Please solve this equation system:
x^2+xy=1/9
y^2+xy=1/16
Thank you!
First i tried to square everything, however could not continue to solve it after that.
Later on I tried to factorize the first two terms in both equations, but I could not do this. My third attempt was to divide every term my x and use the substitution method. The last try was to use the substitution method directly.
All my attempts gave an incorrect answer or I was not able to continue.
Can you please help me?
Thank you!
Rearrange equation 1:
$\displaystyle \displaystyle xy = \frac{1}{9} - x^2$
$\displaystyle \displaystyle xy = \frac{1 - 9x^2}{9}$
$\displaystyle \displaystyle y = \frac{1 - 9x^2}{9x}$.
Now substitute this into the second equation and solve for $\displaystyle \displaystyle x$...
$\displaystyle x^2+xy=\frac{1}{9}$ - - - - - - - - - - 1.
$\displaystyle y^2+xy=\frac{1}{16}$ - - - - - - - - - - 2.
From 1., we get $\displaystyle x(x+y)=\frac{1}{9}$
Therefore, $\displaystyle x+y=\frac{1}{9x}$
From 2., we get $\displaystyle y(x+y)=\frac{1}{16}$
Therefore, $\displaystyle x+y=\frac{1}{16y}$
So, $\displaystyle \frac{1}{9x}=\frac{1}{16y}$
$\displaystyle x:y=16:9$
Now, let's say that $\displaystyle x=16k$, $\displaystyle y=9k$.
From 1,. $\displaystyle 16k(16k+9k)=\frac{1}{9}$
$\displaystyle k^2=\frac{1}{9\times16\times25}$
$\displaystyle k=\pm\frac{1}{60}$
So, $\displaystyle x=\pm\frac{4}{15}$, $\displaystyle y=\pm\frac{3}{20}$.