<s>[INST] Have I done something wrong in solving the following pair of equations? Question given:
Solve,$3x^2-5y^2-7=0\\3xy-4y^2-2=0$
What I have done so far:
$$
3xy-4y^2-2=0\\\frac{3xy}{y}-\frac{4y2}{y}-\frac{2}{y}=0\\3x-4y-\frac{2}{y}=0\\3x=4y+\frac{2}{y}\\x=\frac{(4y+\frac{2}{y})}{3}
$$By substituting $x=\frac{(4y+\frac{2}{y})}{3}$ in the first equation,$$3x^2-5y^2-7=0\\3(\frac{4y+\frac{2}{y}}{3})^2-5y^2-7=0\\3(\frac{16y^2+\frac{4}{y^2}+16}{9})-5y^2-7=0\\16y^2+\frac{4}{y^2}+16-15y^2-21=0\\y^2+\frac{4}{y^2}-5=0\\y^4-5y^2+4=0$$Let $a=y^2$,$$(y^2)^2-5(y^2)+4=0\\a^2-5a+4=0\\a=\frac{5\pm\sqrt{25-4.1.4}}{2}\\a=\frac{5\pm\sqrt{25-16}}{2}\\a=\frac{5\pm3}{2}\\y^2=\frac{5\pm3}{2}\\y=\pm\sqrt{\frac{5\pm3}{2}}$$
Have I done something wrong when solving these pair of equations ? If so, please correct me.
Best Regards !
 [/INST] Hint: Heroic! Let us work more simply. Our two equations are $3x^2-5y^2=7$ and $3xy-4y^2=2$. Multiply the first through by $2$, the second by $7$, and subtract. We get $6x^2-21xy +18y^2=0$, or more simply $2x^2-7xy+6y^2=0$. Now we can solve for $y$ in terms of $x$. One could use the Quadratic Formula, but we get lucky, the thing factors simply. Carry on!
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