1. Problem: A line passes through points $(-2, \rho)$ and $(1, 5)$ with slope $4$. Find $\rho$. Step 1: Use slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. Step 2: Substitute $m=4$, $(x_1, y_1)=(-2, \rho)$, $(x_2, y_2)=(1, 5)$: $$4 = \frac{5 - \rho}{1 - (-2)} = \frac{5 - \rho}{3}$$ Step 3: Multiply both sides by $3$: $$12 = 5 - \rho$$ Step 4: Solve for $\rho$: $$\rho = 5 - 12 = -7$$ 2. Problem: Find equations of lines through $(2, -1)$: a) parallel to line $2x-3y=5$ b) perpendicular to line $2x-3y=5$ Step 1: Put $2x - 3y = 5$ in slope form: $$ -3y = -2x + 5 \implies y = \frac{2}{3}x - \frac{5}{3}$$ Slope $m = \frac{2}{3}$. Step 2a: Parallel line has slope $m=\frac{2}{3}$. Use point-slope form: $$ y - (-1) = \frac{2}{3}(x - 2) \implies y + 1 = \frac{2}{3}x - \frac{4}{3} $$ $$ y = \frac{2}{3}x - \frac{4}{3} - 1 = \frac{2}{3}x - \frac{7}{3}$$ Step 2b: Perpendicular slope is negative reciprocal: $$ m_{\perp} = -\frac{3}{2}$$ Use point-slope form: $$ y + 1 = -\frac{3}{2}(x - 2) = -\frac{3}{2}x + 3$$ $$ y = -\frac{3}{2}x + 3 - 1 = -\frac{3}{2}x + 2$$ 3. Problem: Temperature of coffee starts at $80^{\circ}C$ and decreases $5^{\circ}C$ per hour. A) Define temperature $y$ after $x$ hours. Step 1: The temperature decreases linearly, slope = $-5$. Step 2: At $x=0$, $y=80$. Step 3: Equation: $$ y = 80 - 5x$$ B) Find $x$ when $y=57.5$: $$ 57.5 = 80 - 5x \implies 5x = 80 - 57.5 = 22.5 \implies x = \frac{22.5}{5} = 4.5 $$ 4. Problem: Linear function $f$ with $f(-1) = \frac{3}{3} = 1$ and $f(3) = -5$. Find $f(x) = mx + b$. Step 1: Find slope $m$: $$ m = \frac{f(3) - f(-1)}{3 - (-1)} = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} $$ Step 2: Use point-slope form with point $(-1,1)$: $$ y - 1 = -\frac{3}{2}(x + 1) $$ Step 3: Solve for $y$: $$ y = -\frac{3}{2}x - \frac{3}{2} + 1 = -\frac{3}{2}x - \frac{1}{2} $$ So $$ f(x) = -\frac{3}{2}x - \frac{1}{2} $$ 5. Problem: Solve system: $$ \begin{cases} 3x + y = 2 \\ x - 2y = 3 \end{cases} $$ Step 1: Solve first equation for $y$: $$ y = 2 - 3x $$ Step 2: Substitute into second equation: $$ x - 2(2 - 3x) = 3 $$ $$ x - 4 + 6x = 3 $$ $$ 7x - 4 = 3 $$ $$ 7x = 7 \implies x = 1 $$ Step 3: Substitute $x=1$ back: $$ y = 2 - 3(1) = 2 - 3 = -1 $$ Final answers: 1) $\rho = -7$ 2a) $y = \frac{2}{3}x - \frac{7}{3}$ 2b) $y = -\frac{3}{2}x + 2$ 3a) $y = 80 - 5x$ 3b) $x = 4.5$ 4) $f(x) = -\frac{3}{2}x - \frac{1}{2}$ 5) $x=1$, $y=-1$