| -3х+1|<х-2

To solve the inequality \( |-3x + 1| < x - 2 \), we need to consider two cases for the absolute value expression.

Case 1: \( -3x + 1 \geq 0 \)

In this case, the absolute value \( |-3x + 1| = -3x + 1 \). Therefore, we have: \[ -3x + 1 < x - 2 \] Rearranging this inequality: \[ 1 + 2 < x + 3x \] \[ 3 < 4x \] \[ \frac{3}{4} < x \] This gives us \( x > \frac{3}{4} \).

Now we need to find the range of \( x \) in this case. Since we assumed \( -3x + 1 \geq 0 \): \[ -3x + 1 \geq 0 \implies 1 \geq 3x \implies x \leq \frac{1}{3} \] Thus, in this case, there are no values of \( x \) that satisfy both conditions \( x > \frac{3}{4} \) and \( x \leq \frac{1}{3} \).

Case 2: \( -3x + 1 < 0 \)

In this case, the absolute value \( |-3x + 1| = 3x - 1 \). Thus, we have: \[ 3x - 1 < x - 2 \] Rearranging this inequality: \[ 3x - x < -2 + 1 \] \[ 2x < -1 \] \[ x < -\frac{1}{2} \] Now we need to find the range of \( x \) in this case. Since we assumed \( -3x + 1 < 0 \): \[ -3x + 1 < 0 \implies 1 < 3x \implies x > \frac{1}{3} \] In this case, there are no values of \( x \) that satisfy both \( x < -\frac{1}{2} \) and \( x > \frac{1}{3} \).

Conclusion

Since both cases lead to contradictions, we conclude that there are no real solutions for \( |-3x + 1| < x - 2 \).