Use the method of substitution to solve the system

Use the method of substitution to solve the system

25Y^2-16X^2=400
9Y^2-4X^2=36

Thanks

Quote:

Originally Posted by Amy

Use the method of substitution to solve the system

25Y^2-16X^2=400
9Y^2-4X^2=36

Thanks

check your question. this system has no real solutions

Quote:

Originally Posted by Jhevon

check your question. this system has no real solutions

the question is correct. and you are right that the system has no real roots.

Quote:

Originally Posted by Amy

Use the method of substitution to solve the system

25Y^2-16X^2=400
9Y^2-4X^2=36

Quote:

Originally Posted by Amy

the question is correct. and you are right that the system has no real roots.

okie dokie. so we use the same method as if it had real roots, no surprises here.

$25y^2 - 16x^2 = 400$ ..................(1)

$9y^2 - 4x^2 = 36$ .......................(2)

Solving for $y^2$ from (2) we see that $y^2 = 4 + \frac {4}{9} x^2$
Substitute $4 + \frac {4}{9} x^2$ for $y^2$ in equation (1), we get:

$25 \left( 4 + \frac {4}{9} x^2 \right) - 16x^2 = 400$

$\Rightarrow 100 + \frac {100}{9} x^2 - 16x^2 = 400$

$\Rightarrow - \frac {44}{9} x^2 = 300$

$\Rightarrow x^2 = - \frac {675}{11}$

$\Rightarrow x = \pm \sqrt { \frac {675}{11}} ~i \approx 7.8335 ~i$

Now we simply substitue $\pm \sqrt { \frac {675}{11}} ~i$ for $x$ in either of the original equations to find $y$

You should get $y = \pm \sqrt { \frac {256}{11}} ~i \approx 4.8242 ~i$