1. **Problem 1:** Identify if the expression $\frac{2x+3}{x-1}$ is a rational function, rational equation, or rational inequality. A rational function is a ratio of two polynomials. Since $\frac{2x+3}{x-1}$ is a ratio of polynomials, it is a rational function. 2. **Problem 2:** Solve the rational equation $\frac{1}{x} + 2 = 5$. Step 1: Isolate the fraction: $\frac{1}{x} = 3$. Step 2: Multiply both sides by $x$: $1 = 3x$. Step 3: Solve for $x$: $x = \frac{1}{3}$. 3. **Problem 3:** Solve the rational inequality $\frac{x-2}{x+1} > 0$. Step 1: Find critical points where numerator or denominator is zero: $x=2$ and $x=-1$. Step 2: Test intervals: $(-\infty,-1)$, $(-1,2)$, $(2,\infty)$. Step 3: Check signs: - For $x=-2$: $\frac{-2-2}{-2+1} = \frac{-4}{-1} = 4 > 0$ (True) - For $x=0$: $\frac{0-2}{0+1} = \frac{-2}{1} = -2 < 0$ (False) - For $x=3$: $\frac{3-2}{3+1} = \frac{1}{4} > 0$ (True) Step 4: Solution: $(-\infty,-1) \cup (2,\infty)$ excluding $x=-1$ where denominator is zero. 4. **Problem 4:** Simplify the rational expression $\frac{x^2-9}{x^2-6x+9}$. Step 1: Factor numerator and denominator: - Numerator: $x^2-9 = (x-3)(x+3)$ - Denominator: $x^2-6x+9 = (x-3)^2$ Step 2: Cancel common factor $(x-3)$: Result: $\frac{x+3}{x-3}$, with $x \neq 3$. 5. **Problem 5:** Solve the rational equation $\frac{2}{x+1} = \frac{3}{x-2}$. Step 1: Cross multiply: $2(x-2) = 3(x+1)$ Step 2: Expand: $2x - 4 = 3x + 3$ Step 3: Rearrange: $-4 - 3 = 3x - 2x$ $-7 = x$ Step 4: Check restrictions: $x \neq -1, 2$; $x=-7$ is valid. 6. **Problem 6:** Determine the domain of $f(x) = \frac{5}{x^2 - 4}$. Step 1: Set denominator not equal to zero: $x^2 - 4 \neq 0$ Step 2: Factor: $(x-2)(x+2) \neq 0$ Step 3: Domain is all real numbers except $x=2$ and $x=-2$. 7. **Problem 7:** Solve the rational inequality $\frac{3x}{x-5} \leq 6$. Step 1: Bring all terms to one side: $\frac{3x}{x-5} - 6 \leq 0$ Step 2: Write as single fraction: $\frac{3x - 6(x-5)}{x-5} \leq 0$ Step 3: Simplify numerator: $3x - 6x + 30 = -3x + 30$ Step 4: Inequality: $\frac{-3x + 30}{x-5} \leq 0$ Step 5: Critical points: numerator zero at $x=10$, denominator zero at $x=5$. Step 6: Test intervals: - $(-\infty,5)$: pick $x=0$, fraction $\frac{30}{-5} = -6 \leq 0$ True - $(5,10)$: pick $x=6$, fraction $\frac{-18+30}{1} = 12 > 0$ False - $(10,\infty)$: pick $x=11$, fraction $\frac{-33+30}{6} = -3/6 = -0.5 \leq 0$ True Step 7: Solution: $(-\infty,5) \cup [10,\infty)$ excluding $x=5$ where denominator zero. 8. **Problem 8:** Simplify $\frac{4x^2 - 1}{2x - 1}$. Step 1: Factor numerator: $4x^2 - 1 = (2x - 1)(2x + 1)$ Step 2: Cancel common factor: Result: $2x + 1$, with $x \neq \frac{1}{2}$. 9. **Problem 9:** Solve the rational equation $\frac{1}{x+2} + \frac{2}{x-1} = 0$. Step 1: Find common denominator: $(x+2)(x-1)$. Step 2: Multiply both sides by denominator: $(x-1) + 2(x+2) = 0$ Step 3: Expand: $x - 1 + 2x + 4 = 0$ Step 4: Combine like terms: $3x + 3 = 0$ Step 5: Solve for $x$: $3x = -3$ so $x = -1$ Step 6: Check restrictions: $x \neq -2, 1$; $x=-1$ valid. 10. **Problem 10:** Determine if $\frac{x^2 + x - 6}{x^2 - 4}$ is defined at $x=2$. Step 1: Factor denominator: $x^2 - 4 = (x-2)(x+2)$ Step 2: At $x=2$, denominator is zero, so expression is undefined. **Answer Key:** 1. Rational function 2. $x=\frac{1}{3}$ 3. $(-\infty,-1) \cup (2,\infty)$ 4. $\frac{x+3}{x-3}$ 5. $x=-7$ 6. Domain: $x \neq \pm 2$ 7. $(-\infty,5) \cup [10,\infty)$, $x \neq 5$ 8. $2x + 1$ 9. $x=-1$ 10. Undefined at $x=2$