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I realize this rather rudimentary but it has been over a decade since my algebra classes and now I have problem that I can't figure out. I would like someone to walk me through the steps in solving the "X" in this problem: $$ 1.20 \cdot 10^6 = \frac{1}{\left(\dfrac{1}{6 \cdot 10^6}\right) X} $$ So what I've done so far is this:

$$1,200,000 = \frac{1}{166,666.67X}$$

$$(166,666.67X) * 1,200,000 = \frac{1}{166,666.67X} * (166,666.67X)$$

$$200,000,004,000x = 1$$ But this doesn't make sense.

I supposed to get $x = 5$.

Thanks!

EDIT: The formula is this:

                   1  
MTTF(system) = ---------------
                 n     1
                 E ----------
                i=1   MTTFi

This is my best as I can't upload the image and not sure how else to post this. So the "E" is actually the greek letter for sumation. The "n" would be 1 since this only one system.

MTTF(system) = 1,200,000;
MTTFi = 6,000,000
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  • $\begingroup$ i think the bottom of your fraction on the RHS should be 1/(6*10^6) not (1/6)*10^6. $\endgroup$ – SomethingWitty Feb 7 '13 at 15:03
  • $\begingroup$ Yes, @something witty, that's correct -that's what was written by the OP, I've edited to use the parentheses the OP used. $\endgroup$ – amWhy Feb 7 '13 at 15:08
  • $\begingroup$ @SomethingWitty The original RHS was 1 / [(1/6 * 10^6) * X]. I took the literal interpretation, guided by the subsequent calculations by OP, and edited accordingly. But (see my answer), it was definitely meant to be 1 / [1/(6 * 10^6) * X], you are right. $\endgroup$ – Andreas Caranti Feb 7 '13 at 15:14
  • $\begingroup$ I wanted to post the formula but this site is not letting me upload images. $\endgroup$ – user6722 Feb 7 '13 at 15:16
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    $\begingroup$ ALL: I'll check out the LaTeX but in the mean time I've added the formula I was using. And thanks for all your input!!! $\endgroup$ – user6722 Feb 7 '13 at 15:30
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$$1.20 \cdot 10^6 = \dfrac{1}{\left(\frac{1}{6\cdot10^6}\right)X} \iff 1.2\cdot 10^6 = \frac{6\cdot 10^6}{X}$$ $$\iff X =\frac{6\cdot 10^6}{1.2\cdot 10^6} = \frac {6}{1.2} = 5$$

The nice thing about the problem is that there is no need to evaluate numerically until the very end, where we need only compute $X =\dfrac{6}{1.2}$ (factors of $10^6$ cancel out, in the process)

Another way to think of the problem is as follows (multiplying both sides of the equation by $\dfrac{1}{6\cdot 10^6}$ to cancel out the fraction in the denominator of the right-hand side: $$1.20 \cdot 10^6 = \dfrac{1}{\left(\frac{1}{6\cdot10^6}\right)X} \iff 1.2\cdot 10^6 \cdot \frac{1}{6\cdot 10^6} = \frac{1}{6\cdot 10^6} \cdot \frac{1}{\frac{1}{6\cdot 10^6}X}$$ $$\iff \frac{1.2}{6} = \frac{1}{X} \iff 1.2x = 6\iff X = \frac{6}{1.2} = 5 $$

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  • $\begingroup$ Thank you, Thank you, Thank you $\endgroup$ – user6722 Feb 7 '13 at 15:37
  • $\begingroup$ user6722 You're welcome! $\endgroup$ – amWhy Feb 7 '13 at 15:39
  • $\begingroup$ $\star {}{}{}{}{}$ $\endgroup$ – Mikasa Feb 25 '13 at 13:33
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You can make an exact calculation, without resorting to approximations. Multiply both sides by $\dfrac{1}{6} \cdot 10^6$ to get $$ 1 = 1.20 \cdot \dfrac{1}{6} \cdot 10^{12} X = \frac{1}{5} \cdot 10^{12} X. $$ This gives $X = \dfrac{5}{10^{12}}$.

Since you expect the solution to be $X = 5$, I suppose there is a misprint in your statement, and one of the $10^6$ is actually a $10^{-6}$.

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$1.2 \times 10^6 = \displaystyle \frac{1}{ \frac{1}{6} \times 10^6 x}$

$\Rightarrow (1.2 \times 10^6) \times( \frac{1}{6} \times 10^6 x) = 1$

$\Rightarrow (1.2 \times \frac{1}{6}) \times 10^{12} \times x = 1$

$\Rightarrow (\frac{1}{5} \times 10^{12}) \times x = 1$

$\Rightarrow x = 5 \times 10^{-12}$

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