# Extract a variable from a formula

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 ?

• 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). Aug 16, 2014 at 13:24
• ygosteli Do you have the equation: $$\left(\frac{D-C}{B-A} \right )\cdot (E-A)+C=0$$ ?or something else? Aug 16, 2014 at 13:34
• Sorry, I didn't wrote entirely the equation, my fault... It was : (D−C/B−A)⋅(E−A)+C=X Aug 16, 2014 at 19:04

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.
$$\left(\frac{D-C}{B-A} \right )\cdot (E-A)+C=0 \Rightarrow \left(\frac{D-C}{B-A} \right )\cdot (E-A)=-C \\ \Rightarrow E-A=-C \cdot \frac{B-A}{D-C}, D\neq C \\ \Rightarrow E=A-C \cdot \frac{B-A}{D-C} \\ \Rightarrow E=\frac{A(D-C)-CB+CA}{D-C}=\frac{AD-CA-CB+CA}{D-C}=\frac{AD-CB}{D-C}$$
• $=0$ based on what? (you know what they say about assumption, right?) Aug 16, 2014 at 13:30
• As you said,it should be a equation,and I supposed that he means $=0$..What do they say about assumption? :/ Aug 16, 2014 at 13:31