How do you solve #-(x+2)-2x=-2(x+1)#?

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Answer 1 · 2017-02-14T16:27:32

#x=0#

Firstly, you need to get the #(x+2)# on its own, so that means #+ 2x# on both sides, leaving you with

#-x - 2 = -2 (x + 1) + 2x#

which can be simplified by expanding the brackets then simplifying

#-x - 2 = -2#

which in turn can be also simplified by adding 2 to both sides, giving you

#-x = 0#

which is the same as

#x = 0#

Hopefully this was helpful...

Answer 2 · 2017-02-14T16:31:21

See the entire solution process below:

First, expand the terms within parenthesis. Take special care to ensure you manage the signs of the individual terms correctly:

#-x - 2 - 2x = (-2 xx x) - (2 xx 1)#

#-x - 2 - 2x = -2x - 2#

We can next group and combine like terms on the left side of the equation:

#-x - 2x - 2 = -2x - 2#

#-3x - 2 = -2x - 2#

Now, we can add #color(red)(3x)# and #color(blue)(2)# to each side of the equation to solve for #x# while keeping the equation balanced:

#color(red)(3x) - 3x - 2 + color(blue)(2) = color(red)(3x) - 2x - 2 + color(blue)(2)#

#0 - 0 = 1x - 0#

#0 = x#

#x = 0#