A = 2BD + 2BC + 2DC for C

Solve it for C?

Ok, so A = 2BD + 2BC + 2DC

First, by trying to isolate the C alone on one side, we get rid of any common terms that aren't C. All terms on the right have a 2, so we factor that 2 out: A = 2BD + 2BC + 2DC

--> A = 2 (BD + BC + DC)

Next divide both sides by that 2:

A/2 = 2 (BD+BC+DC) / 2 - - > A/2 = BD + BC + DC

Now, we can subtract the BD from the right since it contains no C. Remember again that in "solving for" a variable, that really translates into "isolate it on one side":

(A/2) - BD = BD - BD + BC + DC

--> (A/2) - BD = BC + DC

We're almost there! So on the right we can finally factor out our C, in which [BC+DC] becomes [C (B+D) ]. Let's put it all together:

(A/2) - BD = BC + DC - - > (A/2) - BD = C (B+D)

Now we can divide out the (B+D) to get our C alone: [ (A/2) - BD]: (B+D) = C (B+D) : (B+D)

--> [ (A/2) - BD] / (B+D) = C

Start by subtracting 2BD from both sides:

A-2BD=2BC+2DC

Factor the C out of the equation:

A-2BD=C (2B+2D)

Divide both sides of the equation by 2B+2D:

(A-2BD) / (2B+2D) = C