How do you rewrite the inequality #abs(11-2x)=13# as a compound inequality?

Question

How do you rewrite the inequality #abs(11-2x)>=13# as a compound inequality?

Abigail Allen · Accepted Answer

#x le -1, or, x ge 12"#.

#"Equivalently, "x !in (-1,12).# Recall the Defn. of Absolute Value , reproduced below for ready reference : #|X|=X, if X ge 0# #=-X, if X lt 0#. Accordingly, we have to consider #2" Mutually Exclusive Cases :"# #"Case 1 : "(11-2x) ge 0.# #(11-2x) ge 0 rArr |11-2x|=11-2x ("by Defn.")# #:. |11-2x|ge13 rArr 11-2xge13rArr11-13ge2x #rArr-2ge2x.# Multiplying this last inequality by #1/2#, &, keeping in mind that as #1/2 gt0#, the sign of the inequality will not be reversed, we get, #-1 ge x.............(1)#. #"Also, "11-2x ge 0 rArr 11 ge 2x rArr 11/2 ge x, i.e., x le 11/2.# #(1)# is in accordanace with this. #"Case 2 : "(11-2x) lt 0#. #:. |11-2x|=-(11-2x)=2x-11..."[Defn.]"# #:. |11-2x| ge 13 rArr 2x-11 ge 13 rArr 2x ge 11+13=24# #rArr x ge 12............(2)# As #11-2x lt 0 rArr 11 lt 2x rArr 11/2 lt x.# #(2) is in accordance with this. Combining #(1), &, (2)" we have, "x le -1, or, x ge 12"#. #"Equivalently, "x !in (-1,12).# Enjoy Maths.!

Abigail Allen · Answer

#|(11-2x)| ge 13 iff -1 ge x, or, xge 12;" i.e., equivalently, "x !in (-1,12).#

Let us have a Second Method to solve this Problem.

We know that,

#(ast)...|x-a| lt delta iff a-delta lt x lt a+delta...[x,a in RR, delta in RR^+]#.

Now, #|11-2x| ge 13 iff |2(11/2-x)| ge 13#

#iff |2||(11/2-x)\ ge 13 iff |(x-11/2)| ge 13/2,.............................[because, |x|=|-x|#

Hence, #|11-2x| ge 13 iff |(x-11/2)| ge 13/2, "or, equivalently,"#

# |(11-2x)| cancel< 13/2 iff |(x-11/2)| cancel< 13/2.#

Using #(ast),# we have,

#|(x-11/2)| lt 13/2 iff (11/2-13/2)lt x lt (11/2+13/2)#

# iff -1 lt x lt 12.#

#:. |(11-2x)| ge 13 iff -1 ge x, or, xge 12," ie., equivalently, "x !in (-1,12).#

Jace Anderson · Answer

undefined To rewrite the inequality |11 - 2x| ≥ 13 as a compound inequality, you split it into two separate inequalities:

1. 11 - 2x ≥ 13
2. 11 - 2x ≤ -13

Then solve each inequality separately to find the range of values for x.

How do you rewrite the inequality #abs(11-2x)&gt;=13# as a compound inequality?

How do you rewrite the inequality #abs(11-2x)>=13# as a compound inequality?