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Mathematics 39 Online
OpenStudy (anonymous):

c+(4-3c)-2=0 help me solve this plez

OpenStudy (lgbasallote):

get rid of the parenthesis \[c + 4 - 3c - 2 = 0\] combine like terms \[4 - 2c - 2 = 0\] add 2c to both sides [4 - 2 = 2c\] simplify

OpenStudy (lgbasallote):

\[4 - 2 = 2c\] simplify

OpenStudy (anonymous):

huh

OpenStudy (anonymous):

can u call me so u can expain it to me i still dont get it

OpenStudy (lgbasallote):

which part dont you get?

OpenStudy (anonymous):

adding 2c to both sides

mathslover (mathslover):

\[\large{c+4-3c-2=0}\] \[\large{c-3c+4-2=0}\] \[\large{c(1-3)+2=0}\] \[\large{c(-2)+2=0}\] \[\large{-2c+2=0}\] Did u get what i did yet?

OpenStudy (anonymous):

You Have to isolate the c

mathslover (mathslover):

@ME39 ... do reply ..

OpenStudy (lgbasallote):

\[4 - 2c + 2 = 0\] add 2c to both sides \[4 - 2c + 2 + 2c = 0 + 2c\] \[4 + 2 + 0 = 2c\] \[4 + 2 = 2c\] do you see it now?

OpenStudy (anonymous):

do i isolate the c start n from the beginning of the problem

OpenStudy (lgbasallote):

no you need to expand first

OpenStudy (lgbasallote):

by that i mean combining like terms

mathslover (mathslover):

\[\large{-2c+2=0}\] \[\large{-2c+2-2=0-2}\] \[\large{-2c=-2}\] \[\large{\frac{-2c}{-2}=\frac{-2}{-2}}\] \[\large{\frac{\cancel{-2}^{1}c}{\cancel{-2}^1}=\frac{\cancel{-2}^1}{\cancel{-2}^1}}\] \[\large{c=1}\]

OpenStudy (anonymous):

8508680042 is there anyway u can call me plz

mathslover (mathslover):

@lgbasallote u understood the quest. wrong i think

OpenStudy (anonymous):

yes i did

OpenStudy (lgbasallote):

yeah i mistyped sorry

OpenStudy (lgbasallote):

what i wrote first was right though lol

mathslover (mathslover):

No problem .. @ME39 Where r u having problems dude .. If you need an online tutor : you can instead go to khan academy videos

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