For $\alpha \in N,$ consider a relation R on N given by $R=\left\{\left(x,y\right):3x+\alpha y is a multiple of 7\right\}.$

The relation R is an equivalence relation if and only if :

$$\begin{gathered} For R to be reflexive \Rightarrow x R x \\ \Rightarrow 3x+\alpha x=7x\Rightarrow \left(3+\alpha \right)x=7K \\ \Rightarrow 3+\alpha =7 \lambda \Rightarrow \alpha =7\lambda -3=7N+4, K, \lambda , N\in 1 \\ \therefore when \alpha divided by 7, remainder is 4. \\ R to be symmetric xRy\Rightarrow yRx \\ 3x+\alpha y=7N_1, 3y+\alpha x=7N_2 \\ \Rightarrow \left(3+\alpha \right)\left(x+y\right)=7\left(N_1+N_2\right)=7N_3 \\ Which holds when 3+\alpha is multiple of 7 \\ \therefore \alpha =7N+4 (as did earlier) \\ R to be transitive \\ xRy \& yRz\Rightarrow xRz \\ 3x+\alpha z=7N_1 \& 3y+\alpha z=7N_2 and \\ 3x+\alpha z=7N_3 \\ \therefore 3x+7N_2-3y=7 N_3 \\ \therefore 7N_1-\alpha y+7N_2-3y=7 N_3 \\ \therefore 7 \left(N_1+N_2\right)-\left(3+\alpha \right)y=7 N_3 \\ \therefore \left(3+\alpha \right) y=7N \\ Which is true again when 3+\alpha divisible by 7, i.e. \\ when \alpha divided by 7, remainder is 4. \end{gathered}$$