Systems of 2 equations in 2 variables: $\sqrt{x}+y=10$ and $x+\sqrt y=5$ I would like to solve the following system:
$$ \sqrt{x}+y=10 \qquad (1)$$
$$ x+\sqrt{y}=5\qquad   (2) $$
I know that this system does not any answer if $ x,y \in \mathbb Z $
Because
by (2) 
$\qquad \sqrt{y}=5-x \qquad \Rightarrow\qquad5-x\ge0\qquad \Rightarrow\qquad x\le5$
and by (1)
$\qquad x\ge0$
then
$\qquad 0 \le x\le 5 $
if $X=0\qquad \Rightarrow\qquad y=10,by (1)\qquad x+\sqrt{y}\neq5$
if $X=1\qquad \Rightarrow\qquad y=9,by (1)\qquad x+\sqrt{y}\neq5$
and Similarly, we continue to $X=5$
my question:
Does this system has any answers if $x,y \in \mathbb R ?$ 
 A: Writing $\sqrt{x}=10-y$, $\sqrt{y}=5-x$ we obtain by squaring
$$
x= y^2 - 20y + 100, \quad y - x^2 + 10x - 25 =0.
$$
Substituting the first equation into the second one gives
$$
 - y^4 + 40y^3 - 590y^2 + 3801y - 9025=0,
$$
which has all roots real. One of these roots leads to a real solution of the original equations, i.e., 
$$
(x,y)=(2.07429337331, 8.5597592655).
$$
A: This system of equations isn't linear because you have square root of a variables. Substituting and simplifyind give us $4$-th degree equation which have only one real root. Approximate value is $(x,y)\approx(2.07429,8.55976)$. Here is the graph of this equations:


A: From (1) : $x = (10-y)^2$.
From (2): $ y = (5-x)^2$. Thus,
$$x = \left(10-(5-x)^2\right)^2,$$ which is a 4th-degree polynomial. Solve it and test the solutions in your system.
A: Your equations are not linear because they have a square root. If you put $x = X^2$ and $y = Y^2$, you will transform your set of equations to
\begin{eqnarray*}
X + Y^2 &=& 10 \\
X^2 + Y &=& 5
\end{eqnarray*}
The first of these equations gives $X = 10 - Y^2 \Rightarrow X^2 = 100 - 20Y^2 + Y^4$. On substituting it in the second one, you will get $Y^4 - 20Y^2 + Y + 95 = 0$. This equation can be easily solved numerically, say on http://www.wolframalpha.com.
A: $\sqrt x + y = 10$ is part of a parabola symmetric about $y = 10$ with vertex at $(0, 10)$ and $x$ intercept at $(100, 0)$
$x + \sqrt y = 5$ is part a parabola symmetric about $x=5$ axis with vertex $(5, 0)$ and $y_$intercept at $(0,25).$  they intersect at four points in the first quadrant, but the solution is the point in the rectangle $[0,5] \times [0,10]$
