91Ó°ÊÓ

We have an equation$\frac{42}{y}=7$ 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.

The 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$\frac{42}{y}=7$

localid="1642045240621" $$\begin{gathered} \frac{42}{6}=7 \\ ⇒7=7 \end{gathered}$$

$y=6$is a solution

Substitute $y=8$in the equation$\frac{42}{y}=7$

$$\begin{gathered} \frac{42}{8}=7 \\ ⇒5\frac{2}{8}≠7 \end{gathered}$$

$y=8$is not a solution

Substitute $y=10$in the equation$\frac{42}{y}=7$

localid="1645195241313" $$\begin{gathered} \frac{42}{10}=7 \\ ⇒4.2≠7 \end{gathered}$$

$y=10$is not solution

Substitute$y=12$in the equation$\frac{42}{y}=7$

$\frac{42}{12}≠7$

$y=12$is not a solution

Substitute$y=14$in the equation$\frac{42}{y}=7$

$$\begin{gathered} \frac{42}{14}=7 \\ ⇒3≠7 \end{gathered}$$

$y=14$is not a solution

Hence, the$y=6$is the solution of the given equation.