Q52. - Algebra 1

Question

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

$4y=32$

Step-by-Step Solution

Answer

The solution of the equation $4y=32$ is $y=8$.

Step 1. Given information.

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 $y$.

Step 2. Apply 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.

Which 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 $4y=32$

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

  $y=6$ 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.