system of equations $\sqrt{x}+y = 11$ and $x+\sqrt{y} = 7$.

Question

If $x,y\in \mathbb{R}$ and $\sqrt{x}+y = 11\;$ and $x+\sqrt{y} = 7$. Then $(x,y) = $

$\underline{\bf{My\;\; Try::}}$ Let $x=a^2$ and $y=b^2$, Then equation is $a+b^2 = 11$ and $a^2+b = 7$.

$(a+b)+(a+b)^2-2ab = 18$ and Now Let $a+b=S$ and $ab=P$, we get $S+S^2-P=18$

Now I did not understand how can I solved it.

Help required

Thanks

Answer 1

HINT:

Assuming $x,y$ are real we have $x=7-\sqrt y\le 7\implies a^2\le 7$

Putting $b=7-a^2,$

$$a+(7-a^2)^2=11\implies a^4-14a^2+a+38=0$$

Observe that $a=2$ is a solution

For the general case, we need to use the formula described here or here

Answer 2

You were on the right track, all you need to do is subtract rather than add. Doing that will give you $(a+b^2)-(a^2+b)=4$

Which can become

$(b^2-a^2)-(b-a)=4=(b-a)(b+a)-(b-a)=(b-a)(b+a-1)$

Since the only way to can obtain 4 by multiplying is either $2*2$ or $4*1$, either

$b-a=2$ and $b+a-1=2$

or

$b-a=1$ and $b+a-1=4$

if $b-a=b+a-1=2$, then $a=1/2$ and $b=5/2$, which means that $(x,y)=(1/4,25/4)$ and that does not satisfy that initial conditions

if $b-a=1$ and $b+a-1=4\implies b+a=5$

then one can easily determine that $b=3$ and $a=2$, which means that $(x,y)=(4,9)$, which satisfies the initial conditions.