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I'm trying to solve the following,from Example 23 of Ratio chapter in Hall and Knight's Higher Algebra, 4th edition:

$ \left\{ \begin{array}{l} 3x^2 - 2y^2 + 5z^2 = 0 \\ 7x^2 - 3y^2 - 15z^2 = 0\\ 5x - 4y + 7z = 6 \\ \end{array} \right. $

Solving equation 1 and 2 by cross multiplication, I got as far as :

$x^2/45 = y^2/80 = z^2/5$.

I'm not able to figure out how to solve the third equation in first degree with this information.

Thanks for the help. Apologies for the formatting...trying to learn Mathjax but am a beginner.

Raffaele
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rajivmedanki
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1 Answers1

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Set $t=\dfrac{x^2}{z^2};\;y=\dfrac{y^2}{z^2}$

$ \left\{ \begin{array}{l} 3 t-2 u+5=0 \\ 7 t-3 u-15=0 \\ \end{array} \right. $

$t=9;\;u=16$

Thus

$\dfrac{x^2}{z^2}=9\lor \dfrac{y^2}{z^2}=16$

$x=\pm 3z;\;y=\pm 4z$

We have two systems

$ \left\{ \begin{array}{l} x=3z \\ y=4z\\ 5 x-4 y+7 z=6 \\ \end{array} \right. $

$x = 3,\;y= 4,\;z= 1$

$ \left\{ \begin{array}{l} x=-3z \\ y=-4z\\ 5 x-4 y+7 z=6 \\ \end{array} \right. $

$x = -\dfrac{9}{4},\;y= -3,\;z= \dfrac{3}{4}$

Hope this helps

Raffaele
  • 26,371