How do you solve #-15\leq - 4x + 5< - 3#?

1 Answer
May 22, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(5)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-15 - color(red)(5) <= -4x + 5 - color(red)(5) < -3 - color(red)(5)#

#-20 <= -4x + 0 < -8#

#-20 <= -4x < -8#

Next, divide each segment by #color(blue)(-4)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operators:

#(-20)/color(blue)(-4) color(red)(>=) (-4x)/color(blue)(-4) color(red)(>) (-8)/color(blue)(-4)#

#5 color(red)(>=) (color(blue)(cancel(color(black)(-4)))x)/cancel(color(blue)(-4)) color(red)(>) 2#

#5 color(red)(>=) x color(red)(>) 2#

Or

#x <= 5#; #x > 2#

Or, in interval notation:

#(2, 5]#