How do you solve and write the following in interval notation: #1/2 x - 1/10 <=x + 1#?

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Answer 1 · 2017-04-12T12:15:07

See the entire solution process below:

First, subtract #color(red)(1/2x)# and #color(blue)(1)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#1/2x - 1/10 - color(red)(1/2x) - color(blue)(1) <= x + 1 - color(red)(1/2x) - color(blue)(1)#

#1/2x - color(red)(1/2x) - 1/10 - color(blue)(1) <= x - color(red)(1/2x) + 1 - color(blue)(1)#

#0 - 1/10 - color(blue)(10/10) <= 2/2x - color(red)(1/2x) + 0#

#-11/10 <= 1/2x#

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

#color(red)(2) xx -11/10 <= color(red)(2) xx 1/2x#

#-22/10 <= cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))x#

#-22/10 <= x#

To state the solution in terms of #x# we can reverse or "flip" the inequality:

#x >= -22/10#

Or, in interval form:

#[-22/10, oo)#