How do you solve #abs(3x) ≤ abs(2x - 5)#?

Question

Genesis Allen · Accepted Answer

#1<= x <= 5#

#|3x|<=|2x-5|#

There are four solutions;

#1st#

#-3x <= -(2x - 5)#

#2nd#

#3x <= 2x - 5#

#3rd#

#3x <=-(2x - 5)#

#4th#

#-3x <= 2x - 5#

From the #1st#

#-3x <= -(2x - 5)#

#-3x <= -2x + 5#

Add #2x# to both sides;

#-3x + 2x <= -2x + 5 + 2x#

#-x <= 5#

Multiply through by Minus #(-)#

#-(-x) <= -(5)#

Note: When you divide or multiplty an inequality sign by a negative value, the sign changes..

#x >= -5#

From the #2nd#

#3x <= 2x - 5#

Subtracting both sides by #2x#;

#3x - 2x <= 2x + 5 - 2x#

#x <= 5#

From the #3rd#

#3x <=-(2x - 5)#

Removing the bracket;

#3x <= -2x + 5#

Add #2x# to both sides;

#3x + 2x <= -2x + 5 + 2x#

#5x <= 5#

Dividing both sides by the coefficient of #x#;

#(5x)/5 <= 5/5#

#(cancel5x)/cancel5 <= cancel5/cancel5#

#x <=5#

From the #4th#

#-3x <= 2x - 5#

Subtracting #2x# from both sides;

#-3x - 2x <= 2x - 5 - 2x#

#-5x <= -5#

Dividing both sides by the coefficient of #x#;

#(-5x)/(-5) <= (-5)/(-5)#

#(cancel(-5)x)/cancel(-5) <= cancel(-5)/cancel(-5)#

#x >= 1#

Hence the possible ranges are;

#x >= -5#

#x <=5#

#x >= 1#

Therefore;

#1<= x <= 5#

Ethan Adkins · Answer

undefined To solve the inequality abs(3x) ≤ abs(2x - 5), you split it into two cases:

1. When 3x ≥ 0 and 2x - 5 ≥ 0:
   In this case, solve 3x ≤ 2x - 5.

2. When 3x < 0 and 2x - 5 < 0:
   In this case, solve -3x ≤ 2x - 5.

Then, find the solutions for each case and combine them.