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I got stuck on this system of equations. Could you help and tell me how should I approach this problem? \begin{align*} x+y^2 &= y^3\\ y+x^2 &= x^3 \end{align*} These are the solutions: \begin{align*} (0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2); \end{align*} |
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Here's your equations: $x+y^2 = y^3$ $y+x^2 = x^3$ I could substitute $x=y^3-y^2$ and get a degree 9 equation in $y$, but I'll try something else. Looking at these, I notice that if I subtract them, I get something in which everything is divisible by $x-y$. Subtracting the second from the first, I get $(x-y)+(y^2-x^2) =y^3-x^3 $ or $(x-y)+(y-x)(y+x) =(y-x)(y^2+xy+x^2) $. If $y \ne x$, dividing by $x-y$ gives $1-(y+x) =y^2+xy+x^2 $ or $0 =y^2+y(x+1)+x^2+x+1 $. Solving this, $\begin{array}\\ y &=\dfrac{-(x+1)\pm \sqrt{(x+1)^2-4(x^2+x+1)}}{2}\\ &=\dfrac{-(x+1)\pm \sqrt{x^2+2x+1-4x^2-4x-4}}{2}\\ &=\dfrac{-(x+1)\pm \sqrt{-3x^2-2x-3}}{2}\\ &=\dfrac{-(x+1)\pm \sqrt{-2x^2-x^2-2x-1-2}}{2}\\ &=\dfrac{-(x+1)\pm \sqrt{-2x^2-(x+1)^2-2}}{2}\\ \end{array} $ Since the discriminant is negative, there are no real values of $y$. Therefore the only real solution is $x=y$. |
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Notice that $x=0 \implies y=0$, hence $(0,0)$ is a trivial solution. We will consider the case where $x \neq 0$. $$x=y^2(y-1)$$ $$y=x^2(x-1)$$ Suppose on the contrary that $x=1$, then from the second equation $y=0$ which contradicts the first equation $x=0$. Hence $x \neq 1$. Suppose on the contrary that $x \in (0,1)$, $$y=x^2(x-1) \in (-1,0)$$ and hence $$x=y^2(y-1)<0$$ which is a contradiction. Hence $x \notin (0,1).$ By symmetry, $y \notin (0,1)$. Also, notice if $x>1$, then $y=x^2(x-1)>0$, and hence $y>1$. and if $x<0$ then $y=x^2(x-1)<0$. $$x=y^2(y-1)$$ $$x^2(x-1)=y$$ Multiply the two equations together, we have $$x^3(x-1)=y^3(y-1)$$ Consider the function, $$f: (1,\infty) \rightarrow (0,\infty),\text{ where }f(t)= t^3(t-1).$$ We can easily see that this function is increasing and it is an injective function. Hence, if $x>1$, we have $f(x)=f(y)$ and hence $x=y$. Similarly, we can consider the function, $$g: (-\infty,0) \rightarrow (0,\infty),\text{ where }g(t)= t^3(t-1).$$ We can easily see that this function is decreasing and it is an injective function. Hence, if $x<0$, we have $g(x)=g(y)$ and hence $x=y$. Hence, we always have $x=y$. $$x+x^2=x^3$$ $$x(1-x-x^2)=0$$ $x=0$ or $1-x-x^2=0$ |
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We have \begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}This curve is symmetric about the line $y=x$.On putting $x=y$ ,we have $x^3-x^2-x=0$ i.e $x=0,-0.618,1.618.$So, the solution pairs are $(0,0),(-0.618,-0.618),(1.618,1.618)$ |
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