How do you solve #2x-5=-5#?

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Answer 1 · 2017-07-21T02:16:42

See a solution process below:

Step 1) Add #color(red)(5)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x - 5 + color(red)(5) = -5 + color(red)(5)#

#2x - 0 = 0#

#2x = 0#

Step 2) Divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 0/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 0#

#x = 0#

Answer 2 · 2017-07-21T02:18:23

This is relatively a simple problem; since the constant (the number) can be cancelled out.

The simplest way is to solve for x:
which enables you to use the zero property of multiplication; if #3x = 0; x = 0#

#2x - 5=-5#
#2x = 0#
#x = 0#