Question Number 156244 by Eric002 last updated on 09/Oct/21
$${solve}\:{the}\:{pairs}\:{of}\:{simultaneous}\:{equations} \\ $$$${ax}−\mathrm{2}{y}=\mathrm{2} \\ $$$${x}+\mathrm{3}{y}=\mathrm{3} \\ $$$${qy}−{px}=\left({q}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)/{pq} \\ $$$${py}+{qx}=\mathrm{2} \\ $$
Answered by MJS_new last updated on 10/Oct/21
$$\left(\mathrm{1}\right)\:{x}=\frac{\mathrm{12}}{\mathrm{3}{a}+\mathrm{2}}\wedge{y}=\frac{\mathrm{3}{a}−\mathrm{2}}{\mathrm{3}{a}+\mathrm{2}}\wedge{a}\neq−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:{x}=\frac{{p}^{\mathrm{2}} +\mathrm{2}{pq}−{q}^{\mathrm{2}} }{\mathrm{2}{pq}^{\mathrm{2}} }\wedge{y}=−\frac{{p}^{\mathrm{2}} −\mathrm{2}{pq}−{q}^{\mathrm{2}} }{\mathrm{2}{p}^{\mathrm{2}} {q}}\wedge{p}\neq\mathrm{0}\wedge{q}\neq\mathrm{0} \\ $$