How to solve for x and y $x^2+y^2-xy-1 = 0$ and $x^2+1=y^2 $?

Question

$$ \Rightarrow x^2+y^2-xy-1=0 $$

$$ \Rightarrow x^2 + 1 = y^2 $$

What I tried was to substitute value of x from one equation to another but $xy$ caused problem in solving it .

How to solve this ?

Answer 1

I would write $$y=\pm\sqrt{1+x^2}$$ then we get $$x^2+x^2+1-x(\pm\sqrt{1+x^2})-1=0$$so $$2x^2=x(\pm\sqrt{1+x^2})$$ Now we have that $$x=0$$ is one solution and then after dividing by $$x\neq 0$$ we get $$2x=\pm\sqrt{1+x^2}$$ and you must square the equation. We get $$(x,y)=\{0,-1\},\{0,1\},\{-\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}}\},\{\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}\}$$

Answer 2

Hint: substitution of $\color{blue}{y^2-1}=x^2$ into $x^2+\color{blue}{y^2}-xy\color{blue}{-1}=0$ gives: $$2x^2-xy = 0 \iff x\left(2x-y\right)=0 \iff \boxed{x = 0 \,\vee\, y=2x}$$

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Addition after comment:

• plugging $x=0$ into $y^2-1=x^2$ immediately gives $y= \pm 1$;
• plugging $y=2x$ into $y^2-1=x^2$ gives $x= \pm \tfrac{1}{\sqrt{3}}$; then $y$ follows from $y=2x$.

Answer 3

Use substitution. Substitute $y = \pm\sqrt{x^2+1}$. $$x^2+(x^2+1)-x\big(\pm\sqrt{x^2+1}\big)-1 = 0$$ Simplify completely. $$x^2+x^2\mp x\sqrt{x^2+1} = 0 \implies 2x^2\mp x\sqrt{x^2+1} = 0 \implies 2x^2 = \pm x\sqrt{x^2+1}$$ Square both sides. $$\implies 4x^4 = x^4+x^2 \implies 3x^4-x^2 = 0$$ Factor. $$x^2(3x^2-1) = 0$$ Now, just set either factor equal to $0$ and solve. Plug in the values of $x$ obtained in either function to get the $y$-coordinates. The solutions become $(0, \pm1)$ and $\big(\pm \frac{1}{\sqrt{3}}, \pm \frac{2}{\sqrt{3}}\big)$. The latter can also be rationalized and written as $\big(\pm \frac{\sqrt{3}}{3}, \pm \frac{2\sqrt{3}}{3}\big)$.

Answer 4

$$\begin{cases}x^2+y^2-xy-1=0\\x^2-y^2+1=0\end{cases}\Rightarrow x(2x-y)=0$$ from which $$x^2+4x^2-2x^2-1=0\iff x=\pm\frac{1}{\sqrt3}\approx\pm 0.57735$$ There are four solutions $(0,\pm1),(\pm0.557,\pm1.155)$

Answer 5

From

$$x^2+y^2-xy-1 = 0,\\x^2+1=y^2,$$ you can eliminate $x$ by $$2y^2-xy-2=0, \\ x=2\frac{y^2-1}{y}. $$

and the second equation yields a biquadratic

$$4(y^2-1)^2+y^2=y^4$$

or

$$(3y^2-4)(y^2-1)=0.$$