How do you solve $(c + 3) - 2 c - (1 - 3 c) = 2$ $(c+3)-2c-(1-3c)=2$ ?

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Answer 1 · 2018-05-13T05:26:04

$c = 0$ $c=0$

1) Remove brackets as they are the same operation (addition / subtraction)

$c + 3 - 2 c - 1 - 3 c = 2$ $c+3-2c-1-3c=2$

2) Simplify the coefficient of c

$- 4 c + 3 - 1 = 2$ $-4c+3-1=2$

3) Simplify the number

$- 4 c + 2 = 2$ $-4c+2=2$

4) Subtract two from both sides

$- 4 c = 0$ $-4c=0$

5) Divide both sides by -4

$c = 0$ $c=0$

How do you solve (c+3)-2c-(1-3c)=2(c+3)−2c−(1−3c)=2(c+3)-2c-(1-3c)=2?