How do you solve the inequality #abs(2x+1)

Question

How do you solve the inequality #abs(2x+1)<=6-x# and write your answer in interval notation?

Hazel Anglin · Accepted Answer

#-7 \leq x \leq 5/3# Since #|a|\leq b \Leftrightarrow -b \leq a \leq b#, the inequality #|2x+1|\leq 6-x#becomes #-(6-x) \leq 2x+1 \leq 6-x# #\Leftrightarrow x-6 \leq 2x+1 \leq 6-x# #\Leftrightarrow x-7 \leq 2x \leq 5-x# Solve the inequality on the left: #x-7 \leq 2x \Leftrightarrow -7 \leq x# Solve the inequality on the right: #2x \leq 5-x \Leftrightarrow 3x \leq 5 \Leftrightarrow x \leq 5/3# Combine the two intervals: #-7 \leq x \leq 5/3#

Eva Abramson · Answer

undefined To solve the inequality $ |2x + 1| \leq 6 - x $ and write the answer in interval notation, follow these steps:

1. Split the inequality into two cases based on the absolute value:
   - Case 1: $ 2x + 1 \geq 0 $ (when $ |2x + 1| = 2x + 1 $)
   - Case 2: $ 2x + 1 < 0 $ (when $ |2x + 1| = -(2x + 1) $)

2. Solve each case separately:
   - For Case 1: $ 2x + 1 \leq 6 - x $
     $ 3x \leq 5 $
     $ x \leq \frac{5}{3} $

- For Case 2: $ -(2x + 1) \leq 6 - x $
     $ -2x - 1 \leq 6 - x $
     $ -x \leq 7 $
     $ x \geq -7 $

3. Combine the solutions for both cases:
   - Case 1: $ x \leq \frac{5}{3} $
   - Case 2: $ x \geq -7 $

So, the solution in interval notation is $ (-\infty, \frac{5}{3}] $.

How do you solve the inequality #abs(2x+1)&lt;=6-x# and write your answer in interval notation?

How do you solve the inequality #abs(2x+1)<=6-x# and write your answer in interval notation?