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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 !

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The only thing missing is the final simplification: $$\pm \sqrt{\frac{5 \pm 3}{2}} = \{ \pm 1, \pm 2 \}.$$ Then $x$ is easily obtained from $y$. –  heropup Feb 22 at 9:18
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Best way to check is to substitute the values back in the original equations. –  Mark Bennet Feb 22 at 9:20
    
@MarkBennet I should have done that before asking this question. :) –  David Sebastian Feb 22 at 9:33
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2 Answers 2

up vote 2 down vote accepted

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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That's a nice way. Thanks Andre! –  David Sebastian Feb 22 at 9:32
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You are welcome. –  André Nicolas Feb 22 at 9:33
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What you did is perfectly correct. You only missed to continue the simplifcations since $a$ is either $4$ or $1$ and then $y$ is ...

Good job !

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Thank you Claude Leibovici. –  David Sebastian Feb 22 at 9:34
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@DavidSebastian. You are welcome. –  Claude Leibovici Feb 22 at 9:36
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