Irrational equation system

Question

I need to solve an equation system : \begin{eqnarray*} \sqrt{7x+y}+\sqrt{x+y}&=&6\\ \sqrt{x+y}+x-y&=&2 \end{eqnarray*} I know that $x=2, y=2$ is the right solution I don't see the way to get there, and I am a bit lazy to do all the calculations with the squares..is there any other way? Thanks in advance!

Answer 1

Let $\sqrt{x+y}=t$. Thus, $t\geq0$, $$t+t^2-2y=2,$$ which gives $$y=\frac{t^2+t-2}{2}$$ and we obtain: $$\sqrt{7t^2-3(t^2+t-2)}+t=6$$ or $$4t^2-3t+6=(6-t)^2,$$ where $0\leq t\leq6$, which gives $t=2$, $y=2$ and $x=2$.

Answer 2

solving your second equation for $$\sqrt{x+y}$$ and plugging this in the first equation we get $$\sqrt{7x+y}+2-x+y=6$$ isolate the square root and square the equation and you will get $$-16-x-x^2+9y+2xy-y^2=0$$ solve this for $x$ or $y$ we get $$x=2,y=2$$, check it!

Answer 3

When you have two radicals, an often convenient technique is to consider $$ (\sqrt{7x+y}+\sqrt{x+y})(\sqrt{7x+y}-\sqrt{x+y})=6(\sqrt{7x+y}-\sqrt{x+y}) $$ that easily yields $$ \sqrt{7x+y}-\sqrt{x+y}=x $$ Subtracting this from the first equation, we get $$ 2\sqrt{x+y}=6-x\tag{*} $$ and substituting in the second equation we obtain $$ 6-x+2x-2y=4 $$ or $x=2y-2$. Now, substituting in (*), $$ \sqrt{3y-2}=4-y $$ We can square, getting $$ 3y-2=16-8y+y^2\qquad y^2-11y+18=0 $$ The roots are $y=2$ and $y=9$, but the second would make $4-y<0$, so it should be discarded.