Use the method of substitution to solve the system
9Y^2-4X^2=36
- 25Y^2-16X^2=400
Thanks
okie dokie. so we use the same method as if it had real roots, no surprises here.
$\displaystyle 25y^2 - 16x^2 = 400$ ..................(1)
$\displaystyle 9y^2 - 4x^2 = 36$ .......................(2)
Solving for $\displaystyle y^2$ from (2) we see that $\displaystyle y^2 = 4 + \frac {4}{9} x^2$
Substitute $\displaystyle 4 + \frac {4}{9} x^2$ for $\displaystyle y^2$ in equation (1), we get:
$\displaystyle 25 \left( 4 + \frac {4}{9} x^2 \right) - 16x^2 = 400$
$\displaystyle \Rightarrow 100 + \frac {100}{9} x^2 - 16x^2 = 400$
$\displaystyle \Rightarrow - \frac {44}{9} x^2 = 300$
$\displaystyle \Rightarrow x^2 = - \frac {675}{11}$
$\displaystyle \Rightarrow x = \pm \sqrt { \frac {675}{11}} ~i \approx 7.8335 ~i$
Now we simply substitue $\displaystyle \pm \sqrt { \frac {675}{11}} ~i$ for $\displaystyle x$ in either of the original equations to find $\displaystyle y$
You should get $\displaystyle y = \pm \sqrt { \frac {256}{11}} ~i \approx 4.8242 ~i$