Solve the inequality |3x + 1| - 2x < |x - 4| + 5.

Step-by-step solution

1. 1

   Find the breakpoints. Each absolute value changes formula where its inside vanishes: $3x+1=0$ at $x=-\frac13$ and $x-4=0$ at $x=4$. These two points cut the real line into three ranges, and on each of them both bars can be removed with fixed signs.

2. 2

   Case 1: $x<-\frac13$. Both insides are negative, so $|3x+1|=-3x-1$ and $|x-4|=-x+4$:

   $(-3x-1)-2x<(-x+4)+5\quad\Rightarrow\quad -5x-1<-x+9\quad\Rightarrow\quad -10<4x\quad\Rightarrow\quad x>-\frac52$

   Intersecting with $x<-\frac13$ gives $-\frac52<x<-\frac13$.

3. 3

   Case 2: $-\frac13\le x<4$. Now $3x+1\ge 0$ but $x-4<0$:

   $3x+1-2x<-x+4+5\quad\Rightarrow\quad x+1<-x+9\quad\Rightarrow\quad 2x<8\quad\Rightarrow\quad x<4$

   That is exactly the range we are already in, so the entire interval $\left[-\frac13,4\right)$ works.

4. 4

   Case 3: $x\ge 4$. Both insides are non-negative:

   $3x+1-2x<x-4+5\quad\Rightarrow\quad x+1<x+1$

   The two sides are identically equal, and a number is never strictly less than itself. So this case contributes nothing — note that with $\le$ instead of $<$ the whole ray $x\ge 4$ would have qualified.

5. 5

   Union the cases.

   $\left(-\tfrac52,-\tfrac13\right)\cup\left[-\tfrac13,4\right)=\left(-\tfrac52,4\right)$

   The breakpoint $-\frac13$ is interior to the union, so nothing is lost there.

6. 6

   Test the boundaries and one interior point. At $x=-\frac52$ both sides equal $\frac{23}{2}$, so the strict inequality fails and the endpoint is excluded. At $x=4$ both sides equal $5$ — also excluded, consistent with Case 3. At $x=0$: left $=1$, right $=9$, and $1<9$ holds ✓. The solution is $-\frac52<x<4$.