Let (x,y) be a pair of real numbers satisfying 56x+33y=x2+y2−y,and33x−56y=x2+y2x.Determine the value of ∣x∣+∣y∣.
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Observe that x+yi1=x2+y2x−yi=33x−56y+(56x+33y)i=(33+56i)(x+yi).So (x+yi)2=33+56i1=(7+4i)21=(657−4i)2.It follows that (x,y)=±(657,−654), so ∣x∣+∣y∣=6511.