The problem requires solving two inequalities: a) $3(2x-1) < 2x + 5$ c) $-4(x-3) > -3(x-5)$

1. Problem Description

The problem requires solving two inequalities:

a) $3(2x-1) < 2x + 5$

c) $-4(x-3) > -3(x-5)$

2. Solution Steps

a) $3(2x-1) < 2x + 5$

First, distribute the $3$ on the left side:

$6x - 3 < 2x + 5$

Subtract $2x$ from both sides:

$6x - 2x - 3 < 2x - 2x + 5$

$4x - 3 < 5$

Add $3$ to both sides:

$4x - 3 + 3 < 5 + 3$

$4x < 8$

Divide both sides by $4$:

$\frac{4x}{4} < \frac{8}{4}$

$x < 2$

b) $-4(x-3) > -3(x-5)$

First, distribute the $-4$ on the left side and the $-3$ on the right side:

$-4x + 12 > -3x + 15$

Add $3x$ to both sides:

$-4x + 3x + 12 > -3x + 3x + 15$

$-x + 12 > 15$

Subtract $12$ from both sides:

$-x + 12 - 12 > 15 - 12$

$-x > 3$

Multiply both sides by $-1$. Remember to flip the inequality sign when multiplying by a negative number:

$(-1)(-x) < (-1)(3)$

$x < -3$

3. Final Answer

a) $x < 2$

c) $x < -3$