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I'm given the following equation: $y^2 + 3xy - 10x^2 + y + 5x = 0$ and asked to solve $y$ in terms of $y$.

My attempt:

$y^2 + (3x+1)\times y - 10x^2 + 5x = 0$

$\Rightarrow (y+(3x+1)/2)^2 - ((3x+1)/2)^2 = (x - (\frac 12)\times x)^2 - (\frac 12)^2 $

But I cant seem to get any further and it's probably in the wrong direction.

psmears
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user262493
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2 Answers2

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$$y^2 + 3xy - 10x^2 + y + 5x = 0\Rightarrow y^2+(3x+1)y-(10x^2-5x) = 0$$

So $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(3x+1)^2+4(10x^2-5x)}}{2}$$

$$\displaystyle = \frac{-(3x+1)\pm \sqrt{49x^2-14x+1}}{2}$$

so we get $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(7x-1)^2}}{3} = \frac{-(3x+1)\pm (7x-1)}{2}$$

So we get $\displaystyle y = 2x-1\;\;,-5x$

juantheron
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  • Good old quadratic formula, the thoughtless route that requires some algebra but always works! I love it :) – CompuChip Aug 23 '15 at 17:57
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$0 = y^2+3xy-10x^2 + y+5x= (y+5x)(y-2x)+(y+5x) = (y+5x)(y-2x+1)$

Now this is much simpler to solve

Brenton
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  • How did you see the first step? @Brenton – user262493 Aug 23 '15 at 15:39
  • @user262493 When I saw the $y+5x$ term in the original equation, I was trying to see if there was a nice way to factor it with the $y^2+3xy−10x^2$ term. Usually factoring out the GCF makes these problems simpler, so that was my first try at what to do -- try to factor a $y+5x$ out of $y^2+3xy−10x^2$ – Brenton Aug 23 '15 at 15:53
  • At the risk of stating the obvious, it helps if you can notice that $3=(5)+(-2)$ and $-10=(5)\times(-2)$. – Scott - Слава Україні Aug 23 '15 at 19:03