Question 1209082
<pre>

{{{system(a + ab^2 = 250b,
a + b = 102)}}}

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

{{{a + a(102-a)^2 = 250(102-a)}}}

{{{a + a(10404-204a+a^2)=25500-250a}}}

{{{a + 10404a-204a^2+a^3=25500-250a}}}

{{{a^3-204a^2+10655a-25500=0}}}

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
    |<u>    100 -10400  25500</u>
      1 -104    255      0

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

{{{(a-100)(a^2-104a+255)=0}}}
a-100=0;  a<sup>2</sup>-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:
{{{system(a + ab^2 = 250b,
a + b = 102)}}}
{{{system(100 + (100)(2)^2 = 250(2),
100 + 2 = 102)}}}
{{{system(100 + 400 = 500,
102= 102)}}}
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:

{{{a^2-104a+255=0}}}

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</pre>