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This may be too basic a question for this site, in which case I'm sure you'll all let me know.

Here is my problem: I have a job worth $\$$40,000 composed of 6 units of x and 3 units of y. I'm trying to determine how much to charge for x and y, where y is worth 30% of x.

I'd really like to be able to solve these sorts of questions myself in the future, but my recollection of algebra is far too hazy.

Could you work it out longhand for me so I can see how to do it?

Preemptive edit: this is not a school question. I'm putting together a SOW for a video job and the 30%(y) is the cost of a French translated version of the video, which is about 30% of the effort since only the audio track changes.

2 Answers 2

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You have: $6x + 3y = 40000$ and $y = 0.3x$ Plugging in the second into the first, we have $\begin{align*} 6x + 3(0.3x) &= 40000\\ 6x + .9x &=40000\\ 6.9x &= 40000\\ x &= \frac{40000}{6.9} \end{align*}$ So: $\begin{align*} x &= \frac{40000}{6.9} \approx 5797.10\\ y &= 0.3x = 0.3\left(\frac{40000}{6.9}\right) \approx 1739.13 \end{align*}$

If each $x$ is $\$5797.10 and each $y$ is $\$1739.13, then a job with six $x$ and 3 $y$ would be $6(5797.10) + 3(1739.13) = 34782.60 + 5217.39 = 39999.99$ with the missing cent due to rounding errors.

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    Awesome. Thanks - I'm embarrassed that the substitution of y for 0.3x didn't occur to me, but glad sites like this exist to set me straight!2012-02-06
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Your job is worth 40000, composed of 6 units of $x$ and 3 units of $y$:

$40,000 = 6x + 3y$

Also, $y$ is worth 30% of $x$:

$y = 0.3 x$

Substituting the second equation into the first equation:

$40,000 = 6x + 3\times 0.3x = 6x + 0.9x = 6.9x$

Therefore

$x = 40,000 / 6.9 = 5791.1$

which means that $y$ is

$y = 0.3x = 0.3 \times 5797.1 = 1739.13$

and you're done. As a final check:

$6x + 3y = 6 \times 5797.1 + 3\times 1739.13 = 34 782.6 + 5217.39 = 39,999.99$

(the slight discrepancy is a rounding error - I rounded all answers to two decimal places because they were expressed in dollars. If you wanted to avoid this error, you could keep more decimal places in your intermediate calculations).

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    Thanks for the detailed answer! +12012-02-06