Question 859033
Use a substitution, {{{u=sqrt(x)}}}, {{{u^2=x}}}, {{{v=sqrt(y)}}},{{{v^2=y}}}
1.{{{u+v^2=7}}}
2.{{{v+u^2=11}}}
From eq. 1,
{{{u=7-u^2}}}
{{{u^2=49-14v^2+v^4}}}
Substitute into eq. 2,
{{{v+49-14v^2+v^4=11}}}
{{{v^4-14v^2+v+38=0}}}
{{{(v-2)(v^3+2v^2-10v-19)=0}}}
Unfortunately the cubic cannot be factored any further. 
You need to use numerical methods to find the roots. 
{{{v=-3.28}}}
{{{v=-1.85}}}
{{{v=3.13}}}
and from above 
{{{v=2}}}
We need to verify that each of these are actual solutions.
{{{sqrt(y)=2}}}
{{{y=4}}}
and
{{{sqrt(x)+4=7}}}
{{{sqrt(x)=3}}}
{{{x=9}}}
Verifying,
{{{x+sqrt(y)=11}}}
{{{9+sqrt(4)=11}}}
{{{9+2=11}}}
{{{11=11}}}
True, good solution is {{{x=9}}}, {{{y=4}}}
Next,
{{{v=3.13}}}
{{{sqrt(y)=3.13}}}
{{{y=9.797}}}
{{{sqrt(x)+9.797=7}}}
{{{sqrt(x)=-2.797}}}
Doesn't lead to a valid solution.
Same for the other two possible solutions.
Only one solution.