Question 1111308
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<pre>
Let x be the larger number, y be the lesser number.


Then

x - y = 6,                (1)
xy = x^2 + y^2 - 223.     (2)


Square equation (1) (both sides).  Keep equation (2) as is:

x^2 - 2xy + y^2 =  36,    (3)
x^2 - xy  + y^2 = 223     (4)    (<<<---=== it is transformed eq(2) )

----------------------------------Subtract eq(3) from eq(4). You will get

      xy        = 187.


Now you have system of two equations

x - y =   6,
xy    = 187.


It is reduced to the quadratic equation

x*(x-6) = 187

x^2 -6x - 187 = 0

{{{x[1,2]}}} = {{{(6 +- sqrt(36 + 4*187))/2}}} = {{{(6 +- 28)/2}}},

{{{x[1]}}} = {{{(6+28)/2}}} = 17,  {{{y[1]}}} = 11.

{{{x[2]}}} = {{{(6-28)/2}}} = -11,  {{{y[2]}}} = -17.


<U>Answer</U>.  There are TWO solutions:   a) (x,y) = (17,11);  b) (x,y) = (-11,-17).      

         Since the problem asks about positive numbers, only first pair satisfies this requirement.
</pre>

Solved.


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