Solve A=2BD+2BC+2DC for C. C=? This makes no sense to me! Another one is if n(A*B)=84 and n(A)=7, find n(B)

I don't know why these are so foreign to me, I think my brain is fried! I can't find a reference to them on the web, so thought I would ask here...thanks!

Solve A=2BD+2BC+2DC for C. C=? This makes no sense to me! Another one is if n(A*B)=84 and n(A)=7, find n(B)
A=2BD+2BC+2DC


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


C = (A - 2BD)/2(B +D)
Reply:Your goal is to isolate C using as many legal algebra steps as needed.





A = 2BD+2BC+2DC subtract 2BD from both sides





A-2BD = 2BC + 2DC factor 2C out of the right side





A-2BD = 2C (B+D) divide both sides by (B+D)





A-2BD / (B+D) = 2C divide both sides by 2





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





As to your second question:





n(A) * n(B) = n(A*B)





n(A) = 7, so n(7*B) = 84





n(B) = 84/7





n(B) = 12

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