1. ## Solve for x

Solve for x.

(y/x) + p(xy) = (q/s)

2. Originally Posted by magentarita
Solve for x.

(y/x) + p(xy) = (q/s)

Hello again. Another monster, I see.

$\displaystyle \frac{y}{x}+p(xy)=\frac{q}{s}$

First, multiply through by your LCD which is xs.

$\displaystyle sy+psyx^2=qx$

$\displaystyle (psy)x^2-qx=-sy$

Now on to completing the square. Divide all terms by (psy)

$\displaystyle x^2-\frac{q}{psy}x=\frac{-sy}{psy}$

$\displaystyle x^2-\frac{q}{psy}x+\left(\frac{q}{2psy}\right)^2=-\frac{1}{p}+\left(\frac{q}{2psy}\right)^2$

$\displaystyle \left(x-\frac{q}{2psy}\right)^2=\frac{q^2}{4p^2s^2y^2}-\frac{1}{p}$

$\displaystyle \left(x-\frac{q}{2psy}\right)^2=\frac{q^2-4ps^2y^2}{4p^2s^2y^2}$

$\displaystyle x-\frac{q}{2psy}=\pm\sqrt{\frac{q^2-4ps^2y^2}{4p^2s^2y^2}}$

$\displaystyle x=\pm\sqrt{\frac{q^2-4ps^2y^2}{4p^2s^2y^2}}+\frac{q}{2psy}$

$\displaystyle x=\frac{\pm\sqrt{q^2-4ps^2y^2}}{2psy}+\frac{q}{2psy}$

$\displaystyle x=\frac{\pm\sqrt{q^2-4ps^2y^2}+q}{2psy}$

Well, that's my solution and I'm stickin' with it.

3. ## wow..........

Originally Posted by masters
Hello again. Another monster, I see.

$\displaystyle \frac{y}{x}+p(xy)=\frac{q}{s}$

First, multiply through by your LCD which is xs.

$\displaystyle sy+psyx^2=qx$

$\displaystyle (psy)x^2-qx=-sy$

Now on to completing the square. Divide all terms by (psy)

$\displaystyle x^2-\frac{q}{psy}x=\frac{-sy}{psy}$

$\displaystyle x^2-\frac{q}{psy}x+\left(\frac{q}{2psy}\right)^2=-\frac{1}{p}+\left(\frac{q}{2psy}\right)^2$

$\displaystyle \left(x-\frac{q}{2psy}\right)^2=\frac{q^2}{4p^2s^2y^2}-\frac{1}{p}$

$\displaystyle \left(x-\frac{q}{2psy}\right)^2=\frac{q^2-4ps^2y^2}{4p^2s^2y^2}$

$\displaystyle x-\frac{q}{2psy}=\pm\sqrt{\frac{q^2-4ps^2y^2}{4p^2s^2y^2}}$

$\displaystyle x=\pm\sqrt{\frac{q^2-4ps^2y^2}{4p^2s^2y^2}}+\frac{q}{2psy}$

$\displaystyle x=\frac{\pm\sqrt{q^2-4ps^2y^2}}{2psy}+\frac{q}{2psy}$

$\displaystyle x=\frac{\pm\sqrt{q^2-4ps^2y^2}+q}{2psy}$

Well, that's my solution and I'm stickin' with it.
You really went all the way this tme around.