Q18. - Algebra 1

Question

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

$111=z^2+11$

Step-by-Step Solution

Answer

The solution set is {10}

Step 1. Definition.

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

Empty Set: The empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero.

Replacement Set: A set, from which the values of the variable of given equation, are chosen, is known as 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 z 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 the replacement set one by one in the equation-

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

$\begin{array}{l}111=z^2+11\\ \Rightarrow 111={10}^2+11\\ \Rightarrow 111=100+11\\ \Rightarrow 111=111\end{array}$

After solving, we get 111 = 111, which is true.

Hence z = 10 belongs to the solution set.

 

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

$\begin{array}{l}111=z^2+11\\ \Rightarrow 111={12}^2+11\\ \Rightarrow 111=144+11\\ \Rightarrow 111=155\end{array}$

After solving, we get 111 = 155, which is false,

Hence z= 12 doesn’t belong to the solution set.

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

$\begin{array}{l}111=z^2+11\\ \Rightarrow 111={14}^2+11\\ \Rightarrow 111=196+11\\ \Rightarrow 111=207\end{array}$

After solving, we get 111 = 207, 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.

$\begin{array}{l}111=z^2+11\\ \Rightarrow 111={16}^2+11\\ \Rightarrow 111=256+11\\ \Rightarrow 111=267\end{array}$

After solving, we get 111 = 267, 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.

$\begin{array}{l}111=z^2+11\\ \Rightarrow 111={18}^2+11\\ \Rightarrow 111=324+11\\ \Rightarrow 111=335\end{array}$

 

After solving, we get 111 = 335, which is false.

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

We can observe that only values for z = 10 are in the solution set.

Therefore, the solution set is {10}