Blericker74

2022-07-14

Solve the system of equations:
$\left\{\begin{array}{l}\left(x-1\right)\sqrt{x-{y}^{2}}=y\left(x-2y+1\right)\\ y\sqrt{x-1}+3\sqrt{x-{y}^{2}}=2x+y-1\end{array}$
This is my try
Condition: $x\ge 1;x\ge {y}^{2}$
We have:
$x-2y+1\ge x-2|y|+1\ge x-\left({y}^{2}+1\right)+1=x-{y}^{2}\ge 0⇒y\ge 0$
Applying AM-GM inequality, we have:
$2\left(2x+y-1\right)=2y\sqrt{x-1}+6\sqrt{x-{y}^{2}}\le \left({y}^{2}+x-1\right)+3\left(1+x-{y}^{2}\right)=4x-2{y}^{2}+2$
$⇒{y}^{2}+y-2\le 0⇒0\le y\le 1$
Squaring both side of first equation, we get:
$\left(x-2{y}^{2}+2y-1\right)\left({x}^{2}-x+2{y}^{2}-2xy\right)=0$
If $x-2{y}^{2}+2y-1=0$, we get $\frac{1}{2}\le x\le 1$. Thus, we get $x=1$ and $y=0$ or $y=1$. Two pairs of $\left(x,y\right)$ don't satisfy the system of equations.
If ${x}^{2}-x+2{y}^{2}-2xy=0$...

owerswogsnz

Expert

since you have
${x}^{2}-x+2{y}^{2}-2xy=0⟹x-{y}^{2}={x}^{2}-2xy+{y}^{2}=\left(x-y{\right)}^{2}$
Note $x>y$,so have
$\sqrt{x-{y}^{2}}=x-y$
take the second equation we have
$y\sqrt{x-1}+3\left(x-y\right)=2x+y-1$
so we have
$y=\frac{x+1}{4-\sqrt{x-1}}\in \left[0,1\right]⟹1\le x\le 2$
take
${x}^{2}-x+2{y}^{2}-2xy=0⟹{x}^{2}-x+2{\left(\frac{x+1}{4-\sqrt{x-1}}\right)}^{2}-2x\frac{x+1}{4-\sqrt{x-1}}=0,1\le x\le 2$
$⟹\frac{{x}^{3}-6\sqrt{x-1}{x}^{2}+8{x}^{2}+10\sqrt{x-1}x-19x+2}{\left(\sqrt{x-1}-4{\right)}^{2}}=0$
take $\sqrt{x-1}=t$,then we have
$⟹\left({t}^{2}+1{\right)}^{3}-6t\left({t}^{2}+1{\right)}^{2}+8\left({t}^{2}+1{\right)}^{2}+10t\left({t}^{2}+1\right)-19\left({t}^{2}+1\right)+2=0$
$⟹\left(t-1\right)\left({t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8\right)=0,0\le t\le 1$
since
$f\left(t\right)={t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8={t}^{5}+6{t}^{3}+4{t}^{2}+4t+\left(8-5{t}^{4}\right)>0,0\le t\le 1$
so $t=1$ then you have $x=2$

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