How do you solve #5x - 3+ x = 3x - 2- 1#?

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Answer 1 · 2017-07-03T15:04:26

See a solution process below:

First, group and combine like terms on each side of the equation:

#5x + x - 3 = 3x - 2 - 1#

#5x + 1x - 3 = 3x - 2 - 1#

#(5 + 1)x - 3 = 3x + (-2 - 1)#

#6x - 3 = 3x + (-3)#

#6x - 3 = 3x - 3#

Next, add #color(red)(3)# and subtract #color(blue)(3x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(blue)(3x) + 6x - 3 + color(red)(3) = -color(blue)(3x) + 3x - 3 + color(red)(3)#

#(-color(blue)(3) + 6)x - 0 = 0 - 0#

#3x = 0#

Now, divide each side of the equation by #color(red)(3)# to solve for #x#:

#(3x)/color(red)(3) = 0/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 0#

#x = 0#