Krdytkyu

$$begin{cases}
x+dfrac{3x-y}{x^2+y^2}=3 \
y-dfrac{x+3y}{x^2+y^2}=0
end{cases}$$
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?

Make the substitution $x = rcos(theta)$ and $y = rsin(theta)$ in order to get
begin{cases}
r^{2}cos(theta) + 3cos(theta) - sin(theta) = 3r\
r^{2}sin(theta) - cos(theta) - 3sin(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 Longleftrightarrowsin(theta) = frac{3r}{r^{4}-10}
end{align*}

Consequently, we have
begin{align*}
&cos^{2}(theta) + sin^{2}(theta) = 1 Leftrightarrowleft[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 rin{sqrt{2},sqrt{5}}quadtext{since}quad rgeq 0
end{align*}

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

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

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

So we can assume $yneq 0$, set $ x=ty $ and substitute.

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

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 $begin{cases}2y+frac 1y=3\y=frac 1yend{cases}iff begin{cases}x=2\y=1end{cases}$




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

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

Thus, $x+3yneq0$ and since $$frac{(x^2+y^2)y}{x+3y}=1,$$ from the first equation we obtain:
$$x+frac{3x-y}{x^2+y^2}cdotfrac{(x^2+y^2)y}{x+3y}=3sqrt{frac{(x^2+y^2)y}{x+3y}}$$ or
$$x^2+6xy-y^2=3sqrt{y(x^2+y^2)(x+3y)}$$ or with $x^2+6xy-y^2geq0$
$$(x^2+6xy-y^2)^2=9y(x^2+y^2)(x+3y)$$ or
$$(x+y)(x-2y)(x^2+4xy+9y^2)=0,$$ which gives $y=-x$ or $x=2y$.

Can you end it now?

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 $a = frac{3}{1-10t^2}binom{1-3t}{t}$:
  $$Rightarrow begin{cases} t= frac{1}{2} Rightarrow boxed{a= binom{1}{-1}} \ t= frac{1}{5} Rightarrow boxed{a= binom{2}{1}} end{cases}$$
  

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

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

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

On solving that equation we have

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

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