How to solve the system of equations: $x+ \sqrt{y} =6$, $2 \sqrt{x} +y=10 $?

Question

how to solve system of equations of the following?

$$x+ \sqrt{y} =6\\2 \sqrt{x} +y=10 $$

Here is what I tried:

Answer 1

$$\sqrt{y}=6-x$$ $$y=(6-x)^2$$

Also:

$$2\sqrt{x}=10-y$$ $$4x=(10-y)^2$$

Put these together:

$$4x=(10-(6-x)^2)^2$$

$$4x=(10-(36-12x+x^2))^2$$

$$4x=(-x^2+12x-26)^2$$

$$4x=x^4-24x^3+196x^2-624x+676$$

$$0=x^4-24x^3+192x^2-624x+676$$

This does not factorize nicely so use technology to get approximations for your solutions:

$$x\approx2.3824,3.4981,8.0775,10,042$$

Double check to see if any lead to square roots of negative numbers.

The only valid solution is $x\approx3.4981$ and $y\approx6.2593$

Answer 2

$$ \begin{cases} \text{x}+\sqrt{\text{y}}=6\\ 2\sqrt{\text{x}}+\text{y}=10 \end{cases}\space\space\space\Longleftrightarrow\space\space\space \begin{cases} \text{x}+\sqrt{\text{y}}=6\\ \text{y}=10-2\sqrt{\text{x}} \end{cases}\space\space\space\Longleftrightarrow\space\space\space \begin{cases} \text{x}+\sqrt{10-2\sqrt{\text{x}}}=6\\ \text{y}=10-2\sqrt{\text{x}} \end{cases} $$

So, we get:

$$\text{x}+\sqrt{10-2\sqrt{\text{x}}}=6\space\Longleftrightarrow\space \text{x}\approx3.49813250456484$$

Answer 3

Squaring the second equation gives $x=(y^2 - 20y + 100)/4$. Substituting this into the first equation and squaring gives $$ y^4 - 40y^3 + 552y^2 - 3056y + 5776=0. $$ This has $4$ real solutions (use Sturm); approximately $y=6.25934096471, y=4.31582661552, y=16.3378253655, y=13.0870070542$. Now we have to see which solutions are valid. It turns out, that it is $y=6.25934096471$, with $x=3.49813250456$.

Answer 4

From the first equation we have
$$\sqrt{y} = 6-x$$
And so
$$y=(x-6)^2$$
Plugging this into the second equation:
$$2\sqrt{x}=10-(x-6)^2$$
$$4x=(10-x^2+12x-36)^2$$
$$x^4-24x^3+196x^2-628x+676=0$$
With numerical methods or the quartic equation you find approximately $x=2.3824,3.4981,8.0775,10.042$.
When plugging back into the system the only non extraneous solution is the 3.4981, so the solution is approximately $x=3.4981,y=6.2595$.

Answer 5

The trick here is to organize the argument in a way that minimizes the amount of messy arithmetic and numerical examination of cases. Here's one approach.

Assuming we are looking for real solutions, we need $x,y\ge0$, so we can write $x=u^2$ and $y=v^2$ with $u,v\ge0$. The equations become

$$\begin{align} u^2+v&=6\\ 2u+v^2&=10 \end{align}$$

which, by solving for $v=6-u^2$ and substituting into $2u+(6-u^2)^2=10$ gives

$$u^4-12u^2+2u+26=0$$

Letting $P(u)$ denote the quartic and evaluating $P(0)=26$, $P(2)=16-48+4+26=-2$, and $P(-2)=16-48-4+26=-6$, we see that the quartic has two negative roots, which are irrelevant for our purposes, and two positive roots, one of which is less than $2$ and the other greater than $2$.

The root that is less than $2$ definitely gives a solution to the original equation, because if $u\lt2$ then $v=6-u^2\gt6-4\ge0$. The other positive root does not, because

$$P(\sqrt6)=36-72+2\sqrt6+26=2\sqrt6-10\lt0$$

implies the other root is greater than $\sqrt6$, which would make $v=6-u^2$ negative.

In sum, the only real solution to the original equation is $(x,y)=(u^2,10-2u)$, where $u$ is the unique positive root of $u^4-12u^2+2u+26$ that is less than $2$. Numerically one gets $u\approx1.87033$, leading to $(x,y)\approx(3.49813,6.25934)$. (The quartic doesn't factor. You can, of course, find an exact expression for its roots in terms of radicals, but trust me, you don't want to.)