b. \( \frac{3 x+1}{7}-\frac{2-4 x}{3} \geq \frac{-5 x-4}{14}+\frac{7 x}{6} \)
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To solve the inequality \( \frac{3 x+1}{7}-\frac{2-4 x}{3} \geq \frac{-5 x-4}{14}+\frac{7 x}{6} \), the first step is to find a common denominator for all the fractions involved. The least common multiple of 7, 3, 14, and 6 is 42. Multiplying all terms by 42 to eliminate the fractions gives: \[ 42 \cdot \left(\frac{3x + 1}{7}\right) - 42 \cdot \left(\frac{2 - 4x}{3}\right) \geq 42 \cdot \left(\frac{-5x - 4}{14}\right) + 42 \cdot \left(\frac{7x}{6}\right) \] Calculating each term: - \( 42 \cdot \frac{3x + 1}{7} = 6(3x + 1) = 18x + 6 \) - \( 42 \cdot \frac{2 - 4x}{3} = 14(2 - 4x) = 28 - 56x \) - \( 42 \cdot \frac{-5x - 4}{14} = 3(-5x - 4) = -15x - 12 \) - \( 42 \cdot \frac{7x}{6} = 7(7x) = 49x \) Substituting these back into the inequality: \[ 18x + 6 - (28 - 56x) \geq -15x - 12 + 49x \] Now, simplifying both sides: \[ 18x + 6 - 28 + 56x \geq -15x - 12 + 49x \] \[ (18x + 56x) + 6 - 28 \geq (-15x + 49x) - 12 \] Combining like terms gives: \[ 74x - 22 \geq 34x - 12 \] Next, isolate \( x \): \[ 74x - 34x \geq -12 + 22 \] \[ 40x \geq 10 \] Finally, divide both sides by 40: \[ x \geq \frac{1}{4} \] The solution to the inequality is \( x \geq \frac{1}{4} \).