How do you solve #abs(2x)

Question

How do you solve #abs(2x)<=abs(x-3)#?

Eva Abramson · Accepted Answer

When dealing with moduli, it is often helpful to split into cases at values where the sign of the enclosed value changes. For our example, #2x# changes sign at #x=0# and #x-3# changes sign at #x=3#. So split into cases: (a) #x < 0# (b) #x = 0# (c) #0 < x < 3# (d) #x = 3# (e) #x > 3# In case (a): #|2x| = -2x# and #|x - 3| = -x+3# So the original inequality is equivalent to #-2x <= -x+3# Adding #2x-3# to both sides we get #-3 <= x# Since this is case (a), we have #-3 <= x < 0# In case (b): #|2x| = |0| = 0# and #|x - 3| = |-3| = 3# So the inequality #|2x| < |x-3|# is satisfied. So #x=0# is also a solution. In case (c): #|2x| = 2x# and #|x - 3| = -x+3# So the original inequality is equivalent to #2x <= -x+3# Add x to both sides and divide both sides by 3 to get: #x <= 1# Since this is case (c), we also require #0 < x < 3#, so this gives us solutions: #0 < x <= 1#. In case (d): #|2x| = 6# and #|x - 3| = 0#, so the original inequality is not satisfied. In case (e); #|2x| = 2x# and #|x - 3| = x - 3# So the original inequality is equivalent to #2x <= x - 3# Subtracting #x# from both sides we get #x <= -3#. Since this is case (e), we also require #x > 3#, which cannot be satisfied at the same time. The union of our solutions from cases (a)-(c) gives us: #-3 <= x <= 1#

Genesis Allen · Answer

undefined To solve the inequality |2x| ≤ |x - 3|, we need to consider two cases:

1. When x - 3 ≥ 0, the inequality becomes 2x ≤ x - 3.
2. When x - 3 < 0, the inequality becomes 2x ≤ -(x - 3).

Solve each case separately to find the solution set.

How do you solve #abs(2x)&lt;=abs(x-3)#?

How do you solve #abs(2x)<=abs(x-3)#?