Math Examples of Relations and Functions

For finding the coordinate plane   for the given equation y = 3x +5 Where value of x =4 we need to follow the steps given below:
Step 1: Given equation is
y =3x +5,
As we know, value of x so we put value of x in the equation and find value of y
y = 3x 4 +5 = 17 (here we have given x =4)
y =17 After solving we get value of y
Step 2: Now, find the Coordinate Plane for given equation:
By above step we get x =4 and y = 17,
It means coordinate plane is (x, y) = (4, 17).

For finding the coordinate plane we need to follow the steps given below:
Step 1: Given equation is,
y = 3x+6 , As we know , we put value of y in the equation and find the value of x.
Step 2: We will find the Ordered Pair of given equation.
Now we substitute the value of x =-1,-2,-3, 0, 1, 2, 3 in given equation.
Here we check all x value for given equation,
Here, we put x =-1 in given equation
y= (3)(-1) +6 =  3
Now we get order pair(x, y)= (-1,3)
Step 3: Now we substitute x= -2 in given equation.
For x =-2
y = (3)(-2)+6 = 0
In this way we get order pair(x1,y1)= (-2 ,0)
Step 4: We substitute x= -3 in given equation.
For x =-3
y = 3(-3) + 6 = -3
Now we get order pair(x2,y2)= (-3,-3)
Step 5:  We substitute x= 0 in given equation.
For x =0
y = 3(0) +6 = 6
Now we get order pair(x3, y3)= (0, 6)
Step 6: We substitute x= 1 in given equation.
For x =1
y = 3(1) +6 = 9
Now we get order pair(x4,y4)= (1,9)
Step 7: We substitute x= 2 in given equation.
Like above we find value of y at x =2
y = 3(2) +6 = 12
Now we get order pair(x5, y5)= (2, 12)
 
Step 8: We substitute x= 3 in given equation.
Here we find value of y , at x =3
y = 3(3) +6 = 15
Now we get order pair(x5, y5)= (3, 15)
Step 9: At last we write all ordered pairs which are Coordinate Plane for given equation.
Coordinate planes are (-1,3) (-2,0)(-3,-3), (0,6), (1,9), (2,12) and (3,15) .

For finding the coordinate plane we need to follow the steps given below:
Step 1: Given equation is:
y = 4x+1,
As we know, we put value of y in the equation and find the value of x
Step 2: Now we will find the Ordered Pair of given equation.
Now we substitute the value of x =-1,-2,0, 1,2  in given equation .
Here we check all x value for given equation
So we find the value y at x =-1
y= (4) (-1) +1 =  -3
Now we get order pair(x, y)= (-1,-3)
Step 3: Then we substitute x= -2 in given equation.
For x =-2
y = (4)-2 +1 = -7
Now we get order pair(x1,y1)= (-2 , -7)
Step 4: Then we substitute x= 0 in given equation.
For x =0
y = 4(0) +1 = 1
Now we get order pair(x2,y2)= (0,1)
Step 5: Then we substitute x= 1 in given equation.
For x =1
y = 4(1) +1 = 5
Now we get order pair(x3,y3)= (1,5)
Step 6: Then we substitute x= 2 in given equation.
For x =2
y = 4(2) +1 = 9
Now we get order pair(x4,y4)= (2,9)
Step 5: At last we write all ordered pairs which are Coordinate Plane for given equation.
Coordinate planes are (-1,-3), (-2,-7), (0,1), (1,5) and (2,9) .

For finding the coordinate plane we need to follow the steps given below:
Step 1: Given equation is
Y = x +1 , we know , we put value of y in the equation and find the value of x
Step 2: We will  find the Ordered Pair of given equation.
Now we substitute the value of x =0 , 1, 2  in given equation .
So we find the value y at x =0
y= 0 +1 = 1
Now we get order pair(x,y)= (0,1)
Step 3: Now we substitute x= 1 in given equation:
For x =1,
y = 1 +1 = 2,
Now we get order pair(x1,y1)= (1,2)
Step 4: Then we substitute x= 2 in given equation:
For x = 2,
y = 2 + 1 = 3,
Now we get order pair(x3, y3)= (2, 3)
Step 5: In this step we write all ordered pairs which are Coordinate Plane for given equation
Coordinate planes are (0,1), (1,2)and (2,3) .

For finding the coordinate plane we need to follow the steps given below:
Step 1: Given equation is
X = 3y +4  As we know , we put value of x in the equation and find the value of y .
Step 2: In this step, we have to find the Ordered Pair of given equation.
Now we substitute the value of  y =1,2  in given equation .
Here we check all y value for given equation
So we find the value x at y =1
X = 3x1 +4 = 7
Now we get order pair (7,1)
 
Step 3: In this step, we substitute y= 2 in given equation.
Now we again find the value of x at y =2
X = 3x2 +4 = 10
Now we get order pair (10, 2)
Step 4: Now we write all ordered pairs which are Coordinate Plane for given
equation.
Coordinate planes (x1,y1)and(x2,y2) are (7,1) and (10,2) .

Given linear equation is: 6.1u – 4v = -3.9
This linear equation can be written as:
⇒6.1u – 4v = -3.9
⇒ 4v = 6.1u + 3.9,
If we further solve this linear equation we find the value of v.
⇒ 4v = 6.1u + 3.9
⇒ v = 6.1u + 3.9
               4
After solving the value of ‘v’ coordinate, now we put different value of ‘u’ so that we can find the other coordinates of ‘v’ for plotting the graph.
If we put the value of ‘u’ is -2 then we get:
⇒ v = 6.1u + 3.9
               4
On putting value of ‘u’ is -2 we get:
⇒ v = 6.1(-2) + 3.9
               4
⇒ v = -12.2 + 3.9
               4
⇒ v = -16.1
             4
If we divide -16.1 by 4 we get:
⇒ v = -4.02.
So when we put value of ‘u’ is -2 then we get value of ‘v’ is -4.02.
If we put the value of ‘u’ is -4 then we get:
⇒ v = 6.1u + 3.9
               4
On putting value of ‘u’ is -4we get:
⇒ v = 6.1(-4) + 3.9
               4
⇒ v = -24.4 + 3.9
              4
⇒ v = -28.3
             4
If we divide -28.3 by 4 we get:
⇒ v = -7.07.
So when we put value of ‘u’ is -4 then we get value of ‘v’ is -7.07.
If we put the value of ‘u’ is 2 then we get:
⇒ v = 6.1u + 3.9
               4
On putting value of ‘u’ is 2 we get:
⇒ v = 6.1(2) + 3.9;
               4
⇒ v = 12.2 + 3.9
              4
⇒ v = 16.1
           4
If we divide 16.1 by 4 we get:
⇒ v = 4.02,
So when we put value of ‘u’ is 2 then we get value of ‘v’ is 4.02.
If we put the value of ‘u’ is 6 then we get:
⇒ v = 6.1u + 3.9
               4
On putting value of ‘u’ is 6 we get:
⇒ v = 6.1(6) + 3.9
               4
⇒ v = 36.6 + 3.9
              4
⇒ v = 40.5
           4
If we divide 40.5 by 4 we get:
⇒ v = 10.1.
So when we put value of ‘u’ is 6 then we get value of ‘v’ is 10.1.
So the coordinates of linear equation is: (-2, -4.02), (-4, -7.07), (2, 4.02), and (6, 10.1).
We see how to plot a graph with help of these coordinate. 
Step1: We take two axes, one axis is in vertical direction and other axis is in horizontal direction.
Step2: We Set all coordinates of ‘u’ on horizontal axis.
Step3: Then set the coordinates of ‘v’ on vertical axis.
Step4: Then mark the coordinates in horizontal and vertical direction.
Step5: Suppose we mark the number -2 in horizontal and -4.02 in vertical, and join both the points and we apply same procedure for all remaining coordinates.
Step6: At last we obtained linear equation graph.

Given the inequality equality is:
⇒ 5u + 6v = 13
When we solve the inequality equation then first we change this inequality equation by equality so that we can find out the coordinates for find the graph.
On changing this equation by equality then the equation becomes:
So we can write the equation as:
⇒ 5u + 6v = 13
⇒ 5u + 6v = 13
Now we find the value of one coordinate, so here in the given equation we can find the coordinates of ‘v’;
⇒ 5u + 6v = 13,
⇒ 6v = -5u +13,
On further solving we get the value of v is:
⇒ v = -5u + 13
               6
For further solving we put the different values of “u” so that we can find the “v” coordinates for plotting the graph.
If we put the value of u is -6 then we get:
Put u = -6;
⇒ v = -5u + 13
               6
On putting the value of u is -6 then we get:
⇒ v = -5(-6) + 13
               6
⇒ v = 30 +13
              6
⇒ v = 43
           6
⇒ v = 7.16
So when we put value of “u” is -6 then we get value of v is 7.16.
If we put the value of u is -2 then we get:
Put u = -2;
⇒ v = -5u + 13
               6
On putting the value of u is -2 then we get:
⇒ v = -5(-2) + 13
               6
⇒ v = 10 +13
              6
⇒ v = 23
          6
⇒ v = 3.83;
So when we put value of “u” is -2 then we get value of v is 3.83.
If we put the value of u is 5 then we get:
Put u = 5;
⇒ v = -5u + 13
               6
On putting the value of u is 5 then we get:
⇒ v = -5(5) + 13
               6
⇒ v = -25 +13
              6
⇒ v = -12
            6
⇒ v = -2
So when we put value of “u” is 5 then we get value of v is -2.
If we put the value of u is 9 then we get:
Put u = 9;
⇒ v = -5u + 13
               6
On putting the value of u is 9 then we get:
⇒ v = -5(9) + 13
               6
⇒ v = -45 +13
              6
⇒ v = -32
            6
⇒ v = -5.33
So when we put value of “u” is 9 then we get value of v is -5.33.
So coordinates are: (-6, 7.16), (-2, 3.83), (5, -2) and (9, -5.33).
With the help of above coordinates we see how to plot the graph. For plotting the graph we have to follow some steps which are given below:
Step1: We take two axes, one axis is in vertical direction and other axis is in horizontal direction.
Step2: We Set all coordinates of ‘u’ on horizontal axis.
Step3: Then set all coordinates of ‘v’ on vertical axis.
Step4: Then mark coordinates in horizontal and vertical direction.
Step5: Suppose we mark the number -6 in horizontal and 7.16 in vertical, and join both the points and we apply same procedure for all remaining coordinates.
Step6: At last we obtained linear equation graph.

Given linear equation is: 5u – 2v = -3;
This linear equation can be written as:
⇒5u – 2v = -3;
⇒ 2v = 5u + 3;
If we further solve this linear equation we find the value of 'v'.
⇒ 2v = 5u + 3;
⇒ v = 5u + 3;
              2
After solving the value of ‘v’ coordinate, now we put different value of ‘u’ so that we can find other coordinates of ‘v’ for plotting the graph.
If we put the value of ‘u’ is -3 then we get:
⇒ v = 5u + 3;
               2
On putting value of ‘u’ is -3 we get:
⇒ v = 5(-3) + 3;
               2
⇒ v = -15 + 3,
              2
⇒ v = -12,
           2
If we divide -12 by 2 we get:
⇒ v = -6;
So when we put value of ‘u’ is -3 then we get value of ‘v’ is -6.
If we put the value of ‘u’ is -1 then we get:
⇒ v = 5(u) + 3;
              2
On putting value of ‘u’ is -1we get:
⇒ v = 5(-1) + 3;,
               2
⇒ v = -5 + 3,
             2
⇒ v = -2,
           2
If we divide 2 by 2 we get:
⇒ v = -1;
So when we put value of ‘u’ is -1 then we get value of ‘v’ is -1.
If we put the value of ‘u’ is 5 then we get:
⇒ v = 5u + 3;
              2
On putting value of ‘u’ is 5 we get:
⇒ v = 5(5) + 3;,
               2
⇒ v = 25 + 3,
              2
⇒ v = 28,
           2
If we divide 28 by 2 we get:
⇒ v = 14;
So when we put value of ‘u’ as 5 then we get value of ‘v’ as 14.
If we put the value of ‘u’ as 7 then we get:
⇒ v = 5u + 3;
               2
On putting value of ‘u’ as 7 we get:
⇒ v = 5(7) + 3;
               2
⇒ v = 35 + 3,
              2
⇒ v = 38,
           2
If we divide 38 by 2 we get:
⇒ v = 16;
 
So when we put value of ‘u’ as 7 then we get value of ‘v’ as 16.
So the coordinates of linear equation is: (-3, -6), (-1, -1), (5, 14), and (7, 16).
With the help of above coordinates we see how to plot the graph. In order to plot a graph we follow some steps which are given below:
Step1: Firstly take two axes, one axis is in vertical direction and other axis is in horizontal direction.
Step2: We Set all coordinates of ‘u’ on the horizontal axis.
Step3: Then set coordinates of ‘v’ on the vertical axis.
Step4: Then mark the coordinates in horizontal and vertical direction.
Step5: Suppose we mark the number -3 in horizontal and -6 in vertical, and join both the points and we apply same procedure for all coordinates.
Step6: At last we obtained linear equation graph.

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6),
(5, 4), (5, 6)}

R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {5, 6, 7, 8, 9, 10}.

(i) Given f(x) = 28 = x² + 3
        x² = 28 – 3 = 25
x =   = 5
(ii)  To find f^-1 (39) and f^-1 (2)
f^-1 (39) = x  f (x) = 39 = x² + 3
x² = 39 – 3 = 36
x =  = 6
f^-1 (2) = x  f (x) = 2 = x² + 3
x² = 2 – 3 = -1
x = . 

x + 3 = 6 and 2x + y = 5
Therefore x = 3 and 6 + y = 5
Therefore y = -1.

Given the relation R on R defined as R = (p, q): p ≤ q,
Here we have R = (p, q): p ≤ q, where p, q ∊
First we check reflexive condition:
Reflexivity: - For any a ∊ R, we have
               P ≤ P
     ⇒ (P, P) ∊ R for all P ∊ R.
     ⇒ So, ‘R’ is reflexive.
 Symmetry: We find that (2, 3) ∊ R but (3, 2) ∉ R. so R is not symmetric.
 Transitivity:  let (P, q) ∊ R and (q, r) ∊ R. then,
         ⇒      P ≤ q and q ≤ r
         ⇒      p ≤ r
         ⇒    (p, r) ∊ R, so ‘R’ is transitive. 

Given that  (1) R₁ =(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),
                                  (2) R₂ = (1, 2), (2, 1), (1, 3), (3, 1),
                                  (3) R₃ = (1, 2), (2, 3), (3, 1).
If we check the condition for reflexive i.e. a relation ‘R’ on Set ‘A’ is said to be reflexive if every element of ‘A’ is related to itself.
(1) Reflexive: - (1, 1), (2, 2), (3, 3) ∊ R₁. So we can say that relation ‘R₁’ is reflexive.
Symmetric: - Here we observe that (1, 2) ∊ R₁ but (2, 1) ∉ R₁. So relation ‘R₁’ is not symmetric on relation ‘A’.
Transitive: - Here we observe that (2, 3) ∊ R₁ and (3, 1) ∊ R₁ but (2, 1) ∉ R₁. So relation R₁ is not transitive on relation ‘A’.

(2) Reflexive: - (1, 1), (2, 2), (3, 3) are not present in R₂. So we can say that relation ‘R₁’ is not reflexive.
Symmetric: - Here we observe that the order pair obtained by interchanging the components of ordered pairs in R₂ is also in R₂. So relation R₁ is symmetric on relation on ‘A’.
Transitive: - Here we observe that (1, 2) ∊ R₂ and (2, 1) ∊ R₂ but (1, 1) ∉ R₂. So relation R₂  is not transitive on relation A.
 
(3) Reflexive :-( 1, 1), (2, 2), (3, 3) are not present in R₃. So we can say that relation R₃ is not reflexive.
 Symmetric: - We observe that the (2, 3) ∊ R₃ but (3, 1) ∉ R₃. So relation R₃ is symmetric on relation on ‘A’.
Transitive: -We find that (1, 2) ∊ R₃ and (2, 3) ∊ R₃ but (1, 3) ∉ R₃. So relation R₃ is not transitive on relation A.

The relation ‘R’ on Set A = 2, 3, 4 where,
                   R = (2, 2), (3, 3), (4, 4), (2, 3), (3, 4)
 Since 2, 3, 4 ∊ A and (2, 2), (3, 3), (4, 4) ∊ R i.e. for each a ∊ A, (a, a) ∊ R. So relation ‘R’ is reflexive.
 And we find that (2, 3) ∊ R but (3, 2) ∉ R.
 So relation ‘R’ is not symmetric.
 Also, (2, 3) ∊ R and (3, 4) ∊ R but (2, 4) ∉ R. So ‘R’ is not transitive.