How do you solve # 3 abs( x + 5 ) < 21 #?

2 Answers
Apr 29, 2015

#x<2# or #x> -12#

Solution

#3abs(x+5)<21#

#abs(x+5)<21/3#

#abs(x+5)<7#

It has two solutions

Either

#x+5<7# ......(i)

Or

#x+5> -7#.....(ii)

Solving (i)

#x+5-5<7-5#

#x+cancel5-cancel5<2#

#x<2#

Now, on solving (ii)

#x+5-5> -7-5#

#x+cancel5-cancel5> -12#

#x> -12#

Apr 29, 2015

We can divide both sides by 3 to get

#abs(x+5) < 21/3#

#->abs(x+5) < 7#

This tells us that #x+5# will lie between #-7 and 7#:

#-7 < x + 5 < 7 #

# - 7 - 5 < x + cancel(5) - cancel(5) < 7 - 5#

#color(green)( -12 < x < 2#

To verify your answer , choose appropriate values of #x# and see if the inequality is satisfied

  • Say #x = -10#

Left hand side = #3*abs(-10+5) = 3*abs(-5) = 3*5 = 15# (#<21#)

  • Say #x = 1#

Left hand side = #3*abs(1+5) = 3*abs(6) = 3*6 = 18# (#<21#)