1. **State the problem:** Solve the system of equations by graphing: $$3x + y = 3$$ $$3x - y = 3$$ 2. **Rewrite each equation in slope-intercept form $y = mx + b$:** For the first equation: $$3x + y = 3 \implies y = 3 - 3x$$ For the second equation: $$3x - y = 3 \implies -y = 3 - 3x \implies y = 3x - 3$$ 3. **Identify the slopes and intercepts:** - First line: slope $m_1 = -3$, y-intercept $b_1 = 3$ - Second line: slope $m_2 = 3$, y-intercept $b_2 = -3$ 4. **What kind of lines are formed?** Since $m_1 = -3$ and $m_2 = 3$, the lines have different slopes, so they are **intersecting lines**. 5. **Find the point of intersection by solving the system algebraically:** Add the two original equations: $$3x + y = 3$$ $$3x - y = 3$$ Adding: $$3x + y + 3x - y = 3 + 3 \implies 6x = 6 \implies x = \frac{6}{6} = 1$$ Substitute $x=1$ into the first equation: $$3(1) + y = 3 \implies 3 + y = 3 \implies y = 0$$ 6. **Point of intersection:** $$\boxed{(1, 0)}$$ 7. **Type of system:** Since the lines intersect at exactly one point, this is a **consistent and independent system**. 8. **Number of solutions:** There is exactly **one solution** to the system, the point $(1,0)$. Summary: - a) The lines are intersecting lines with different slopes. - b) Yes, there is a point of intersection. - c) The point of intersection is $(1,0)$. - d) The system is consistent and independent. - e) There is exactly one solution.