How do you solve and write the following in interval notation: #28 - 3x1/2#?

Question

How do you solve and write the following in interval notation: #2>8 - 3x>1/2#?

Julia Adams · Accepted Answer

See a solution process below:

First, subtract #color(red)(8)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced: #2 - color(red)(8) > 8 - color(red)(8) - 3x > 1/2 - color(red)(8)# #-6 > 0 - 3x > 1/2 - (2/2 xx color(red)(8))# #-6 > -3x > 1/2 - color(red)(16)/2# #-6 > -3x > (1 - color(red)(16))/2# #-6 > -3x > -15/2# Next, divide each segment by #color(blue)(-3)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators: #(-6)/color(blue)(-3) color(red)(<) (-3x)/color(blue)(-3) color(red)(<) (-15/2)/color(blue)(-3)# #2 color(red)(<) (color(blue)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(<) 15/6# #2 color(red)(<) x color(red)(<) 5/2# Or #x > 2#; #x < 5/2# Or, in interval notation: #(2, 5/2)#

Hazel Anglin · Answer

undefined The solution to $2 > 8 - 3x > \frac{1}{2}$ in interval notation is $(2, 2.5)$.

How do you solve and write the following in interval notation: #2&gt;8 - 3x&gt;1/2#?

How do you solve and write the following in interval notation: #2>8 - 3x>1/2#?