1. **State the problem:** We are given the system of linear equations: $$x + 3y = -3$$ $$x - 3y = -3$$ We need to graph these lines and determine the number of solutions (points of intersection). 2. **Recall the method:** The solution to a system of linear equations corresponds to the point(s) where the lines intersect. If they intersect at exactly one point, there is one solution. If they coincide, there are infinitely many solutions. If they are parallel and distinct, there are no solutions. 3. **Rewrite each equation in slope-intercept form $y=mx+b$:** For the first equation: $$x + 3y = -3 \implies 3y = -x - 3 \implies y = \frac{-x - 3}{3} = -\frac{1}{3}x - 1$$ For the second equation: $$x - 3y = -3 \implies -3y = -x - 3 \implies 3y = x + 3 \implies y = \frac{x + 3}{3} = \frac{1}{3}x + 1$$ 4. **Analyze slopes and intercepts:** - First line slope: $m_1 = -\frac{1}{3}$, intercept $b_1 = -1$ - Second line slope: $m_2 = \frac{1}{3}$, intercept $b_2 = 1$ Since $m_1 \neq m_2$, the lines are not parallel and will intersect at exactly one point. 5. **Find the intersection point by solving the system:** Set the right sides equal: $$-\frac{1}{3}x - 1 = \frac{1}{3}x + 1$$ Multiply both sides by 3 to clear denominators: $$-x - 3 = x + 3$$ Bring all terms to one side: $$-x - 3 - x - 3 = 0 \implies -2x - 6 = 0$$ Add 6 to both sides: $$-2x = 6$$ Divide both sides by $-2$: $$x = \frac{6}{\cancel{-2}}\cancel{-1} = -3$$ 6. **Find $y$ by substituting $x = -3$ into one of the equations:** Using $y = \frac{1}{3}x + 1$: $$y = \frac{1}{3}(-3) + 1 = -1 + 1 = 0$$ 7. **Conclusion:** The lines intersect at the point $(-3, 0)$, so the system has exactly one solution. **Final answer:** The system has one solution at $\boxed{(-3, 0)}$.