0
$\begingroup$

If I have

$$ 60 = 40/X + X $$

How do I remove $40$ and find $X$?

As I understand it, I have to multiply both sides by $X$. But if I do that then I'm left with

$$ \begin{eqnarray*} &60X = 40 + X &\Rightarrow \\ &60X - X = 40 &\Rightarrow \\ &X - X = 40 / 60 &\Rightarrow \\ & 1 = 2/3 & \end{eqnarray*} $$

And I think I ended at the wrong spot...

$\endgroup$
3
  • 2
    $\begingroup$ Minor slip. When you multiply both sides by $X$, you should end up with $60X=40+X^2$. You may have forgotten to multiply the $X$ at the right end by $X$. $\endgroup$ – André Nicolas Oct 17 '11 at 22:26
  • $\begingroup$ @Andre Nicolas - Do I have to do that to all other variables in the equation? for instance if that equation was instead 60 = 40 / X + x^2 + X it would come out to be 60X = 40 + X^3 + x^2? $\endgroup$ – ahodder Oct 17 '11 at 23:14
  • 2
    $\begingroup$ @Aedon, yes -- you do it for each term, whether or not they contain variables. It's not a special magical rule, it's just the ordinary $a(b+c)=ab+ac$ applied with $a=x$, $b=\frac{40}{x}$ and $c=x$. $\endgroup$ – hmakholm left over Monica Oct 17 '11 at 23:18
2
$\begingroup$

You know $X\ne 0$ so you can multiply by it. That will leave you a quadratic equation for $X$.

$\endgroup$
3
  • $\begingroup$ I'm not sure I follow, how will that leave me with a quadratic equation for X? $\endgroup$ – ahodder Oct 17 '11 at 22:22
  • $\begingroup$ Multiplying both sides by $X$ gives $60X=40+X^2\implies X^2-60X+40=0$. $\endgroup$ – yunone Oct 17 '11 at 22:27
  • $\begingroup$ Ok I see now, thank for the help :) $\endgroup$ – ahodder Oct 17 '11 at 22:30

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.