How do you graph and solve: #color(white)("d") 4-|5/3y+4|<2/5#?

1 Answer
Jan 29, 2018

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(4)# from each side of the inequality to isolate the absolute value function while keeping the inequality balanced:

#4 - color(red)(4) - abs(5/3y + 4) < 2/5 - color(red)(4)#

#0 - abs(5/3y + 4) < 2/5 - (5/5 xx color(red)(4))#

#-abs(5/3y + 4) < 2/5 - 20/5#

#-abs(5/3y + 4) < -18/5#

Next, multiply each side of the inequality by #color(red)(-1)# to make each side positive while keeping the inequality balanced. However, because we are multiplying an inequality by a negative number we must reverse the inequality operator.

#color(red)(-1) xx -abs(5/3y + 4) > color(red)(-1) xx -18/5#

#abs(5/3y + 4) > 18/5#

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

#-18/5 > 5/3y + 4 > 18/5#

Then, subtract #color(red)(4)# from each segment of the system of inequalities to isolate the #y# term while keeping the system balanced:

#-18/5 - color(red)(4) > 5/3y + 4 - color(red)(4) > 18/5 - color(red)(4)#

#-18/5 - (5/5 xx color(red)(4)) > 5/3y + 0 > 18/5 - (5/5 xx color(red)(4))#

#-18/5 - 20/5 > 5/3y > 18/5 - 20/5#

#-38/5 > 5/3y > -2/5#

Now, multiply each segment by #color(red)(3)/color(blue)(5)# to solve for #y# while keeping the system balanced:

#color(red)(3)/color(blue)(5) xx -38/5 > color(red)(3)/color(blue)(5) xx 5/3y > color(red)(3)/color(blue)(5) xx -2/5#

#-114/25 > cancel(color(red)(3))/cancel(color(blue)(5)) xx color(blue)(cancel(color(black)(5)))/color(red)(cancel(color(black)(3)))y > -6/25#

#-114/25 > y > -6/25#

Or

#y < -114/25#; #y > -6/25#

Or, in interval notation

#(-oo, -114/25)#; #(-6/25, +oo)#