How do you solve #6( 4+ c ) = 5( c - 5)#?

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Answer 1 · 2018-03-16T03:53:28

#c=-49#

Well, to start off, you would first use the distributive property to get rid of the parentheses from both sides.

#6(4+c) = 5(c-5)#

becomes

#24+6c = 5c -25#

Next, you would want to isolate the variable by getting all of them to one side of the equal sign and the non-variables on the other.

So from

#24+6c = 5c-25#

you would subtract #5c# from both sides and #24# from both sides

#24-24 +6c -5c = 5c-5c -25 -24#

The #24#'s cancel out and so do the #5c#'s. You are then left with

#6c-5c = -25-24#

Subtract #5c# from #6c#, add #-24# and #-25# together to get #-49#, and that is your answer!

#c = -49#