The solution above is incomplete for there are two solutions.

Let the smaller number be x and the larger number be x+k where k > 0



















Believe it or not, that factors!:



  <--[Remember that k > 0]


Using k = 6, substituting in




smaller = 7, larger = 7+6 = 13   <--one answer

Using 








      <--second answer

Edwin

First equation is

    x^2 + y^2 = 218.    (1)


Second equation is

    y*(x-y) = 42.       (2)


We do not see an algebra way to solve it; so, will assume the numbers integer
and will try to solve the problem using the properties of integer numbers.


Using elementary number theory arguments, from equation (1) it is easy to derive
that the numbers x and y should both be odd.


The number 42 has the following odd divisors 1, 3, 7, 21.

So, we write

    x^2 = 218 - y^2

and look, which of these values of y will give us the square of integer x

    218 - 1^2 = 217 is not a square of an integer

    218 - 3^2 = 218 - 9 = 209 is not a square of an integer;

    218 - 7^2 = 218 - 49 = 169 is the square of the integer number 13.

    218 - 21^2 is negative, so it is not a square, at all.


So, only y= 7 is the appropriate candidate, and with x= 13 it really gives a solution.

After seeing the solution by Edwin, I quickly got that there is 
a straightforward Algebra way to do the same without any tricks.


This way is to express  x =   from equation (2) and substitute it into equation (1)
written in the form

     = 218 - .


It will give a biquadratic equation with four roots for y, that will bring four routs for x.

Two of the above are incomplete because there are two solutions. 
Mine (the second one) is the only complete one.
Ikleyn mistook the problem as requiring that the solution be integers.
That was not necessary and caused the second solution to be lost.

The trick to making the algebra simpler was to use the smaller as x
and the larger as x+k where k > 0

Edwin