1. **Simplify the expression** $\frac{x-1}{x+3} - \frac{2(x-1)}{x+3}$. Since the denominators are the same, subtract the numerators: $$\frac{x-1 - 2(x-1)}{x+3} = \frac{x-1 - 2x + 2}{x+3} = \frac{-x + 1}{x+3} = \frac{-(x-1)}{x+3}$$ 2. **Simplify the expression** $\frac{x^2 + 9x + 20}{x^2 + 5x} \div \frac{x-3}{x+4}$. Rewrite division as multiplication by reciprocal: $$\frac{x^2 + 9x + 20}{x^2 + 5x} \times \frac{x+4}{x-3}$$ Factor where possible: $$\frac{(x+4)(x+5)}{x(x+5)} \times \frac{x+4}{x-3}$$ Cancel common factor $x+5$: $$\frac{\cancel{(x+4)}\cancel{(x+5)}}{x\cancel{(x+5)}} \times \frac{\cancel{(x+4)}}{x-3} = \frac{(x+4)}{x} \times \frac{(x+4)}{x-3} = \frac{(x+4)^2}{x(x-3)}$$ 3. **Simplify the expression** $\frac{x-4}{x+9} \times \frac{x+7}{x+9}$. Multiply numerators and denominators: $$\frac{(x-4)(x+7)}{(x+9)(x+9)} = \frac{(x-4)(x+7)}{(x+9)^2}$$ 4. **Solve the equation** $\frac{18}{x-3} = 2$. Multiply both sides by $x-3$: $$18 = 2(x-3)$$ Distribute: $$18 = 2x - 6$$ Add 6 to both sides: $$18 + 6 = 2x$$ $$24 = 2x$$ Divide both sides by 2: $$x = \frac{24}{2} = 12$$ Check domain restriction: $x \neq 3$ because denominator cannot be zero. Solution: $x = 12$ 5. **Solve the equation** $\frac{8}{3(x+4)} = \frac{x}{3} + \frac{1}{x+4}$. Multiply both sides by $3(x+4)$ to clear denominators: $$8 = x(x+4) + 3$$ Expand: $$8 = x^2 + 4x + 3$$ Subtract 8 from both sides: $$0 = x^2 + 4x + 3 - 8 = x^2 + 4x - 5$$ Factor quadratic: $$0 = (x+5)(x-1)$$ Solutions: $$x = -5 \text{ or } x = 1$$ Check domain restrictions: $x \neq -4$ and $x \neq 0$ (from denominators), both solutions valid. 6. **Find $x$ such that flow rate from first pipe equals net flow from second pipe:** Given: First pipe flow rate: $\frac{x+1}{x^2 + 6}$ Second pipe net flow: $\frac{2}{x-2} - \frac{3}{x+3}$ Set equal: $$\frac{x+1}{x^2 + 6} = \frac{2}{x-2} - \frac{3}{x+3}$$ Find common denominator on right side: $$\frac{2(x+3) - 3(x-2)}{(x-2)(x+3)} = \frac{2x + 6 - 3x + 6}{(x-2)(x+3)} = \frac{-x + 12}{(x-2)(x+3)}$$ Equation becomes: $$\frac{x+1}{x^2 + 6} = \frac{-x + 12}{(x-2)(x+3)}$$ Cross multiply: $$ (x+1)(x-2)(x+3) = (-x + 12)(x^2 + 6) $$ Expand left side: First, $(x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6$ Then: $$(x+1)(x^2 + x - 6) = x(x^2 + x - 6) + 1(x^2 + x - 6) = x^3 + x^2 - 6x + x^2 + x - 6 = x^3 + 2x^2 - 5x - 6$$ Expand right side: $$(-x + 12)(x^2 + 6) = -x(x^2 + 6) + 12(x^2 + 6) = -x^3 - 6x + 12x^2 + 72 = -x^3 + 12x^2 - 6x + 72$$ Set equal: $$x^3 + 2x^2 - 5x - 6 = -x^3 + 12x^2 - 6x + 72$$ Bring all terms to left: $$x^3 + 2x^2 - 5x - 6 + x^3 - 12x^2 + 6x - 72 = 0$$ Combine like terms: $$2x^3 - 10x^2 + x - 78 = 0$$ Divide entire equation by 2: $$\cancel{2}x^3 - \cancel{10}x^2 + \cancel{1}x - \cancel{78} = 0 \Rightarrow x^3 - 5x^2 + \frac{1}{2}x - 39 = 0$$ (Actually, dividing 78 by 2 is 39, so correct is:) $$x^3 - 5x^2 + \frac{1}{2}x - 39 = 0$$ This cubic can be solved by trial or numerical methods. Try integer roots using Rational Root Theorem: possible roots are factors of 39 (±1, ±3, ±13, ±39). Test $x=3$: $$3^3 - 5(3)^2 + \frac{1}{2}(3) - 39 = 27 - 45 + 1.5 - 39 = -55.5 \neq 0$$ Test $x=13$: $$13^3 - 5(13)^2 + \frac{1}{2}(13) - 39 = 2197 - 845 + 6.5 - 39 = 1319.5 \neq 0$$ Test $x=1$: $$1 - 5 + 0.5 - 39 = -42.5 \neq 0$$ Test $x=6$: $$216 - 180 + 3 - 39 = 0$$ So $x=6$ is a root. Divide cubic by $x-6$ to find quadratic factor: Using synthetic division: Coefficients: 1, -5, 0.5, -39 Bring down 1 Multiply 1*6=6; add to -5 = 1 Multiply 1*6=6; add to 0.5 = 6.5 Multiply 6.5*6=39; add to -39 = 0 Quotient: $x^2 + x + 6.5$ Solve quadratic: $$x^2 + x + 6.5 = 0$$ Discriminant: $$\Delta = 1^2 - 4(1)(6.5) = 1 - 26 = -25 < 0$$ No real roots. **Real solution:** $x = 6$ **Check domain restrictions:** Denominators cannot be zero: $x^2 + 6 \neq 0$ always true. $x-2 \neq 0 \Rightarrow x \neq 2$ $x+3 \neq 0 \Rightarrow x \neq -3$ $x=6$ is valid. **Verification:** First pipe flow at $x=6$: $$\frac{6+1}{6^2 + 6} = \frac{7}{36 + 6} = \frac{7}{42} = \frac{1}{6}$$ Second pipe net flow: $$\frac{2}{6-2} - \frac{3}{6+3} = \frac{2}{4} - \frac{3}{9} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$ Matches first pipe flow. **Interpretation:** At 6 minutes, the flow rates match, which makes sense in the real-world context. 7. **Lauren and Jim data entry problem:** Jim's rate: $\frac{1}{7}$ job per hour. Combined rate: $\frac{1}{4}$ job per hour. Let Lauren's rate be $\frac{1}{L}$. Equation: $$\frac{1}{L} + \frac{1}{7} = \frac{1}{4}$$ Subtract $\frac{1}{7}$ from both sides: $$\frac{1}{L} = \frac{1}{4} - \frac{1}{7} = \frac{7 - 4}{28} = \frac{3}{28}$$ Invert both sides: $$L = \frac{28}{3} = 9\frac{1}{3} \text{ hours}$$ Lauren takes 9 hours and 20 minutes to do the job alone. --- **Summary of solutions:** 1. $\frac{-(x-1)}{x+3}$ 2. $\frac{(x+4)^2}{x(x-3)}$ 3. $\frac{(x-4)(x+7)}{(x+9)^2}$ 4. $x=12$ 5. $x = -5, 1$ 6. $x=6$ 7. Lauren takes $\frac{28}{3}$ hours or 9 hours 20 minutes.