Blericker74

Answered

2022-07-14

Solve the system of equations:

$\{\begin{array}{l}(x-1)\sqrt{x-{y}^{2}}=y(x-2y+1)\\ y\sqrt{x-1}+3\sqrt{x-{y}^{2}}=2x+y-1\end{array}$

This is my try

Condition: $x\ge 1;x\ge {y}^{2}$

We have:

$x-2y+1\ge x-2|y|+1\ge x-({y}^{2}+1)+1=x-{y}^{2}\ge 0\Rightarrow y\ge 0$

Applying AM-GM inequality, we have:

$2(2x+y-1)=2y\sqrt{x-1}+6\sqrt{x-{y}^{2}}\le ({y}^{2}+x-1)+3(1+x-{y}^{2})=4x-2{y}^{2}+2$

$\Rightarrow {y}^{2}+y-2\le 0\Rightarrow 0\le y\le 1$

Squaring both side of first equation, we get:

$(x-2{y}^{2}+2y-1)({x}^{2}-x+2{y}^{2}-2xy)=0$

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 $(x,y)$ don't satisfy the system of equations.

If ${x}^{2}-x+2{y}^{2}-2xy=0$...

$\{\begin{array}{l}(x-1)\sqrt{x-{y}^{2}}=y(x-2y+1)\\ y\sqrt{x-1}+3\sqrt{x-{y}^{2}}=2x+y-1\end{array}$

This is my try

Condition: $x\ge 1;x\ge {y}^{2}$

We have:

$x-2y+1\ge x-2|y|+1\ge x-({y}^{2}+1)+1=x-{y}^{2}\ge 0\Rightarrow y\ge 0$

Applying AM-GM inequality, we have:

$2(2x+y-1)=2y\sqrt{x-1}+6\sqrt{x-{y}^{2}}\le ({y}^{2}+x-1)+3(1+x-{y}^{2})=4x-2{y}^{2}+2$

$\Rightarrow {y}^{2}+y-2\le 0\Rightarrow 0\le y\le 1$

Squaring both side of first equation, we get:

$(x-2{y}^{2}+2y-1)({x}^{2}-x+2{y}^{2}-2xy)=0$

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 $(x,y)$ don't satisfy the system of equations.

If ${x}^{2}-x+2{y}^{2}-2xy=0$...

Answer & Explanation

owerswogsnz

Expert

2022-07-15Added 12 answers

since you have

${x}^{2}-x+2{y}^{2}-2xy=0\u27f9x-{y}^{2}={x}^{2}-2xy+{y}^{2}=(x-y{)}^{2}$

Note $x>y$,so have

$\sqrt{x-{y}^{2}}=x-y$

take the second equation we have

$y\sqrt{x-1}+3(x-y)=2x+y-1$

so we have

$y={\displaystyle \frac{x+1}{4-\sqrt{x-1}}}\in [0,1]\u27f91\le x\le 2$

take

${x}^{2}-x+2{y}^{2}-2xy=0\u27f9{x}^{2}-x+2{\left({\displaystyle \frac{x+1}{4-\sqrt{x-1}}}\right)}^{2}-2x{\displaystyle \frac{x+1}{4-\sqrt{x-1}}}=0,1\le x\le 2$

$\u27f9{\displaystyle \frac{{x}^{3}-6\sqrt{x-1}{x}^{2}+8{x}^{2}+10\sqrt{x-1}x-19x+2}{(\sqrt{x-1}-4{)}^{2}}}=0$

take $\sqrt{x-1}=t$,then we have

$\u27f9({t}^{2}+1{)}^{3}-6t({t}^{2}+1{)}^{2}+8({t}^{2}+1{)}^{2}+10t({t}^{2}+1)-19({t}^{2}+1)+2=0$

$\u27f9(t-1)({t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8)=0,0\le t\le 1$

since

$f(t)={t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8={t}^{5}+6{t}^{3}+4{t}^{2}+4t+(8-5{t}^{4})>0,0\le t\le 1$

so $t=1$ then you have $x=2$

${x}^{2}-x+2{y}^{2}-2xy=0\u27f9x-{y}^{2}={x}^{2}-2xy+{y}^{2}=(x-y{)}^{2}$

Note $x>y$,so have

$\sqrt{x-{y}^{2}}=x-y$

take the second equation we have

$y\sqrt{x-1}+3(x-y)=2x+y-1$

so we have

$y={\displaystyle \frac{x+1}{4-\sqrt{x-1}}}\in [0,1]\u27f91\le x\le 2$

take

${x}^{2}-x+2{y}^{2}-2xy=0\u27f9{x}^{2}-x+2{\left({\displaystyle \frac{x+1}{4-\sqrt{x-1}}}\right)}^{2}-2x{\displaystyle \frac{x+1}{4-\sqrt{x-1}}}=0,1\le x\le 2$

$\u27f9{\displaystyle \frac{{x}^{3}-6\sqrt{x-1}{x}^{2}+8{x}^{2}+10\sqrt{x-1}x-19x+2}{(\sqrt{x-1}-4{)}^{2}}}=0$

take $\sqrt{x-1}=t$,then we have

$\u27f9({t}^{2}+1{)}^{3}-6t({t}^{2}+1{)}^{2}+8({t}^{2}+1{)}^{2}+10t({t}^{2}+1)-19({t}^{2}+1)+2=0$

$\u27f9(t-1)({t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8)=0,0\le t\le 1$

since

$f(t)={t}^{5}-5{t}^{4}+6{t}^{3}+4{t}^{2}+4t+8={t}^{5}+6{t}^{3}+4{t}^{2}+4t+(8-5{t}^{4})>0,0\le t\le 1$

so $t=1$ then you have $x=2$

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