Solve the following system of non-linear equations: $$\begin{align}x^2+4xy+y^2&=13\\ 2x^2+3xy&=8.\end{align}$$
I started off this problem by rearranging and substituting but I’m stuck. I think that I have to change this system to a simple equation in one variable, but I don't know how to do it, please help me.
On
Hint: $$y=\frac{8-2x^2}{3x},\quad x\neq 0 \tag 1$$
for the second equation. After substitute into first equation $$x^2+4xy+y^2=13$$ this value of $y$ you will found these solutions:
$$x=i\frac{8\sqrt{11}}{11},\:x=-i\frac{8\sqrt{11}}{11},\:x=1,\:x=-1$$ Take these solutions and substitute them on $2x^2+3xy=8$ to find the $y$ values.
On
As @Michael Hoppe stated in his answer, if we multiply the first equation by $−8$ and the second one by $13$, we can add the two equations and factor to get $(9x+8y)(2x-y) = 0$. We now have that $$2x = y$$ $$\text{or}$$ $$9x = -8y.$$
Quick side note: When you can't go any further try multiplying each equation by a certain number so that when you add or subtract, you cancel out the constant. Then you have a chance of factoring.
Plugging into the second equation gets us $2x^2+6x^2 = 8$, so $8x^2 = 8$ and $x = \pm 1$. Substituting each again into the second equation gets us two solution pairs, $(1, 2)$ and $(-1, -2)$.
Now we plug in $y = -\frac{9}{8}x$ and get $2x^2-\frac{27}{8}x^2 = 8$, so $-\frac{11}{8}x^2 = 8$. $x^2 = -\frac{64}{11}$, so $x = \pm i\cdot \frac{8}{\sqrt{11}}$. We get two more solution pairs, $(\frac{8i}{\sqrt{11}}, -\frac{9i}{\sqrt{11}})$ and $(-\frac{8i}{\sqrt{11}}, \frac{9i}{\sqrt{11}})$. (Of course, you can rationalize the denominator on your own.)
Final answer: $(1, 2), (-1, -2), (\frac{8i}{\sqrt{11}}, -\frac{9i}{\sqrt{11}}), (-\frac{8i}{\sqrt{11}}, \frac{9i}{\sqrt{11}})$.
-FruDe
Get rid of $13$ and $8$ by multiplying the first equation by $-8$ and the second one by $13$. Add both. Factorizing gives $(9x+8y)(2x-y)=0$.