How do you solve #1< 3x - 2\leq 10#?

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Answer 1 · 2018-06-16T12:04:05

See a solution process below:

First, add #color(red)(2)# to each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#1 + color(red)(2) < 3x - 2 + color(red)(2) <= 10 + color(red)(2)#

#3 < 3x - 0 <= 12#

#3 < 3x <= 12#

Now, divide each segment by #color(red)(3)# to solve for #x# while keeping the system balanced:

#3/color(red)(3) < (3x)/color(red)(3) <= 12/color(red)(3)#

#1 < (color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) <= 4#

#1 < x <= 4#

Or

#x > 1#; #x <= 4#

Or, in interval notation:

#(1, 4]#