# Thread: Solve for (x; y) in the system

1. ## Solve for (x; y) in the system

can anybody please Solve for (x; y) in the system?

2. Equation 1 - Equation 2:

$\displaystyle \displaystyle [(e^x + 2)^2 - y] - [4(e^x + 2) - y] = 3 - (-1)$

$\displaystyle \displaystyle (e^x + 2)^2 - 4(e^x + 2) = 4$

$\displaystyle \displaystyle (e^x + 2)^2 - 4(e^x + 2) - 4 = 0$.

Now let $\displaystyle \displaystyle X = e^x + 2$ so that you get the quadratic

$\displaystyle \displaystyle X^2 - 4X - 4 = 0$

$\displaystyle \displaystyle X^2 - 4X + (-2)^2 - (-2)^2 - 4 = 0$

$\displaystyle \displaystyle (X-2)^2 - 8 = 0$

$\displaystyle \displaystyle (X-2)^2 = 8$

$\displaystyle \displaystyle X-2 = \pm 2\sqrt{2}$

$\displaystyle \displaystyle X = 2\pm 2\sqrt{2}$

$\displaystyle \displaystyle e^x + 2 = 2\pm 2\sqrt{2}$

$\displaystyle \displaystyle e^x = \pm 2\sqrt{2}$.

Since $\displaystyle \displaystyle e^x > 0$ for all $\displaystyle \displaystyle x$

$\displaystyle \displaystyle e^x = 2\sqrt{2}$

$\displaystyle \displaystyle x = \ln{(2\sqrt{2})}$

$\displaystyle \displaystyle x = \ln{(2^{\frac{3}{2}})}$

$\displaystyle \displaystyle x = \frac{3}{2}\ln{2}$.

Now back-substitute to find $\displaystyle \displaystyle y$.

3. amazing! the best!
y = 9 + 8 sq rt. 2

thanks