Q54. - Algebra 1

Question

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

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

Step-by-Step Solution

Answer

The solution of the equation $\frac{42}{y}=7$ is $y=6$.

Step 1. Given information.

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 $y$.

Step 2. Apply the concept.

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.

Step 3. Calculation.

Substitute $y=6$ in the equation $\frac{42}{y}=7$

       $$\begin{gathered} \frac{42}{6}=7 \\ \Rightarrow 7=7 \end{gathered}$$

  $y=6$ is a solution 

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

           $$\begin{gathered} \frac{42}{8}=7 \\ \Rightarrow 5\frac{2}{8}\ne 7 \end{gathered}$$

  $y=8$ is not a solution 

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

           $$\begin{gathered} \frac{42}{10}=7 \\ \Rightarrow 4.2\ne 7 \end{gathered}$$

  $y=10$ is not solution 

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

        $\frac{42}{12}\ne 7$

  $y=12$ is not a solution 

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

        $$\begin{gathered} \frac{42}{14}=7 \\ \Rightarrow 3\ne 7 \end{gathered}$$

  $y=14$ is not a solution

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