Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$

Question

Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$.

I tried to solve this question as follows:

$x-y=x^2+y^2+49+2xy-14x-14y$

$x+y=x^2+y^2+1-2xy-2x+2y$

Here I realized that I can't transform one equation into $x=...$ and then replace it into the other, as that would give a polynomial of too high a degree. Hence I attempted to combine them:

$-2y=48+4xy-12x-16y$

$48+4xy-12x-14y=0$

And I got stuck here.

Could you please explain to me how to solve the question, as I feel I'm very close?

Answer 1

Try $1≤a=x-y, 7≤b=x+y$

$$\begin{cases} a=(b-7)^2 \\ b=(a-1)^2 \end{cases} \implies a=((a-1)^2-7)^2$$

$$a^4 - 4 a^3 - 8 a^2 + 23 a + 36 = 0$$

$$(a - 4) (a^3 - 8 a - 9) = 0$$

$$ \implies a=4≥1$$

$$\implies b=(a-1)^2=9≥7.$$

Then you can find $x$ and $y.$

The last cubic equation doesn't have rational roots.

Answer 2

We see there are $(x+y)$ and $(x-y)$ in the equations and both are positive (since under square root). So we make the transformations $$u^2=x-y \quad , \quad v^2=x+y$$

The new equations are $$u=v^2-7 \quad , \quad v=u^2-1$$ Eliminating $v$, we get $$u^4-2u^2-u-6=0$$ which has an easy root $u=2$. Can you now finish?

Answer 3

I would suggest assuming $x+y=a$ and $x-y=b$. This forms a 2x2 system that given $a$ and $b$ can quickly be solved. After that you have

\begin{cases} \sqrt{b}=a-7 \\ \sqrt{a}=b-1 \end{cases} $$\Leftrightarrow \begin{cases} b=(a-7)^2 \\ a=(b-1)^2 \end{cases} $$ $$\Leftrightarrow b=((b-1)^2-7)^2 $$ $$\Leftrightarrow b=(b^2 - 2b +1 -7)^2 $$ $$\Leftrightarrow b=(b^2 - 2b -6)^2 $$ $$\Leftrightarrow b=b^4 -4b^3 -8b^2 +24b +36 $$ $$\Leftrightarrow b^4 -4b^3 -8b^2 +23b +36 =0 $$ $$\Leftrightarrow (b-4)(b^3-8b-9) =0 $$

which has two solutions. One is obviously 4 and the other on can be approximated numerically or computed using the cubic formula found here: https://math.vanderbilt.edu/schectex/courses/cubic/

After computing the two possible solutions you can substitute to find a here $a=(b-1)^2$ and then solve the system.

Moreover the solutions for b are $b=4$ and $b \approx 3.27803\dots $ . For b =4 we have that $a=(4-1)^2 = 9$ so \begin{cases} x+y=9 \\ \ x-y=4 \end{cases} $$\Leftrightarrow \begin{cases} x=13/2 \\ y=5/2 \end{cases} $$

And you do the same for $b \approx 3.27803\dots $