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I have the equation $$x = \frac{y+y}{\frac{y}{70} + \frac{y}{90}} $$ and I need to solve for x. My calculator has already shown me that it's not necessary to know y to solve this equation, but I can't seem to figure it out. This is how I try to solve it: $$ x = \frac{y+y}{\frac{y}{70} + \frac{y}{90}} = 2y\left(\frac{70}{y} + \frac{90}{y}\right) = 2y\left(\frac{90+70}{y}\right) = 2y\cdot\frac{160}{y} = \frac{320y}{y} = 320 $$ But according to my calculator, this is not correct. The answer should be 78.75, but I don't know why. Any help would be much appreciated.

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  • $\begingroup$ Format in latex form, it's hard to read fractions in ASCII. $\endgroup$
    – UserX
    Commented Aug 20, 2014 at 23:43
  • $\begingroup$ You went wrong when you did $$\frac{1}{\frac{y}{70}+\frac{y}{90}}=\frac{70}{y}+\frac{90}{y}$$ You cannot take the reciprocals like that. $\endgroup$ Commented Aug 20, 2014 at 23:44
  • $\begingroup$ @user148432 fixed, apologies $\endgroup$
    – bream
    Commented Aug 20, 2014 at 23:47
  • $\begingroup$ Also, no apologizing for "simple" questions. This isn't about measuring your skill level: it's about increasing it. $\endgroup$ Commented Aug 20, 2014 at 23:49
  • $\begingroup$ I think your error is at the third equality, it should be $x=\frac{2y}{\frac{(70+90)y}{70*90}}$. $\endgroup$
    – UserX
    Commented Aug 20, 2014 at 23:50

3 Answers 3

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When manipulating expressions, the most important thing is to think small. Find the smallest part of the expression you can do something with. In this case,

$$\frac{2y}{\frac{y}{70}+\frac{y}{90}}$$

has as the smallest available task add the fractions on the bottom, which gives (only showing that part)

$$\frac{y}{70}+\frac{y}{90}=\frac{90y+70y}{70\cdot 90}=\frac{160y}{6300}=\frac{8y}{315}$$

Putting that back in, we now have

$$\frac{2y}{\left(\frac{8y}{315}\right)}=2y\left(\frac{315}{8y}\right)=\frac{630y}{8y}=\frac{315}{4} \text{for } y\ne 0$$

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  • $\begingroup$ Thank you! I understand now how to solve the problem, but I'm just curious as to why the nested fraction in your method $\frac {1}{\frac{8y}{315}}$ can be turned into $\frac {315}{8y}$ but $\frac {1}{\frac{y}{70} + \frac {y}{90}}$cannot be turned into $\frac {70}{y} + \frac {90}{y}$. $\endgroup$
    – bream
    Commented Aug 21, 2014 at 21:01
  • $\begingroup$ The addition gets in the way; you have to take the reciprocal of the whole thing. Remember: the reciprocal of $2+3$ is not $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$, but $\frac{1}{5}$. $\endgroup$ Commented Aug 21, 2014 at 21:06
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Dividing numerator and denominator by $y$ (assuming of course that $y \ne 0$),and multiplying them by $70$ and $90$, $$ x = \dfrac{2}{1/70 + 1/90} = \dfrac{2 \times 70 \times 90}{90 + 70} = \dfrac{12600}{160} = \dfrac{315}{4}$$

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This is my solution:

$$ x=\frac{y+y}{\frac{y}{70}+\frac{y}{90}}={y}\cdot\frac{2}{{y}\cdot\left(\frac{1}{70}+\frac{1}{90}\right)}=\frac{2}{\frac{1}{70}+\frac{1}{90}}=\frac{2\cdot6300}{90+70}=12600/160=\boxed{78.75}$$

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