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.

  • 2
    $\begingroup$ I'd probably let $2u=372-x$. Then the equation is:$$\frac{13}{150+u} + \frac{12}{u} =\frac{25}{54+u}$$ That might be easier to solve. $\endgroup$ – Thomas Andrews Feb 5 '16 at 2:43
  • 1
    $\begingroup$ It becomes a linear equation because 13+12=25, so the quadratic terms cancel. $\endgroup$ – Thomas Andrews Feb 5 '16 at 2:48
  • $\begingroup$ $x=48$ of course $\endgroup$ – GEdgar Feb 5 '16 at 2:52

$$\frac{26}{672-x} + \frac{24}{372-x} = \frac{50}{480-x}$$
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?

  • $\begingroup$ Okay, thank you so much. I can't believe I forgot the basic algebra skills. Hopefully I'll remember them on the AMC 12 next week! Thanks again! $\endgroup$ – Junlin Yi Feb 10 '16 at 2:41
  • $\begingroup$ @JunlinYi Sure thing, glad I could help! Just remember on these problems to always get rid of fractions if you can... it's not the fastest strategy in most cases (there are faster ways to solve the problem than I showed), but it works pretty much every time. The substitution is also a bit unnecessary, and serves only to help with the mental math. If you have time to multiply the 3-4 digit number by hand then finding a substitution may not be worth it. If you do, remember to plug back in the solution for $u$ (or whatever variable you chose) to get an answer in terms of the original variable! $\endgroup$ – Brevan Ellefsen Feb 10 '16 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.