Answer: The required solution of the given system is
(x, y) = (3, 1) and (4, 0).
Step-by-step explanation: We are given to solve the following system of equations :

From equation (i), we have

Substituting the value of y from equation (iii) in equation (ii), we get

When, x = 3, then from (iii), we get

And, when x = 4, then from (iii), we get

Thus, the required solution of the given system is
(x, y) = (3, 1) and (4, 0).