Question 831255
I'm changing y(first) to {{{y}}} and y(second) to {{{z}}} just for clarity.
1.{{{y=2x/(x+3)}}}
2.{{{z=3/(x+5)}}}
3.{{{y+z=1}}}
From eq. 3, substitute using eq. 1 and eq. 2,
{{{2x/(x+3)+3/(x+5)=1}}}
{{{2x(x+5)+3(x+3)=(x+3)(x+5)}}}
{{{2x^2+10x+3x+9=x^2+8x+15}}}
{{{2x^2+13x+9=x^2+8x+15}}}
{{{x^2+5x-6=0}}}
{{{(x+6)(x-1)=0}}}
To possible solutions:
{{{x+6=0}}}
{{{x=-6}}}
Check the original equation:
{{{y=2(-6)/(-6+3)=-12/-3=4}}}
{{{z=3/(-6+5)=3/(-1)=-3}}}
{{{y+z=1}}}
True
.
.
.
{{{x-1=0}}}
{{{x=1}}}
Check the original equation:
{{{y=2(1)/(1+3)=2/4=1/2}}}
{{{z=3/(1+5)=3/6=1/2}}}
{{{y+z=1}}}
True
(x,y,z)=(-6,4,-3)
(x,y,z)=(1,1/2,1/2)