How do you solve #abs(3-2y)>=8#?

1 Answer
Jul 15, 2017

Answer:

See a solution process below:

Explanation:

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-8 >= 3 - 2y >= 8#

First, subtract #color(red)(3)# from each segment of the system of inequalities to isolate the #y# term while keeping the system balanced:

#-color(red)(3) - 8 >= -color(red)(3) + 3 - 2y >= -color(red)(3) + 8#

#-11 >= 0 - 2y >= 5#

#-11 >= -2y >= 5#

Now, divide each segment by #color(blue)(-2)# to solve for #y# while keeping the system balanced. However, because we are dividing or multiplying an inequality by a negative number we must reverse the inequality operators:

#(-11)/color(blue)(-2) color(red)(<=) (-2y)/color(blue)(-2) color(red)(<=) 5/color(blue)(-2)#

#11/2 color(red)(<=) (color(red)(cancel(color(black)(-2)))y)/cancel(color(blue)(-2)) color(red)(<=) -5/2#

#11/2 color(red)(<=) y color(red)(<=) -5/2#

Or

#y <= -5/2#; #y >= 11/2#

Or, in interval notation:

#(-oo, -5/2]#; #[11/2, +oo)#