# Thread: Solve an equation system

1. ## Solve an equation system

x^2+xy=1/9
y^2+xy=1/16
Thank you!

2. What ideas have you had so far?

3. 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.
Thank you!

4. Rearrange equation 1:

$\displaystyle xy = \frac{1}{9} - x^2$

$\displaystyle xy = \frac{1 - 9x^2}{9}$

$\displaystyle y = \frac{1 - 9x^2}{9x}$.

Now substitute this into the second equation and solve for $\displaystyle x$...

5. Thank you prove it!
I substituted the value into the equation, however I did not get the correct answer. I got x=4/21, but the answer should be 4/15 or -4/15. Can you please show me how to solve it?

6. $x^2+xy=\frac{1}{9}$ - - - - - - - - - - 1.

$y^2+xy=\frac{1}{16}$ - - - - - - - - - - 2.

From 1., we get $x(x+y)=\frac{1}{9}$

Therefore, $x+y=\frac{1}{9x}$

From 2., we get $y(x+y)=\frac{1}{16}$

Therefore, $x+y=\frac{1}{16y}$

So, $\frac{1}{9x}=\frac{1}{16y}$

$x:y=16:9$

Now, let's say that $x=16k$, $y=9k$.

From 1,. $16k(16k+9k)=\frac{1}{9}$

$k^2=\frac{1}{9\times16\times25}$

$k=\pm\frac{1}{60}$

So, $x=\pm\frac{4}{15}$, $y=\pm\frac{3}{20}$.

7. Thank you very much lanierms!
I understand how to solve it now.

8. I hope you do. However, you should understand that you do not yet have the "solution". The problem does not ask for separate values of x and y but for pairs that satisfy the equation.

9. I do not understand HallsofIvy. Can you please explain. When I look in the key, it shows x and y values as the answers.