Question 258746
<pre><font size = 4 color = "indigo"><b>
{{{system(x+y=-10,
xy=-20)}}}

Solve the first for y

{{{x+y=-10}}}
{{{y=-10-x}}}

Substitute {{{(-10-x)}}} for {{{y}}} in

{{{xy=-20)}}}
{{{x(-10-x)=-20)}}}
{{{-10x-x^2=-20}}}
{{{-x^2-10x+20=0}}}

Multiply through by {{{-1}}}

{{{x^2+10x+20=0}}}

That doesn't factor so we use the quadratic formula:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-(10) +- sqrt( (10)^2-4*(1)*(-20) ))/(2*(1)) }}}

{{{x = (-10 +- sqrt( 100+80) )/2 }}}

{{{x = (-10 +- sqrt( 180) )/2 }}} 

{{{x = (-10 +- sqrt(36*5) )/2 }}}

{{{x = (-10 +- 6*sqrt(5) )/2 }}}

Make two fractions

{{{x = -10/2 +- 6*sqrt(5) /2 }}}

{{{x = -5 +- 3*sqrt(5)}}}

Two solutions for x

{{{x = -5 + 3*sqrt(5)}}} and {{{x = -5 - 3*sqrt(5)}}}

Substituting the first in

{{{y=-10-x}}}

{{{y=-10-(-5 + 3*sqrt(5))}}}

{{{y=-10+5 - 3*sqrt(5)}}}

{{{y=-5 - 3*sqrt(5)}}}

so one solkution is

{{{x = -5 + 3*sqrt(5)}}}, {{{y = -5 - 3*sqrt(5)}}}

----------------------------

Substituting the second in

{{{y=-10-x}}}

{{{y=-10-(-5 - 3*sqrt(5))}}}

{{{y=-10+5 + 3*sqrt(5)}}}

{{{y=-5 + 3*sqrt(5)}}}

so the other solution is

{{{x = -5 - 3*sqrt(5)}}}, {{{y = -5 + 3*sqrt(5)}}}

But there are really only two numbers, it's just a 
matter of which you call x and which you call y,
or which you call "the first numner" and
which you call "the second number".

Edwin</pre>