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
