1. Solve the equation $\frac{x}{3} + \frac{y}{4} = \frac{x}{2}$ for $y$. Step 1: Multiply both sides by 12 (the least common multiple of 3, 4, and 2) to clear denominators: $$12 \times \left(\frac{x}{3} + \frac{y}{4}\right) = 12 \times \frac{x}{2}$$ $$4x + 3y = 6x$$ Step 2: Rearrange to isolate $y$: $$3y = 6x - 4x$$ $$3y = 2x$$ $$y = \frac{2x}{3}$$ --- 2. Solve $\frac{1}{x^2} = 16$. Step 1: Multiply both sides by $x^2$: $$1 = 16x^2$$ Step 2: Divide both sides by 16: $$x^2 = \frac{1}{16}$$ Step 3: Take the square root: $$x = \pm \frac{1}{4}$$ --- 3. Solve the inequality $\frac{x + 3}{x - 2} \leq 1$. Step 1: Subtract 1 from both sides: $$\frac{x + 3}{x - 2} - 1 \leq 0$$ Step 2: Write with common denominator: $$\frac{x + 3 - (x - 2)}{x - 2} \leq 0$$ $$\frac{x + 3 - x + 2}{x - 2} \leq 0$$ $$\frac{5}{x - 2} \leq 0$$ Step 3: Since numerator 5 is positive, inequality depends on denominator: $$x - 2 < 0 \implies x < 2$$ Step 4: Domain excludes $x=2$ (denominator zero), so solution is: $$(-\infty, 2)$$ --- 4. Solve the inequality $\frac{3x + 1}{x - 1} > 2$. Step 1: Subtract 2: $$\frac{3x + 1}{x - 1} - 2 > 0$$ Step 2: Common denominator: $$\frac{3x + 1 - 2(x - 1)}{x - 1} > 0$$ $$\frac{3x + 1 - 2x + 2}{x - 1} > 0$$ $$\frac{x + 3}{x - 1} > 0$$ Step 3: Find critical points where numerator or denominator zero: $$x + 3 = 0 \implies x = -3$$ $$x - 1 = 0 \implies x = 1$$ Step 4: Test intervals: - For $x < -3$, numerator negative, denominator negative, fraction positive. - For $-3 < x < 1$, numerator positive, denominator negative, fraction negative. - For $x > 1$, numerator positive, denominator positive, fraction positive. Step 5: Solution where fraction $> 0$ and denominator $\neq 0$: $$(-\infty, -3) \cup (1, \infty)$$ --- 5. Analyze the functions: Function 1: $$y = \frac{2x + 4}{x + 3}$$ - Domain: All real $x$ except $x = -3$ (denominator zero). - Vertical asymptote: $x = -3$. - Horizontal asymptote: Degree numerator = degree denominator = 1, so divide leading coefficients: $$y = \frac{2x}{x} = 2$$ So horizontal asymptote at $y = 2$. - Intercepts: - $y$-intercept: Set $x=0$: $$y = \frac{0 + 4}{0 + 3} = \frac{4}{3}$$ - $x$-intercept: Set numerator zero: $$2x + 4 = 0 \implies x = -2$$ Function 2: $$f(x) = \frac{4x + 1}{x^2 - 9} = \frac{4x + 1}{(x - 3)(x + 3)}$$ - Domain: All real $x$ except $x = 3$ and $x = -3$. - Vertical asymptotes: $x = 3$ and $x = -3$. - Horizontal asymptote: Degree numerator 1, denominator 2, so horizontal asymptote at $y=0$. - Intercepts: - $y$-intercept: Set $x=0$: $$f(0) = \frac{1}{-9} = -\frac{1}{9}$$ - $x$-intercept: Set numerator zero: $$4x + 1 = 0 \implies x = -\frac{1}{4}$$ --- Summary: - Solutions: 1. $y = \frac{2x}{3}$ 2. $x = \pm \frac{1}{4}$ 3. $x < 2$ 4. $x \in (-\infty, -3) \cup (1, \infty)$ - Domain, intercepts, asymptotes for given rational functions as above.