MHB Solve for x & y: Simultaneous Equations

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The discussion focuses on solving the simultaneous equations involving variables a and b. The equations are transformed using complex numbers, where a is represented as the real part and b as the imaginary part. By manipulating the equations, solutions for z are derived, leading to two sets of values for (a, b): (1, -1) and (2, 1). The use of complex numbers simplifies the problem and provides a clear pathway to the solution. The thread concludes with positive reinforcement for the contributors' efforts.
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Solve the simultaneous equations

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

$b-\left(\dfrac{a+3b}{a^2+b^2}\right)=0$
 
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[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]
 
jacks said:
[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!:)
 
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