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my maths a a bit rusty and I need to extract a variable from a formula. It's needed for a project about air quality in order to convert data from sensors to an index.

The formula is :

$$\left (\frac{D-C}{B-A} \right ) \cdot (E-A)+C$$

And I need to extract $E$.

Would anyone be able to help me to extract it ?

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    $\begingroup$ Extract??? It needs to be an equation if you want to "extract" something out of it (i.e., have it "isolated" on one side of the equation). $\endgroup$
    – barak manos
    Aug 16, 2014 at 13:24
  • $\begingroup$ ygosteli Do you have the equation: $$\left(\frac{D-C}{B-A} \right )\cdot (E-A)+C=0$$ ?or something else? $\endgroup$
    – evinda
    Aug 16, 2014 at 13:34
  • $\begingroup$ Sorry, I didn't wrote entirely the equation, my fault... It was : (D−C/B−A)⋅(E−A)+C=X $\endgroup$
    – ygosteli
    Aug 16, 2014 at 19:04

2 Answers 2

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Ignoring what is on the right hand side, I prefer to suppose that $$\left (\frac{D-C}{B-A} \right ) \cdot (E-A)+C=X$$ and then, doing the same as Evinda did, arrive to $$E=\frac{(B C-A D)+(A-B) X}{C-D}$$ If $X=0$, as Evinda assumed, we have the same result.

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$$\left(\frac{D-C}{B-A} \right )\cdot (E-A)+C=0 \Rightarrow \left(\frac{D-C}{B-A} \right )\cdot (E-A)=-C \\ \Rightarrow E-A=-C \cdot \frac{B-A}{D-C}, D\neq C \\ \Rightarrow E=A-C \cdot \frac{B-A}{D-C} \\ \Rightarrow E=\frac{A(D-C)-CB+CA}{D-C}=\frac{AD-CA-CB+CA}{D-C}=\frac{AD-CB}{D-C}$$

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    $\begingroup$ $=0$ based on what? (you know what they say about assumption, right?) $\endgroup$
    – barak manos
    Aug 16, 2014 at 13:30
  • $\begingroup$ As you said,it should be a equation,and I supposed that he means $=0$..What do they say about assumption? :/ $\endgroup$
    – evinda
    Aug 16, 2014 at 13:31
  • $\begingroup$ Sorry, cannot elaborate on that due to website regulations... though, here is a link that might give you a rough idea (I hope I'm not breaking the regulations): youtube.com/watch?v=G-2NimrRPAQ $\endgroup$
    – barak manos
    Aug 16, 2014 at 13:38
  • $\begingroup$ @barakmanos Seems like the kind of movie I'd like. What's it called? Weird that I missed this for so long. $\endgroup$
    – Git Gud
    Aug 16, 2014 at 14:16
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    $\begingroup$ Thanks, my fault, I was meaning X but forgot to out the entire equation... As Claude said, your equation is also correct if X = 0 Thanks! $\endgroup$
    – ygosteli
    Aug 16, 2014 at 19:07

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