Solve for x & y: Simultaneous Equations

[anemone]

Solve the simultaneous equations

$a+\dfrac{3a-b}{a^2+b^2}=3$

$b-\left(\dfrac{a+3b}{a^2+b^2}\right)=0$

 

[juantheron]

[sp]
My Solution:: Given $\displaystyle a+\left(\frac{3a-b}{a^2+b^2}\right) = 3.....(1)$

and $\displaystyle b-\left(\frac{a+3b}{a^2+b^2}\right) = 0......(2)$

Multiply eqn...$\bf{(2)}$ by $i=\sqrt{-1}$ and Added::

$\displaystyle (a+ib)+\frac{1}{a^2+b^2}\left[3(a-ib)-(b+ia)\right] = 3$

Now Let $z=a+ib$ and $\bar{z}=a-ib$ and $z\cdot \bar{z}=(a^2+b^2)$

So equation is $\displaystyle z+\frac{1}{z\cdot \bar{z}}\left[3\bar{z}-i\bar{z}\right] = 3$

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

So $\displaystyle z = (1-i)\;,(2+i)\Rightarrow a+ib = (1-i)\;,(2+i)$

So $(a,b) = \left\{(1,-1)\;\;,(2,1)\right\}$[/sp]

 

[anemone]

[sp]
My Solution:: Given $\displaystyle a+\left(\frac{3a-b}{a^2+b^2}\right) = 3.....(1)$

and $\displaystyle b-\left(\frac{a+3b}{a^2+b^2}\right) = 0......(2)$

Multiply eqn...$\bf{(2)}$ by $i=\sqrt{-1}$ and Added::

$\displaystyle (a+ib)+\frac{1}{a^2+b^2}\left[3(a-ib)-(b+ia)\right] = 3$

Now Let $z=a+ib$ and $\bar{z}=a-ib$ and $z\cdot \bar{z}=(a^2+b^2)$

So equation is $\displaystyle z+\frac{1}{z\cdot \bar{z}}\left[3\bar{z}-i\bar{z}\right] = 3$

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

So $\displaystyle z = (1-i)\;,(2+i)\Rightarrow a+ib = (1-i)\;,(2+i)$

So $(a,b) = \left\{(1,-1)\;\;,(2,1)\right\}$[/sp]

Good job, jacks! Thanks for participating!:)