We know that while solving equations containing known variables, we use the different properties of equalities. The equation provided to us may contain one or more variables. When we solve an equation we arrive at a particular value of the variable.

It is possible to have more than one value for the variable. In this section let us discuss about solution set of an equation, solution set of an inequality, empty solution, graphing solution set and few examples based on solution sets.

The solution set is the value of the variable in the equation, which when substituted in the given equation, will satisfy both sides of the equation. The solution set consists of one or more possible values of x of an equation.
In the case of a polynomial equation, the solution set is the zero of the polynomial.
When an equation has more than one solution we list them in set notation as x = {  __, __ , __ , . . . . . . }, such that each solution is separated by commas.

Solved Examples

Question 1: Find the solution set of the equation, x + 5 = 16
Solution:
 
We have x + 5 = 16
              =>     x + 5 - 5 = 16 - 5
              =>               x = 11
The Solution Set of the given equation is x = { 11 }
 

Question 2: Verify if x = -3 is the solution set of the equation x2 + 6 x + 9 = 0
Solution:
 
We have x2 +  6 x + 9 = 0
Substituting x = 3, we get
               ( - 3 )2 + 6 ( - 3 ) + 9 = 0
  =>                        9 - 18 + 9 = 0
  =>                            18 - 18 = 0
  =>                                    0 = 0

By substituting x = -3, we obtain the same value on both sides of the equation.
therefore, x = -3 is the solution set of the given equation.

 

Solution set of an equation is the real value of the variable present in the equation.
By substituting the values of the solution set in the original (given) we verify that both sides of the equation remains same.
1. For a given equation we can find the solution set.
2. For the given solution set, we can verify if it forms the correct solution to the given equation.

When we find the solution set of inequality, we need to domain of the variable and then write the possible solutions of the inequality.

Solved Examples

Question 1: Find the solution set of the equation: 3 ( x- 4 ) = 2 ( 9 - x)
Solution:
 
We have 3 ( x - 4 ) = 2 ( 9 - x )
             =>          3 . x - 3(4) = 2(9) - 2 . x
             =>            3 x - 12   = 18 - 2 x
             =>          3 x + 2 x   = 18 + 12
             =>                    5x   = 30

             =>                      x   = $\frac{30}{5}$

                                            = 6

                   The solution set is x = { 6 }
 

Question 2: Write the solution set of x $\le$ 5, and x $\epsilon$ N
Solution:
 
The domain consists of Natural numbers and the inequality is x less than or equal to 5
               ( i. e ) N = { 1, 2, 3, 4, 5,. . . . .       }
               The natural numbers less than or equal to 5 are 1, 2, 3, 4 and 5.

Therefore, The solution set is x = { 1, 2, 3, 4, 5 }
 

Question 3: Write the solution set of x >  - 4 and x $\epsilon$ Z.
Solution:
 
The domain consists of Integers and the inequality is x greater than - 4
                ( i. e ) Z = { . . . . . . - 4,  -3,  -2,  -1,  0 , 1, 2, 3, 4 . . . . . . .}
                The integers greater than -4 are , -3, -2, -1, 0, 1, 2, 3, 4, . . . . . . . .  .

                Therefore, the solution set is { -3, -2, -1, 0, 1, 2, 3, 4, . . . . . . . .  . }
 

Question 4: Write the solution set of  -2 < x  $\le$ 10 and x $\epsilon$ W
Solution:
 
We have -2 < x  $\le$ 10
The domain of x is set of whole numbers.
                ( i. e ) W = { 0, 1, 2, 3, 4, . . . . .  . . . .}
The whole numbers greater than -2 and less than or equal to 10 are,  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 since the whole numbers cannot be negative ( whole numbers are non-negative).

The solution set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
 

The empty solution set is the one which does not contain any solution.

Solved Examples

Question 1: x <  0 , and x $\epsilon$ N
Solution:
 
The domain of the solution set is Natural numbers and there is no natural number less than 0.
Therefore, our solution set is x = {  }

 

Question 2: - 4 < x < -1 and x $\epsilon$ W
Solution:
 
The domain of the solution set is Whole numbers.
Since all the whole numbers are non-negative, there are no whole numbers between -4 and -1
Therefore, the solution set is empty  = { }

 

While graphing solution set we follow the following steps:
Step 1: Solve the equation or inequation for the unknown variable.
Step 2: Draw a number line and mark the negative positive side of it with 0 in the middle.
Step 3: In case of an equation, the solution set will be the integer points on the number line if the domain consists of natural numbers or whole numbers or integers. In the case of rational or irrational numbers we locate the points approximately between the possible integers.
Step 4: In the case of inequality, identify the region which belongs to the solution and mark the solution set on the number line.

Solved Examples

Question 1: Solve x2 - 11 x + 28 = 0 and graph the solution set on the number line.
Solution:
 
We have x2 - 11 x + 28 = 0
Factorising by splitting the middle term, we get,
                       x2 - 4 x - 7 x - 28 = 0
       =>       x ( x - 4 ) - 7 ( x - 4 ) = 0
       =>               ( x- 4 ) ( x - 7 ) = 0
       =>                             x - 4 = 0 ( or ) x - 7 = 0
       =>                                  x = 4 , x = 7
               The solution set is x = { 4, 7 }
The following Graph shows the solution on the one-dimensional and two-dimensional planes.

Graphing Solution Set
 

Question 2: Solve: 2x - 3 $\ge$ 5 , x $\epsilon$ R. Show the solution set on the number line
Solution:
 
We have 2 x - 3 $\ge$ 5
=>      2 x $\ge$ 5 + 3
=>     2 x $\ge$ 8

 =>      x $\ge$ $\frac{8}{2}$ = 4

Therefore, Solution Set = { x : x $\ge$ 4, and x $\epsilon$ R }
How to Graph Solution Set
 


Solved Examples

Question 1: Find the solution set: $\frac{2x-3}{3x+2}$ = $\frac{-2}{3}$
Solution:
 
We have $\frac{2x-3}{3x+2}$ = $\frac{-2}{3}$

Cross Multiplying, we get,
                3 ( 2 x - 3 ) = -2 ( 3 x + 2 )
      =>            6 x - 9  = - 6 x - 4
      =>        6 x + 6 x  = - 4 + 9
      =>               12 x = 5
      =>                   x = $\frac{5}{12}$
 
    The solution set is x = { $\frac{5}{12}$ }
 

Question 2: Find the solution set of 3 x2 + 10 x + 3 = 0
Solution:
 
We have 3 x2 + 10 x + 3 = 0
Factorising by splitting the middle term, we get
                         3 x2 + 9 x + x + 3 = 0
         =>   3 x ( x + 3 ) + 1 ( x + 3 ) = 0
         =>           ( x + 3 ) ( 3 x + 1 ) = 0
         =>                              x + 3 = 0 ( or ) 3 x + 1 = 0

         =>                                   x = - 3 , x = - $\frac{1}{3}$

     The solution set is { - $\frac{1}{3}$, - 3  }
 

Question 3: Find the solution set of the equality 4 x - 5 < 3 x + 10 , x $\epsilon$ N
Solution:
 
We have 4 x - 5 < 3 x + 10
           =>           4 x - 3 x < 10 + 5
           =>                     x < 15
The solution set is x = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 }