Question

Find the solution set for each equation if the replacement sets are$\mathit{y}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{9}\mathbf{\}}$ and$\mathit{z}\mathbf{=}\mathbf{\{}\mathbf{10}\mathbf{,}\mathbf{12}\mathbf{,}\mathbf{14}\mathbf{,}\mathbf{16}\mathbf{,}\mathbf{18}\mathbf{\}}$

role="math" localid="1643607471298" $\mathit{z}\mathbf{+}\mathbf{10}\mathbf{=}\mathbf{22}$

Short answer

The solution set is$\left\{12\right\}$

Step-by-step solution

Step 1. Definition.

Set:- A set is a collection of objects or numbers which is shown using braces.

Replacement Set: A set, from which the values of the variable of the given equation, are chosen, is known as a replacement set.

Solution Set: A solution set is a set of elements chosen from the replacement set which, satisfy the equation or make an open sentence true.

Step 2. Method to find the values of the solution set.

To find the solution set from the replacement set,

We have to follow the given steps:

1. Take value for n from the replacement set.
2. Solve the equation for z.
3. After solving if both sides have equal value.
   Take z as an element of the solution set.

Step 3. Substitute the values of a replacement set in the equation.

Let’s substitute the values of a replacement set one by one in the equation-

Take z = 10 from the replacement set and put it in the given equation (open sentence).

$\begin{array}{l}z+10=22\\ \Rightarrow 10+10=22\\ \Rightarrow 20=22\end{array}$

After solving, we get 20 = 22, which is False.

Hence z = 10 doesn’t belong to the solution set.

Take z = 12 from the replacement set and put it in the given equation.

$\begin{array}{l}z+10=22\\ \Rightarrow 12+10=22\\ \Rightarrow 22=22\end{array}$

After solving, we get 22 = 22, which is True,

Hence z = 12 belongs to the solution set.

Take z = 14 from the replacement set and put it in the equation.

$\begin{array}{l}z+10=22\\ \Rightarrow 14+10=22\\ \Rightarrow 24=22\end{array}$

After solving, we get 24 = 22, which is False.

Hence z = 14 doesn’t belong to the solution set.

Take z = 16 from the replacement set and put it in the equation.

$\begin{array}{l}z+10=22\\ \Rightarrow 16+10=22\\ \Rightarrow 26=22\end{array}$

After solving, we get 26 = 22, which is False.

Hence z = 16 doesn’t belong to the solution set.

Take z = 18 from the replacement set and put it in the equation.

$\begin{array}{l}z+10=22\\ \Rightarrow 18+10=22\\ \Rightarrow 28=22\end{array}$

After solving, we get 28 = 22, which is False.

Hence z = 18 doesn’t belong to the solution set.

We can observe that only the value for z = 12 belongs to the solution set.

Therefore, the solution set is {12}