Two years ago,a father was five times the age of his son.Two years later his age will be eight more than three times the age of his son.Find the present age of father and son

Step 1: identify the two variables 

             Father = x

              Son = y

Step 2 : formulate the equations

Two years ago,a father was five times the age of his son

 

                         x - 2 = 5(y - 2)

                         x - 2 = 5y - 10

                         x - 5y = -10 + 2

Thus, Equation 1 is          

                         x - 5y = - 8

Two years later his age will be eight more than three times the age of his son 

                        x + 2 = 3(y + 2) + 8

                        x + 2 = 3y + 6 + 8

                        x + 2 = 3y + 14

                        x - 3y = 14 - 2 

Thus, Equation 2 is

                       x - 3y = 12

Step 3 : Eleminate one variable

Subtracting both Equations

                       x - 5y = - 8 - (x - 3y = 12)

                       x - x -5y + 3y = - 8 - 12

                            -2y = -20

                               y = (-20) / (-2)

                               y = 10

Step 4: Substitute the value of y in Equation 1

                       x - 5y = - 8

                       x - 5(10) = - 8

                       x - 50 = - 8

                       x = - 8 + 50 

                       x = 42

Therefore father is 42 years old now and son is 10 years old