Can u please help me with my math?

Question

I need help on this problem: 3(4c+3c)-6+-3
I'm in seventh grade and just startin this kinda stuff so can u please help me?:)

Answer 1

Okay. First, you use the distributive property on 3(4c+3c). First, you'd go 3x4c=12c, and then 3x3c=9c. Then, you'd add these together, as you always do when you use the distributive property this way, and then you'd have 12c+9c-6+-3. Then, adding your c terms, you'd get 21c-6+3. Then, adding your coefficients (numbers without variables attached to them, or letters attached to them), you'd have 21c-3. This is your simplest form, because you cannot add coefficients and numbers attached to variables together. This is because you have no idea what 21c equals. C could be any number, so this would be your simplest form, or your answer.
There you go. I'm glad to help. :)

Answer 2

3(4c+3c)-6+-3

=> (3*4c+3*3c)-6+-3 multiply what is inside the bracket by 3
=> 12c+9c-6+-3 now add "like" variables all c's & constants
=> 21c-9 ans

if the initial expression was equal to zero then your ans above is equal to zero.

21c-9=0 passing the nine to the other side changes the sign

therefore 21c=9
c=9/21
c=3/7 ans(2)

Answer 3

First, note that there are two terms with a c in parenthesis. Add those together, and get 7c. Now, note the two integers, -6 and -3. When you add two negative numbers together you create a bigger negative number, in this case, 9. Now you have 3(7c)-9, or 21c-9.

Remember, adding two negative numbers is like having a bill for some amount (which subtracts from your wallet), and then get another bill for some amount which subtracts even more. The total bill is the two added together normally, but then put a negative sign in front of the answer.

Answer 4

At the end, you don't need the addition sign, it can be written as
(- 6 - 3).

First: eliminate the parenthesis > multiply the outer term with the terms in the parenthesis....

(3)(4c) + (3)(3c) - 6 - 3

Sec: combine the terms....

12c + 9c - 9

Third: combine "like" terms - combine the coefficients with the same base (c-variable)....

21c - 9

Answer 5

you follow BEDMAS (brackets, edponents, divide, multiply, add, subtract)

first you work on the brackets (add 4c = 3c ) to get

3(7c) -6 +-3

then you multiply (3*7c) to get

21c -6 +-3

add like terms (6+-3)
21c - 9...

this is where i would stop unless youwere to solve for c then you take it to the next step

say the equation is equal to 0
21c-9=0

move the -9 to the opposite side of the equal sign changing the sign to a postive (you are adding 9 to each side -9 and 9 cancle each other out)

21c=9

then divide each side by 21 to get c by itself

21c/21 = 9/21( the 21s cancle each other out)

c= 9/21

Answer 6

you have to distribute out the 3(4c+3c) you can rewrite this as:
(3*4c)+(3*3c) = 12c+9c = 21c
then you still have the -6+(-3) = -9
combine these both together to get = 21c - 9
again it's:
3(4c+3c)-6+(-3) which is the same as:
=(3*4c)+(3*3c)-6-3
=12c+9c-9
then add like terms
=21c-9
Always remember to add the same like terms together and simplify it as much as possible

Answer 7

3(4c+3c)-6+-3
12c+9c-6+-3 distribute
21c-6+-3 combine like terms
21c-9 combine like terms

Answer 8

3(4c+3c) - 6 + -3 =
3(7c) - 6 + -3 =
(3x7c) - 6 + -3 =
21c - 6 + -3 =
21c - (6+ -3) =
21c - 3


how's that looks? understandable?

Answer 9

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Answer 10

multiply out brackets to give 12c + 9c then -6 + -3 is -9 so answer is

21c - 9