A=1/3(a+b+c)?

Solve for c:


A.


c = 3A - (a - b)





B.


c = 3A - a - b





C.


c = A + 3a + 3b





D.


c = 3Aab

A=1/3(a+b+c)?
A = 1/3 (a + b + c)





Multiply both sides by 3:





3A = 3 × 1/3 (a + b + c)





3A = a + b + c





Subtract a and b from both sides:





3A - a - b = a - a + b - b + c





3A - a - b = c





Which is the same as:





c = 3A - a - b





Answer B
Reply:First: move 1/3 to the opposite side (use the opposite sign) %26gt; multiply the reciprocal of 1/3 (3/1) by both sides...





(A)(3/1) = [1/3(a + b + c)](3/1)





*Cross cancel "like" terms....





3A = a + b + c





SEc: subtract "a" from both sides...





3A - a = a - a + b + c


3A - a = b + c





Third: solve for "c" by isolating it on one side %26gt; subtract "b" from both sides...





3A - a - b = b - b + c


3A - a - b = c





c = 3A - a - b





B. is the correct solution
Reply:A
Reply:First note that 3A = a+b+c





A) 2b+c





B) c





C) 10a/3+10b/3+c/3





D) a^2bc + ab^a = a+b+c2c + abc
Reply:i really think it goes like this:





A = 1/3(a+b+c)


3A = a+b+c


3A - a - b = c =%26gt; answer
Reply:Given:A=1/3(a+b+c)





3A=a+b+c





3A-a=b+c





3A-a-b=c





terefore,c=3A-a-b


correct option is option no.B