SOLUTION: ( Evaluate each expression for the given values of the variables) 3(2c+d)-d; c=5 and d=1 3(2+5+1)-d I dont understand the second step if i do what is in the parenthese fir

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: ( Evaluate each expression for the given values of the variables) 3(2c+d)-d; c=5 and d=1 3(2+5+1)-d I dont understand the second step if i do what is in the parenthese fir      Log On


   

Question 7120: ( Evaluate each expression for the given values of the variables)
3(2c+d)-d; c=5 and d=1
3(2+5+1)-d

I dont understand the second step if i do what is in the parenthese first then distrubute or distrubute first? i just cant get past the second step.
Answer by glabow(165) About Me  (Show Source):
You can put this solution on YOUR website!
I don't undrstand your substitution of c=5 and d=1.
The expression
3(2c+d)-d says to multiply c by 2, then add d. This entire group is then multiplied by 3 and at the very last d is subtracted.
So,
3 ( 2x5 + 1) - 1 = 3 ( 10 + 1) - 1 = 3(11) - 1 = 33 - 1 = 32
Rule is:
Do the operations in all groups first
Do multiplication and division before addition and subtraction.