1. Solve the inequality $5x - 3 \leq 7 - 3x$. Start by adding $3x$ to both sides: $$5x - 3 + 3x \leq 7 - 3x + 3x$$ $$8x - 3 \leq 7$$ Add 3 to both sides: $$8x - 3 + 3 \leq 7 + 3$$ $$8x \leq 10$$ Divide both sides by 8: $$\frac{\cancel{8}x}{\cancel{8}} \leq \frac{10}{8}$$ $$x \leq \frac{5}{4}$$ 2. Solve the inequality $2x - \frac{1}{2} \geq 7x + \frac{7}{6}$. Subtract $7x$ from both sides: $$2x - 7x - \frac{1}{2} \geq 7x - 7x + \frac{7}{6}$$ $$-5x - \frac{1}{2} \geq \frac{7}{6}$$ Add $\frac{1}{2}$ to both sides: $$-5x - \frac{1}{2} + \frac{1}{2} \geq \frac{7}{6} + \frac{1}{2}$$ $$-5x \geq \frac{7}{6} + \frac{3}{6} = \frac{10}{6} = \frac{5}{3}$$ Divide both sides by $-5$ and reverse inequality sign: $$\frac{\cancel{-5}x}{\cancel{-5}} \leq \frac{5/3}{-5}$$ $$x \leq -\frac{1}{3}$$ 3. Solve the inequality $\frac{x + 5}{2} \leq \frac{12 + 3x}{4}$. Multiply both sides by 4 to clear denominators: $$4 \times \frac{x + 5}{2} \leq 4 \times \frac{12 + 3x}{4}$$ $$2(x + 5) \leq 12 + 3x$$ Expand left side: $$2x + 10 \leq 12 + 3x$$ Subtract $2x$ from both sides: $$10 \leq 12 + x$$ Subtract 12 from both sides: $$10 - 12 \leq x$$ $$-2 \leq x$$ Or equivalently: $$x \geq -2$$ 4. Solve the inequality $\frac{x + 1}{x - 1} \geq 4$. Bring all terms to one side: $$\frac{x + 1}{x - 1} - 4 \geq 0$$ Rewrite 4 as $\frac{4(x - 1)}{x - 1}$: $$\frac{x + 1}{x - 1} - \frac{4(x - 1)}{x - 1} \geq 0$$ Combine fractions: $$\frac{x + 1 - 4(x - 1)}{x - 1} \geq 0$$ Expand numerator: $$x + 1 - 4x + 4 = -3x + 5$$ So: $$\frac{-3x + 5}{x - 1} \geq 0$$ Find critical points where numerator or denominator is zero: Numerator zero: $-3x + 5 = 0 \Rightarrow x = \frac{5}{3}$ Denominator zero: $x - 1 = 0 \Rightarrow x = 1$ Test intervals: - For $x < 1$, numerator $-3x + 5 > 0$, denominator $x - 1 < 0$, fraction negative. - For $1 < x < \frac{5}{3}$, numerator positive, denominator positive, fraction positive. - For $x > \frac{5}{3}$, numerator negative, denominator positive, fraction negative. Inequality $\geq 0$ holds where fraction is positive or zero. At $x=\frac{5}{3}$ numerator zero, fraction zero, included. At $x=1$ denominator zero, undefined, excluded. Solution: $$1 < x \leq \frac{5}{3}$$ 5. Solve the inequality $x^2 - x - 2 \leq 0$. Factor quadratic: $$x^2 - x - 2 = (x - 2)(x + 1)$$ Set equal to zero: $$x - 2 = 0 \Rightarrow x = 2$$ $$x + 1 = 0 \Rightarrow x = -1$$ Parabola opens upward (coefficient of $x^2$ positive), so inequality $\leq 0$ holds between roots: $$-1 \leq x \leq 2$$ 6. Solve the inequality $x^2 - 7x > -10$. Bring all terms to one side: $$x^2 - 7x + 10 > 0$$ Factor quadratic: $$x^2 - 7x + 10 = (x - 5)(x - 2)$$ Set equal to zero: $$x = 2, 5$$ Parabola opens upward, so inequality $> 0$ holds outside roots: $$x < 2 \text{ or } x > 5$$