How to solve complex absolute value inequalities like #absx+ abs(x-2)

Question

How to solve complex absolute value inequalities like #absx+ abs(x-2)< 5#?

Genesis Allen · Accepted Answer

The solution is # x in (-3/2, 7/2)#

This is an inequality with absolute values.

#|x|+|x-2|<5#

#|x|+|x-2|-5<0#

Let #f(x)=|x|+|x-2|-5#

#{(x>=0),(x-2>=0):}#, #<=>#, #{(x>=0),(x>=2):}#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaaaa)##0##color(white)(aaaaaaaaa)##2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##|x|##color(white)(aaaaaaaaaa)##-x##color(white)(aaaaaaa)##x##color(white)(aaaaaaaa)##x#

#color(white)(aaaa)##|x-2|##color(white)(aaaaaa)##-x+2##color(white)(aa)##-x+2##color(white)(aaaa)##x-2#

In the interval #(-oo,0)#

#-x-x+2-5<0#, #=>#, #2x > -3#, #=>#, #x> -3/2#

#-3/2 in (-oo, 0)#

In the interval #(0,2)#

#x-x+2-5<0#, #=>#, #0-3<0 #, # => #, solution

In the interval #(2,+oo)#

#x+x-2-5<0#, #=>#, #2x<7#, #=>#, #x<7/2#

#7/2 in (2, +oo)#

The solution is # x in (-3/2, 7/2)#

graph{|x|+|x-2|-5 [-10, 10, -5, 5]}

Ethan Adkins · Answer

undefined To solve complex absolute value inequalities like $|x| + |x - 2| < 5$:
1. Identify the critical points where the expressions inside the absolute value bars change sign. In this case, the critical points are $x = 0$ and $x = 2$.
2. Create intervals on the number line based on these critical points.
3. Test each interval by selecting a test point within it and evaluating the inequality.
4. Determine which intervals satisfy the inequality.
5. Express the solution as a combination of the intervals that satisfy the inequality.

In this specific example:
- Test $x = -1$: $|(-1)| + |(-1 - 2)| = 1 + 3 = 4$, which is less than 5.
- Test $x = 1$: $|1| + |1 - 2| = 1 + 1 = 2$, which is less than 5.
- Test $x = 3$: $|3| + |3 - 2| = 3 + 1 = 4$, which is less than 5.

Thus, the solution is $0 < x < 2$.

How to solve complex absolute value inequalities like #absx+ abs(x-2)&lt; 5#?

How to solve complex absolute value inequalities like #absx+ abs(x-2)< 5#?