How do you solve and write the following in interval notation: # 3x – 2 <7# and #–3x <= 15#?

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Answer 1 · 2017-08-24T18:58:14

Se a solution process below:

Solve First Equation For x:

#3x - 2 < 7#

#3x - 2 + color(red)(2) < 7 + color(red)(2)#

#3x - 0 < 9#

#3x < 9#

#(3x)/color(red)(3) < 9/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) < 3#

#x < 3#

Solve second Equation For x:

#-3x <= 15#

#(-3x)/color(blue)(-3) color(red)(>=) 15/color(blue)(-3)#

#(color(blue)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(>=) -5#

#x color(red)(>=) -5#

The Solution Is:

#x >= -5# and #x < 3#

Or, in interval notation:

#[-5, 3)#