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Example equation:

$$ \frac{a}{\frac{f+hce}{g}}-c=q-g$$

I want to solve for g.

If I want to move the g from one side to the other, and I therefore multiply the left side to get rid of the g under that annoying fraction, do I also have to multiply c by g as well? Then when I put it on the other side, I have to multiply the q by g, as well, don't I? Such that I get qg - gg? or qg - g^2?

Some of this seems like I should know this, but due to all the variables, its confusing me. Not to mention, I think I'm supposed to multiply the c by g on the left side when trying to move g over but that would basically mean I can't move g over to be exclusively on the right side. This is NOT the actual equation I'm trying to make sense of, it would be too complex and confusing. I have somewhat simplified it.

Or is this simply unsolvable for g without g being in the answer?

(I tried to format this correctly with the mathjax stuff but it wasn't showing up in the preview to confirm I had typed it correctly.

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  • $\begingroup$ this equation show now show properly $\endgroup$ – Mine Aug 10 '17 at 0:27
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There are at least three ways to do this, some much better than others:

1) Multiply one term by $1 = \frac gg$

$\frac{a}{\frac{f+hce}{g}}-c=q-g$

$\frac gg\frac{a}{\frac{f+hce}{g}}-c=q-g$

$\frac {ga}{f+hce} - c = q-g$.

You don't have to worry about multiplying other terms be cause you are only multiplying one turn by $1$.

You can develop a "sloth on a diving board" intuition that $\frac {x}{\frac yz} = \frac {zx}{y}$. (The sloth gets tired of hanging off the bottom of the diving board and swings up to the top.)

2) Multiply all times by $\frac {f+hce}g$.

$\frac{a}{\frac{f+hce}{g}}-c=q-g$

$\frac{f+hce}g(\frac{a}{\frac{f+hce}{g}}-c)=\frac {f+hce}g(q-g)$

$a -\frac{c(f+hce)}g = \frac {q(f+hce)}{g} - (f+hce)$

Then multiply again be $g$

$a -\frac{c(f+hce)}g = \frac {q(f+hce)}{g} - (f+hce)$

$g(a -\frac{c(f+hce)}g)= g(\frac {q(f+hce)}{g} - (f+hce))$

$ag - c(f+hce) = q(g+hce) - g(f + hce)$

3) Mutiply both sides by $\frac 1q$

$\frac{a}{\frac{f+hce}{g}}-c=q-g$

$\frac 1g(\frac{a}{\frac{f+hce}{g}}-c)=\frac 1g(q-g)$

$\frac {a}{f+hce} - \frac cg = \frac qg - 1$.

I really recommend 1).

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If you multiply the first term on the left hand side, by $g$, you also need to multiply $c$, $q$ and $g$ by $g$. You always need to multiply both sides of an equation by the same factor if you want it to stay the same. And since multiplication is distributive, you need to multiply every factor by $g$.

I think writing out the fraction in full form would make things easier:

$$ \frac{a}{f+\frac{hce}{g}}-c=q-g$$ What you can do, is multiply the left term by $1=g/g$: $$ \frac{g}{g}\cdot\frac{a}{f+\frac{hce}{g}}-c=q-g $$ $$ \frac{ag}{fg+hce}-c=q-g $$ then multiply the whole equation by $fg+hce$: $$ag-c\cdot (fg+hce) = q\cdot (fg+hce) -g\cdot (fg+hce)$$

and then solve the resulting quadratic equation.

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  • $\begingroup$ I have updated the look so my equation shows properly, its not what you thought. Sorry. I think your method may still help, but the quadratic you suggest still won't result in me getting what "g equals". $\endgroup$ – Mine Aug 10 '17 at 0:28
  • $\begingroup$ When you have a fraction in the denominator of a larger fraction, that means you are dividing the numerator (a) by that fraction ((f +hce)/g). Which is the same as multiplying by the reciprocal of the denominator. So your left-most term is equal to a * ( g / (f+hce) ). $\endgroup$ – Andrew Gray Aug 10 '17 at 2:07
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Updated with the new example: Yes. When you multiply by $g$, you multiply both sides of the equation by $g$. However in your example the $g$ on the right is essentially in the numerator because it is under two fraction bars. To start the simplification we multiply the whole fraction by $1=\frac gg$, doing $$\frac a{\frac {f+hce}g}-c=q-g\\ \frac {ag}{f+hce}-c=q-g\\ $$ Now we have a linear equation in $g$ so we can collect the terms. $$g\left(1+\frac a{f+hce}\right)=c+q\\ g=\frac {c+q}{1+\frac a{f+hce}}$$ If you want, you can clear the compound fraction $$g=\frac {(c+q)(f+hce)}{f+hce+a}$$

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  • $\begingroup$ sorry, i didn't show that right, trying to update it to show the fraction properly $\endgroup$ – Mine Aug 10 '17 at 0:22
  • $\begingroup$ The basic idea is the same. The $g$ is still in the numerator. If I guess properly where you are going, you will get a linear equation in $g$ because the first term will be $\frac {ag}{f+hce}$. I have to go now. $\endgroup$ – Ross Millikan Aug 10 '17 at 0:25

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