# Use the method of substitution to solve the system

• Jun 18th 2007, 10:10 PM
Amy
Use the method of substitution to solve the system
Use the method of substitution to solve the system
1. 25Y^2-16X^2=400
9Y^2-4X^2=36

Thanks
• Jun 18th 2007, 10:24 PM
Jhevon
Quote:

Originally Posted by Amy
Use the method of substitution to solve the system
1. 25Y^2-16X^2=400
9Y^2-4X^2=36

Thanks

check your question. this system has no real solutions
• Jun 18th 2007, 11:12 PM
Amy
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.
• Jun 18th 2007, 11:27 PM
Jhevon
Quote:

Originally Posted by Amy
Use the method of substitution to solve the system
1. 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 $y = \pm \sqrt { \frac {256}{11}} ~i \approx 4.8242 ~i$