Okay, I realize this seems like a really stupid question, but on a math contest (without calculators) I got down to this equation:
$$\frac{26}{672-x} + \frac{24}{372-x} = \frac{50}{480-x}$$
whereupon I spent 15 minutes trying to solve for $x$ until I eventually gave up. Especially, on a math contest, this is not an ideal situation. $x$ ended up being 48, but I don't recall how to do this type of algebra without a calculator. For those curious, the problem appeared on the AMC 12A.
$$\frac{26}{672-x} + \frac{24}{372-x} = \frac{50}{480-x}$$
We could just go ahead and start solving this, but it will make the terms smaller if we let $2u=372-x$ $$\frac{26}{300+2u} + \frac{24}{2u} = \frac{50}{2u+108}$$ factoring a two out of top and bottom, we get $$\frac{13}{u+150} + \frac{12}{u} = \frac{25}{u+54}$$
We then want to get common fraction on the left
$$\frac{13u(u+54)}{u(u+150)(u+54)} + \frac{12(u+150)(u+54)}{u(u+150)(u+54)}-\frac{25u(u+150)}{u(u+150)(u+54)}=0$$ $$\frac{13u(u+54)+12(u+150)(u+54)-25u(u+150)}{u(u+150)}=0$$ $$13u(u+54)+12(u+150)(u+54)-25u(u+150)=0$$
$$13u^2+(13)(54)u+(12u+(12)(150))(u+54)-25u^2-(150)(25)u=0$$ $$13u^2+(13)(54)u+12u^2+(12)(150)u+(54)(12)u+(54)(150)(12)-25u^2-(150)(25)u=0$$ $$(13)(54)u+(12)(150)u+(54)(12)u-(150)(25)u=-(54)(150)(12)$$
Now comes the mental math part; $$u[(13)(54)+(12)(150)+(54)(12)-(150)(25)]=-(54)(150)(12)$$
Can you get it from here?