# Thread: Use the method of substitution to solve the system

1. ## 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

2. 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

3. 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.

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

$\displaystyle 25y^2 - 16x^2 = 400$ ..................(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 $\displaystyle y = \pm \sqrt { \frac {256}{11}} ~i \approx 4.8242 ~i$