Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$

Question

$$\begin{cases} x+\dfrac{3x-y}{x^2+y^2}=3 \\ y-\dfrac{x+3y}{x^2+y^2}=0 \end{cases}$$ Solve in the set of real numbers.

The furthest I have got is summing the equations, and I got $$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$ But I have no idea how to solve this problem. How can I solve it?

Answer 1

Make the substitution $x = r\cos(\theta)$ and $y = r\sin(\theta)$ in order to get \begin{cases} r^{2}\cos(\theta) + 3\cos(\theta) - \sin(\theta) = 3r\\ r^{2}\sin(\theta) - \cos(\theta) - 3\sin(\theta) = 0\\ \end{cases}

From the second equation, it results that $\cos(\theta) = (r^{2}-3)\sin(\theta)$. Therefore we get \begin{align*} (r^{2} + 3)\cos(\theta) - \sin(\theta) & = 3r \Longleftrightarrow (r^{2}+3)(r^{2}-3)\sin(\theta) - \sin(\theta) = 3r \Longleftrightarrow\\ (r^{4}- 10)\sin(\theta) & = 3r \Longleftrightarrow\sin(\theta) = \frac{3r}{r^{4}-10} \end{align*}

Consequently, we have \begin{align*} &\cos^{2}(\theta) + \sin^{2}(\theta) = 1 \Leftrightarrow\left[\frac{3r(r^{2}-3)}{r^{4}-10}\right]^{2} + \left[\frac{3r}{r^{4}-10}\right]^{2} = 1 \Leftrightarrow\\\\ & 9r^{2}(r^{4}-6r^{2}+9) + 9r^{2} = (r^{4}-10)^{2} \Leftrightarrow 9r^{6}-54r^{4} + 90r^{2} = r^{8} - 20r^{4} + 100 \Leftrightarrow\\\\ & r^{8} - 9r^{6} + 34r^{4} - 90r^{2} + 100 = 0 \Leftrightarrow (r^{8} - 5r^{6}) - (4r^{6} - 20r^{4}) + (14r^{4} - 70r^{2}) - (20r^{2} - 100) = 0 \Leftrightarrow\\\\ & r^{6}(r^{2}-5) - 4r^{4}(r^{2}-5) + 14r^{2}(r^{2}-5) - 20(r^{2}-5) = (r^{6} - 4r^{4} + 14r^{2} - 20)(r^{2}-5) = 0 \Leftrightarrow\\\\ & [(r^{6} - 2r^{4}) - (2r^{4} - 4r^{2}) + (10r^{2}-20)](r^{2}-5) = 0 \Leftrightarrow (r^{4}-2r^{2}+10)(r^{2}-2)(r^{2}-5) = 0 \Leftrightarrow\\\\ & (r^{2} - 2)(r^{2} - 5) = 0 \Leftrightarrow r\in\{\sqrt{2},\sqrt{5}\}\quad\text{since}\quad r\geq 0 \end{align*}

Finally, for each value of $r$ there corresponds a solution. More precisely, the solution set is described by $$S = \{(1,-1),(2,1)\}$$

Answer 2

Hint: Write your equations in the form $$\frac{3x-y}{3-x}=\frac{x+3y}{y}$$

Answer 3

Note that if $y=0$ then we get $-\frac 1x=0$ on second equation, so the system has no solutions.

So we can assume $y\neq 0$, set $\ x=ty\ $ and substitute.

We get $\begin{cases}t^3y^2+ty^2+3t-1=3t^2y+3y \\t^2y^2+y^2-t-3=0\end{cases}$

And this allow us to isolate $y^2=\dfrac{t+3}{t^2+1}$

Substituting in first equation rewritten $ty^2(t^2+1)+3t-1=3y(t^2+1)$

gives $t(t+3)+3t-1=3y(t^2+1)\iff 3y=\dfrac{t^2+6t-1}{t^2+1}$

Now be identifying $9y^2$ we get $(t^2+6t-1)^2=9(t+3)(t^2+1)$

We can remark $t=-1$ is a root $6^2=9(2)(2)$ else just graph it, and notice $-1$ and $2$ seems to be roots (so try to factor these terms).

Indeed the factorization is $$(t+1)(t-2)(t^2+4t+13)=0$$

Reporting in initial equations gives us the solutions:

• $t=2$ and $x=2y$ leads to $\begin{cases}2y+\frac 1y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=2\\y=1\end{cases}$

• $t=-1$ and $x=-y$ leads to $\begin{cases}-y-\frac 2y=3\\y=\frac 1y\end{cases}\iff \begin{cases}x=1\\y=-1\end{cases}$

Answer 4

If $x+3y=0$ so from the second equation we obtain $y=0$, which gives $x=0$, which is impossible.

Thus, $x+3y\neq0$ and since $$\frac{(x^2+y^2)y}{x+3y}=1,$$ from the first equation we obtain: $$x+\frac{3x-y}{x^2+y^2}\cdot\frac{(x^2+y^2)y}{x+3y}=3\sqrt{\frac{(x^2+y^2)y}{x+3y}}$$ or $$x^2+6xy-y^2=3\sqrt{y(x^2+y^2)(x+3y)}$$ or with $x^2+6xy-y^2\geq0$ $$(x^2+6xy-y^2)^2=9y(x^2+y^2)(x+3y)$$ or $$(x+y)(x-2y)(x^2+4xy+9y^2)=0,$$ which gives $y=-x$ or $x=2y$.

Can you end it now?

Answer 5

Sketch of a solution rewriting the given system in terms of linear algebra:

• Set $\boxed{a = \binom{x}{y}}$. Then the given system looks as follows: $$\begin{cases} x+\dfrac{3x-y}{x^2+y^2}=3 \\ y-\dfrac{x+3y}{x^2+y^2}=0 \end{cases} \Longleftrightarrow \boxed{a +\frac{1}{||a||^2}\begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}a = \binom{3}{0}}$$
• Set $\boxed{t = \frac{1}{ ||a||^2}}$ and $\boxed{A = \begin{pmatrix}3 & -1 \\ -1 & -3 \end{pmatrix}} \Rightarrow (I + tA)a = \binom{3}{0} \Rightarrow \boxed{a = (I + tA)^{-1}\binom{3}{0}}$.
• So, calculate $a$ by inverting $(I + tA)$: $$\boxed{a =} (I + tA)^{-1}\binom{3}{0} = \begin{pmatrix}1+3t & -t \\ -t & 1-3t \end{pmatrix}^{-1}\binom{3}{0} = \ldots = \boxed{\frac{3}{1-10t^2}\binom{1-3t}{t} }$$
• Find $t >0 $ such that $t = \frac{1}{ ||a||^2} \Rightarrow \boxed{t= \frac{1}{5}, \frac{1}{2}}$
• Plug these values into $a = \frac{3}{1-10t^2}\binom{1-3t}{t}$: $$\Rightarrow \begin{cases} t= \frac{1}{2} \Rightarrow \boxed{a= \binom{1}{-1}} \\ t= \frac{1}{5} \Rightarrow \boxed{a= \binom{2}{1}} \end{cases}$$

Answer 6

$$\displaystyle x+\frac{3x-y}{x^2+y^2}=3\cdots (1)$$

$$\displaystyle y-\frac{x+3y}{x^2+y^2}=0\cdots\cdots(2)\times i $$

Now adding these two equations

and substituting $z=x+iy$ and $\bar{z}=x-iy$

and $$|z|^2=x^2+y^2$$

So we have $$z+\frac{3-i}{z}=3\Rightarrow z^2-3z+(3-i)=0$$

On solving that equation we have

$$z=2+i\;\;,1-i$$

So we get $$(x,y)=(2,1)\;\;,(1,-1)$$