1. **Solve the system of equations using the graphical method:** Given: $$\begin{cases} 3x - y = 7 \\ 2x + 3y = 1 \end{cases}$$ Step 1: Express each equation in slope-intercept form $y = mx + b$. For the first equation: $$3x - y = 7 \implies -y = 7 - 3x \implies y = 3x - 7$$ For the second equation: $$2x + 3y = 1 \implies 3y = 1 - 2x \implies y = \frac{1 - 2x}{3} = -\frac{2}{3}x + \frac{1}{3}$$ Step 2: Plot both lines on the coordinate plane. Step 3: Find the intersection point of the two lines by solving the system algebraically. Set the right sides equal: $$3x - 7 = -\frac{2}{3}x + \frac{1}{3}$$ Multiply both sides by 3 to clear denominators: $$3(3x - 7) = 3\left(-\frac{2}{3}x + \frac{1}{3}\right) \implies 9x - 21 = -2x + 1$$ Add $2x$ to both sides: $$9x + 2x - 21 = 1 \implies 11x - 21 = 1$$ Add 21 to both sides: $$11x = 22$$ Divide both sides by 11: $$x = \cancel{\frac{11x}{11}} = \cancel{\frac{22}{11}} = 2$$ Step 4: Substitute $x=2$ into one of the original equations to find $y$. Using $y = 3x - 7$: $$y = 3(2) - 7 = 6 - 7 = -1$$ **Solution:** $$(x, y) = (2, -1)$$ 2. **Graph the inequality $2x + 3y \leq 1$ to find the solution set:** Step 1: Rewrite the inequality in slope-intercept form: $$2x + 3y \leq 1 \implies 3y \leq 1 - 2x \implies y \leq \frac{1 - 2x}{3} = -\frac{2}{3}x + \frac{1}{3}$$ Step 2: Graph the boundary line $y = -\frac{2}{3}x + \frac{1}{3}$ (solid line because of $\leq$). Step 3: Shade the region below the line because $y$ is less than or equal to the expression. 3. **Graph the system of inequalities:** $$\begin{cases} y \leq 2x - 3 \\ y \geq -3 \\ y \leq -1.25x + 2.5 \end{cases}$$ Step 1: Graph each boundary line: - $y = 2x - 3$ (solid line, shade below) - $y = -3$ (horizontal line, shade above) - $y = -1.25x + 2.5$ (solid line, shade below) Step 2: The solution set is the intersection of all shaded regions. 4. **Solve the system using the elimination method:** $$\begin{cases} \frac{3}{2}x + \frac{1}{2}y = -3 \\ \frac{1}{2}y = x + 2 \end{cases}$$ Step 1: Rewrite the second equation: $$\frac{1}{2}y = x + 2 \implies y = 2x + 4$$ Step 2: Substitute $y = 2x + 4$ into the first equation: $$\frac{3}{2}x + \frac{1}{2}(2x + 4) = -3$$ Simplify: $$\frac{3}{2}x + x + 2 = -3$$ Combine like terms: $$\frac{3}{2}x + x = \frac{3}{2}x + \frac{2}{2}x = \frac{5}{2}x$$ So: $$\frac{5}{2}x + 2 = -3$$ Subtract 2 from both sides: $$\frac{5}{2}x = -5$$ Multiply both sides by $\frac{2}{5}$: $$x = -5 \times \frac{2}{5} = -2$$ Step 3: Substitute $x = -2$ into $y = 2x + 4$: $$y = 2(-2) + 4 = -4 + 4 = 0$$ **Solution:** $$(x, y) = (-2, 0)$$