# $x^2 + y^2 = 13, \, x^2 - 3xy + 2y^2 = 35$

I have been unable to solve the above problem.

I will detail my attempt below:

# My Solution

$$x^2 + y^2 = 13 \tag{1}$$ $$x^2 - 3xy + 2y^2 = 35 \tag{2}$$
Let $$y = mx$$.
Rewrite $$(1)$$ and $$(2)$$.
$$x^2 + m^2x^2 = 13 \tag{1.1}$$ $$x^2 - 3mx^2 + 2m^2x^2 \tag{2.1}$$ Factorise the above equations. $$x^2(m^2 + 1) = 13 \tag{1.2}$$ $$x^2(2m^2 - 3m + 1) = 35 \tag{2.2}$$

Divide $$(1.2)$$ by $$(2.2)$$ $$\frac{x^2(m^2 + 1)}{x^2(2m^2 - 3m + 1)} = \frac{13}{35}$$ $$\frac{(m^2 + 1)}{(2m^2 - 3m + 1)} = \frac{13}{35}$$

Cross multiply. $$35(m^2 + 1) = 13(2m^2 - 3m + 1)$$ $$35m^2 + 35 = 26m^2 - 39m + 13$$ $$9m^2 + 39m + 22 = 0$$ $$9m^2 + 18m + 11m + 22 = 0$$ $$9m(m + 2) +11(m + 2) = 0$$ $$(9m + 11)(m + 2) = 0$$ $$m = -2 \text{ or } m = \frac{-11}{9}$$

When $$m = -2$$ Substitute $$m = -2$$ into $$1.2$$ $$x^2(4 + 1) = 13$$ $$5x^2 = 13$$ $$x^2 = \frac{13}{5}$$ $$x = \sqrt{\frac{13}{5}}$$ $$y = -2\sqrt{\frac{13}{5}}$$

When $$m = -\frac{11}{9}$$ Substitute $$m = -\frac{11}{9}$$ into $$1.2$$ $$x^2(\frac{121}{81} + 1) = 13$$ $$x^2(\frac{202}{81}) = 13$$ $$x^2 = \frac{13 * 81}{202} = \frac{1053}{202}$$ $$x = 2.283717322$$ $$y = \frac{-11}{9} * 2.283717322 = -2.790542727$$.

However, when I plug my values for $$x$$ and $$y$$ back into $$(1)$$ and $$(2)$$, I get $$13$$ and $$39.9$$ respectively.

No matter how many times I go over the equation, I just can't solve it satisfactorily, such that when I plug the values back into the equations, I get $$13$$ and $$35$$.

What mistakes did I make, and please write out the correct solution.

• There's a mistake when you are solving the equation 9m^2 + 39m+22=0 above. Spot it. Jul 28, 2017 at 11:55
• Gotcha @астонвіллаолофмэллбэрг. I've figured it out. My factorisation was wrong. These brain farts are humiliating. Jul 28, 2017 at 12:29
• Exactly. It's fine, it happens to everybody, and now that you have realized, it's inconsequential. Jul 28, 2017 at 16:15

I think the best way is the following: $$35(x^2+y^2)=13(x^2-3xy+2y^2),$$ which is without dividing.

We obtain $$22x^2+39xy+9y^2=0$$ or $$22x^2+6xy+33xy+9y^2=0$$ or $$2x(11x+3y)+3y(11x+2y)=0$$ or $$(11x+3y)(2x+3y)=0,$$ which gives $x=-\frac{3}{11}y$ or $x=-\frac{3}{2}y$ and we have the answer: $$\left\{\left(-\frac{3}{\sqrt{10}},\frac{11}{\sqrt{10}}\right),\left(\frac{3}{\sqrt{10}},-\frac{11}{\sqrt{10}}\right), (-3,2),(3,-2) \right\}.$$ The mistake in your solution is:

$x^2=9$ gives $x=3$ or $x=-3$.

• How did you get four solutions? I only got two solutions when I solved it below... Jul 28, 2017 at 13:11
• @Tobi Alafin What is exactly your question? About which solution do you say? I am ready to explain. Jul 28, 2017 at 13:13
• I made an error in my solution. Someone pointed out the error, so I rectified it below. I got only two solutions. I'm asking how you got four. (the factorisation you did above is magic to me, but that's by the way). Jul 28, 2017 at 13:14
• @Tobi Alafin I added something. See now please. Jul 28, 2017 at 13:19
• Thanks, I've seen it. Jul 28, 2017 at 13:32

Write the second equation as $(x^2+y^2)-3xy +y^2 = 35.$ Then substitute the first equation to get $13 -3xy +y^2 = 35$ which you can solve for $x$ to get

$$x=\frac{y^2-22}{3y}.$$

Plug this into the first equation:

$$\frac{y^4-44y^2+484}{9y^2} + y^2 = 13.$$

Clear fractions and gather like terms:

$$10y^4-161y^2+484 = 0.$$

This is quadratic in $y^2$ so you can use quadratic formula or just factor to get

$$(y^2-4)(10y^2-121) = 0.$$

So $y=\pm 2$ or $y=\pm \frac{11}{\sqrt{10}}.$ Then check for extraneous solutions.

There was an error in my earlier solution. I rectify it here:

## My Solution

$$x^2 + y^2 = 13 \tag{1}$$ $$x^2 - 3xy + 2y^2 = 35 \tag{2}$$
Let $$y = mx$$.
Rewrite $$(1)$$ and $$(2)$$.
$$x^2 + m^2x^2 = 13 \tag{1.1}$$ $$x^2 - 3mx^2 + 2m^2x^2 \tag{2.1}$$ Factorise the above equations. $$x^2(m^2 + 1) = 13 \tag{1.2}$$ $$x^2(2m^2 - 3m + 1) = 35 \tag{2.2}$$

Divide $$(1.2)$$ by $$(2.2)$$ $$\frac{x^2(m^2 + 1)}{x^2(2m^2 - 3m + 1)} = \frac{13}{35}$$ $$\frac{(m^2 + 1)}{(2m^2 - 3m + 1)} = \frac{13}{35}$$

Cross multiply. $$35(m^2 + 1) = 13(2m^2 - 3m + 1)$$ $$35m^2 + 35 = 26m^2 - 39m + 13$$ $$9m^2 + 39m + 22 = 0$$ $$x = \frac{-b \, \pm \sqrt{b^2 - 4ac}}{2a}$$ $$m = \frac{-39 \, \pm \sqrt{39^2 - 4{\times}9{\times}22}}{18}$$ $$m = \frac{-39 \, \pm 27}{18}$$ $$m = \frac{-39 + 27}{18} \text{ or } m = \frac{-39 - 27}{18}$$ $$m = \frac{-2}{3} \text{ or } m = \frac{-11}{3}$$

When $$m = \frac{-2}{3}$$ Substitute $$m = \frac{-2}{3}$$ into $$(1.2)$$ $$x^2(\frac{4}{9} + 1) = 13$$ $$x^2(\frac{13}{9} = 13$$ $$x^2 = \frac{13 * 9}{13}$$ $$x^2 = 9$$ $$x = \sqrt{9} = \pm 3$$ $$y = \frac{-2}{3} * \pm 3 = \pm 2$$

When $$m = -\frac{11}{3}$$ Substitute $$m = -\frac{11}{3}$$ into $$(1.2)$$ $$x^2(\frac{121}{9} + 1) = 13$$ $$x^2(\frac{130}{9}) = 13$$ $$x^2 = \frac{13 * 9}{130} = \frac{9}{10}$$ $$x = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$$ $$x = \frac{3}{\sqrt{10}} * \frac{\sqrt{10}}{\sqrt{10}} = \pm \frac{3\sqrt{10}}{10}$$ $$y = \frac{-11}{3} * \frac{3\sqrt{10}}{10} = \pm \frac{11\sqrt{10}}{10}$$.

$$\therefore (x,y) = (3, -2) \text{ or } (-3, 2) \text{ or } \left(\frac{3\sqrt{10}}{10}, \frac{-11\sqrt{10}}{10}\right) \text{ or } \left(\frac{-3\sqrt{10}}{10}, \frac{11\sqrt{10}}{10}\right)$$

Q.E.D

• $x^2=9$ gives $x=3$ or $x=-3$. There is a similar mistake else. Jul 28, 2017 at 13:23