Solution
(a) Solve $2(3x - 1) \lt 4 - 3x$
$$ \begin{aligned} &\text{Step 1: Expand the brackets.} \\ &2(3x - 1) \lt 4 - 3x \\ &6x - 2 \lt 4 - 3x \\ \\ &\text{Step 2: Combine like terms.} \\ &6x + 3x \lt 4 + 2 \\ &9x \lt 6 \\ \\ &\text{Step 3: Solve for } x. \\ &x \lt \frac{6}{9} \\ &x \lt \frac{2}{3} \\ \\ &\boxed{x \lt \frac{2}{3}} \end{aligned} $$
(b) Solve $3x^2 - 8x - 3 \lt 0$
$$ \begin{aligned} &\text{Step 1: Factorize the quadratic expression.} \\ & 3x^2 - 8x - 3 = 0 \quad \text{(use the quadratic formula)} \\ & x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 3,\; b = -8,\; c = -3 \\ & x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2(3)} \\ & x = \frac{8 \pm \sqrt{64 + 36}}{6} \\ & x = \frac{8 \pm \sqrt{100}}{6} \\ & x = \frac{8 \pm 10}{6} \\ \\ &\text{Step 2: Simplify the roots.} \\ & x = \frac{8 + 10}{6} \quad \text{or} \quad x = \frac{8 - 10}{6} \\ & x = 3 \quad \text{or} \quad x = -\frac{1}{3} \\ \\ &\text{Step 3: Determine the intervals where } 3x^2 - 8x - 3 \lt 0. \\ &\text{The roots divide the number line into three intervals:} \\ &(-\infty, -\frac{1}{3}), \quad \left(-\frac{1}{3}, 3\right), \quad (3, \infty) \\ &\text{Test a point from each interval in } 3x^2 - 8x - 3. \\ \\ &\text{For } x = -1 \quad (-\infty, -\frac{1}{3}): \\ &\quad 3(-1)^2 - 8(-1) - 3 = 3 + 8 - 3 = 8 > 0 \quad \text{(Not valid)} \\ \\ &\text{For } x = 0 \quad \left(-\frac{1}{3}, 3\right): \\ &\quad 3(0)^2 - 8(0) - 3 = -3 \lt 0 \quad \text{(Valid)} \\ \\ &\text{For } x = 4 \quad (3, \infty): \\ &\quad 3(4)^2 - 8(4) - 3 = 48 - 32 - 3 = 13 > 0 \quad \text{(Not valid)} \\ \\ &\text{Thus, the solution is:} \\ &\boxed{-\frac{1}{3} \lt x \lt 3} \end{aligned} $$
(c) Solve both $2(3x - 1) \lt 4 - 3x$ and $3x^2 - 8x - 3 \lt 0$
$$ \begin{aligned} &\text{Step 1: Combine the two inequalities.} \\ & x \lt \frac{2}{3} \quad \text{and} \quad -\frac{1}{3} \lt x \lt 3 \\ \\ &\text{Step 2: Find the intersection of the two solutions.} \\ &\text{The overlap is: } -\frac{1}{3} \lt x \lt \frac{2}{3} \\ \\ &\text{Thus, the solution is:} \\ &\boxed{-\frac{1}{3} \lt x \lt \frac{2}{3}} \end{aligned} $$
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