How do you solve #-x - 29= 13+ 2x#?

1 Answer
Mar 11, 2017

#x=-14#

Explanation:

Your goal here is to isolate #x# on one side of the equation. Notice that you have #x# terms on both sides of the equation, so the first thing to do here will be to get both of them on the same side of the equal sign.

One way to do that is to add #x# to both sides

#- color(red)(cancel(color(black)(x))) + color(red)(cancel(color(black)(x))) - 29 = 13 + 2x + x#

This will get you

#-29 = 13 + 3x#

Next, subtract #13# from both sides

#-29 - 13 = color(red)(cancel(color(black)(13))) + 3x - 13#

to get

#-42 = 3x#

Finally, divide both sides by #3# to get #x# by itself

#(-42)/3 = (color(red)(cancel(color(black)(3)))x)/color(red)(cancel(color(black)(3)))#

You will thus have

#-14 = x#

This is equivalent to saying that

#x = -14#

Notice that you can get the same result by going

#-x - 2x - 29 = 13 + color(red)(cancel(color(black)(2x))) - color(red)(cancel(color(black)(2x)))#

#-3x - 29 = 13#

followed by

#-3x - color(red)(cancel(color(black)(29))) + color(red)(cancel(color(black)(29))) = 13 + 29#

#-3x = 42#

This time, divide both sides by #-3# to get

#(color(red)(cancel(color(black)(-3)))x)/(color(red)(cancel(color(black)(-3)))) = 42/(-3)#

#x = -14#