Steps to solve this system of equations: $\sqrt{x}+y=7$, $\sqrt{y}+x=11$

I want to solve this system of equations, I have been out of Maths for a long!!

$$\sqrt{x}+y=7$$

$$\sqrt{y}+x=11$$

Just wondering easiest step to find values for $x$ and $y$ from the above equations?

Solution 1:

First rearrange the equations like this: $$y-4=3-\sqrt{x} \\x-9=2-\sqrt{y}$$ then multiply two sides of these equations to each other; $$(y-4)(x-9)=(3-\sqrt{x})(2-\sqrt{y})$$ Since $x,y\ge 0$ we can factor LHS : $$(\sqrt{y}-2)(\sqrt{y}+2)(\sqrt{x}-3)(\sqrt{x}+3)=(3-\sqrt{x})(2-\sqrt{y})$$ Implies that$$ (\sqrt{y}-2)(\sqrt{x}-3)((\sqrt{y}+2)(\sqrt{x}+3)-1)=0 $$ at least one of the factors must be zero.

1. $$(\sqrt{y}-2)=0 \Rightarrow y=4$$

by substitution $y=4$ in the first eq. we have $x=9$.

and by checking in the second eq. implies that $$x=9\qquad and \qquad y=4$$ is a solution.

or

2. $$(\sqrt{x}-3)=0 \Rightarrow x=9$$

by substitution in the first eq. we have $y=4$ and by checking in the second eq. implies that this is not a new solution.

or

3. $$(\sqrt{y}+2)(\sqrt{x}+3)-1=0 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)=1$$ but it has no solution for $x$ and $y$ because $$ (\sqrt{y}+2)\ge 2 ; \quad (\sqrt{x}+3) \ge 3 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)\ge 6 $$ as a result $(\sqrt{y}+2)(\sqrt{x}+3)=1$ has no solution for $x$ and $y$.

Solution 2:

Note that : $x,y \geq 0$

$(7-y)^2=11-\sqrt y$

substitute $~\sqrt y = t~$ , so :

$(7-t^2)^2=11-t \Rightarrow t^4-14 \cdot t^2+49=11-t \Rightarrow t^4-14 \cdot t^2+t+38=0$

Note that possible nonnegative integer solutions of equation $~\sqrt x+y=7~$ are :

$(x,y) \in \{(1,6),(4,5),(9,4),(16,3),(25,2),(36,1),(49,0) \}$

Since pair $(x,y)$ has to be solution of equation $\sqrt y +x =11$ also we have that :

$(x,y)=(9,4) \Rightarrow \sqrt y=2 \Rightarrow t=2 ~$

Therefore polynomial $~t^4-14 \cdot t^2+t+38~$ has to be divisible by $(t-2)$

If you divide this polynomial by $(~t-2)~$ using Polynomial long division method you will get polynomial :

$t^3+2t^2-10t-19$

So you should solve equation $~t^3+2t^2-10t-19=0~$ to obtain other solutions .

This can be done using general formula of roots .