Solve the second for b = (102-a) and substitute in the first:









Since the last term is divisible by 100, let's take a chance and
see if 100 is a root with synthetic division:

100 | 1 -204  10655 -25500
    |    100 -10400  25500
      1 -104    255      0

Sure enough, 100 is a root! So the cubic factors as


a-100=0;  a2-104a+255=0
    a=100; 
  b=102-a;
b=102-100=2

So there is one solution: a=100, b=2. (a,b) = (100,2)

Checking:



That checks.  But that cubic equation has too more solutions.

You know how to solve a quadratic by the quadratic formula,

So solve this quadratic equation by the quadratic formula as
it doesn't factor:



and you'll get two more solutions for " a ",

Then substitute them in b = 102-a and you'll have two more 
solutions (a,b).

Edwin