Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x + 4y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): 4x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ 4x + x = 10, which is obviously not true everytime.

Take x = 4.

4x + x = 10

⇒ 16 + 4 = 10

⇒ 20 = 10, which is not true.

That is 20 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 4x + y = 10

Now, replace x by y and y by x. We get

4y + x = 10, which may or may not be true.

Take x = 1 and y = 6

4x + y = 10

⇒ 4(1) + 6 = 10

⇒ 4 + 6 = 10

⇒ 10 = 10

4y + x = 10

⇒ 4(6) + 1 = 10

⇒ 24 + 1 = 10

⇒ 25 = 10, which is not true.

⇒ 4y + x ≠ 10

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, and then (y, x) ∉ R ∀ x, y ∈ N

⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

Then, (x, z) ∈ R

We have

4x + y = 10

⇒ y = 10 – 4x

Where x, y ∈ N

So, put x = 1

⇒ y = 10 – 4(1)

⇒ y = 10 – 4

⇒ y = 6

Put x = 2

⇒ y = 10 – 4(2)

⇒ y = 10 – 8

⇒ y = 2

We can’t take y >2, because if we put y = 3

⇒ y = 10 – 4(3)

⇒ y = 10 – 12

⇒ y = –2

But, y ≠ –2 as y ∈ N

So, only ordered pairs possible are

R = {(1, 6), (2, 2)}

This relation R can never be transitive.

Because if (a, b) ∈ R, then (b, c) ∉ R

⇒ R is not reflexive.

Hence, the relation is neither reflexive nor symmetric nor transitive.