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**HallsofIvy** Another way to solve general inequalities is to first solve the associated **equation**.

|2x+1|= |x- 5| means that either (1) 2x+ 1= x- 5 or (2)2x+ 1= -(x- 5)

In the first case, 2x- x= -1- 5 or x= -6.

In the second case, 2x+ 1= -x+ 5 so 2x+ x= 5- 1, 3x= 4, x= 4/3.

The point is that both |2x+ 1| and |x- 5| are **continuous** functions so they change from ">" to "<" and vice-versa only where they are "=".

The two points, x= -6, and x= 4/3, divide the real numbers into three intervals, x< -6, -6< x< 4/3, and x> 4/3.

-7 < -6 and if x= -7, |2x+1|= 13 while |x- 5|= 12. Yes, 13> 12. x= -7 satisfies the inequality so any x<-6 satisfies it.

-6< 0< 4/3 and if x= 0, |2x+ 1|= 1 while |x- 5|= 5. No, 1< 5. x= 0 does not satisfy the inequality so no value of x between -6 and 4/3 satisfies it.

2> 4/3 and if x= 2, |2x+1|= 5 while |x- 5|= 2. Yes, 13> 2. x= 2 satisfies the inequality so any x> 4/3 satisfies it.