c+(4-3c)-2=0 Can someone help, step by step?

can someone explain to me STEP bySTEP how to solve this equation? thanks

you can write these equation
c+4-3c-2=0
-2c +2 = 0

add two side with (-2)
-2c + 2 +(-2) = 0+ (- 2)
-2c = -2 (multipl both side with (-1)
2c = 2 (then divide both with 2
2c/2 = 2/2
c = 1

You need to isolate the C
Step 1) c -3c + 4 -2 =0 (Note the parentheses are totally unnecessary here)
Step 2) -2c +2 =0 (Simplifying the equation)
Step 3) 2=2c (isolating the C variable)
Step 4) 1=c (solving for C)

i’m not sure but …
1) distributive property: c+ (1×4) +(1x-3c)+-2=0
2) combine like terms c+4+-3c+-2=0
3) add the opposite -2c+2-2=0-2
4) divide both sides of the equation by -2c=-2
5)c=1
hope that helped!

Source(s): Math skills

-3c+c = 2c
4-2c-2=0
Combind 4 and -2 for 2
-2c+2=0
-2 -2
-2c=-2
C=1

Source(s): Brain

thaats algebra (Y)
c+(4-3c)-2=0
c+4-3c-2=0 <- get rid of brackets -2c+2=0 <- add like terms .....-2..-2 <- subtract 2 from each side so your left with.. -2c=-2 divide them using fractions... -2c/-2 ..... -2/-2 the -2's cancel out leaving , c=1 NOW TO CHECK subsitute c with 1

Source(s): math class

Answer 6

c + (4 – 3c) – 2 = 0
c + (4 – 3c) = 2
4 – 3c = -c + 2
-4c = -2
c = 1/2