1. **State the problem:** Solve the system of equations by graphing and find the solution as an ordered pair. Given system: $$3x + 3y = -6$$ $$10x + 5y = -5$$ 2. **Rewrite each equation in slope-intercept form $y = mx + b$ to understand the lines:** For the first equation: $$3x + 3y = -6$$ Subtract $3x$ from both sides: $$3y = -3x - 6$$ Divide both sides by 3: $$y = \cancel{\frac{3}{3}}{-x} - \cancel{\frac{6}{3}}{2}$$ So, $$y = -x - 2$$ For the second equation: $$10x + 5y = -5$$ Subtract $10x$ from both sides: $$5y = -10x - 5$$ Divide both sides by 5: $$y = \cancel{\frac{5}{5}}{-2x} - \cancel{\frac{5}{5}}{1}$$ So, $$y = -2x - 1$$ 3. **Analyze the lines:** - First line slope: $-1$, y-intercept: $-2$ - Second line slope: $-2$, y-intercept: $-1$ Since the slopes are different, the lines intersect at exactly one point. 4. **Find the intersection point by solving the system algebraically:** Set the right sides equal since both equal $y$: $$-x - 2 = -2x - 1$$ Add $2x$ to both sides: $$-x + 2x - 2 = -1$$ Simplify: $$x - 2 = -1$$ Add 2 to both sides: $$x = 1$$ 5. **Substitute $x=1$ into one of the original equations to find $y$:** Using $y = -x - 2$: $$y = -1 - 2 = -3$$ 6. **Solution:** $$(x, y) = (1, -3)$$ **Answer:** The solution to the system is the ordered pair $(1, -3)$. --- **Note on the graph:** The user described two vertical lines at $x = -3$ and $x = 3$ which do not correspond to the given equations. The actual lines from the equations are not vertical and intersect at $(1, -3)$.