered} 4·8=32 \\ \Rightarrow 32=32 \end{gathered}$$

$y=8$is a solution

Substitute $y=10$in the equation$4y=32$

$$\begin{gathered} 4·10=32 \\ \Rightarrow 40=32 \end{gathered}$$

$y=10$is not solution

Substitute $y=12$in the equation$4y=32$

$$\begin{gathered} 4·12=32 \\ \Rightarrow 48=32 \end{gathered}$$

$y=12$is not a solution

Substitute $y=14$in the equation$4y=32$

$$\begin{gathered} 4·14=32 \\ \Rightarrow 56=32 \end{gathered}$$

$y=14$is not a solution.

Hence, the$y=8$ is solution of the given equation.