1. **State the problem:** Solve the system of linear equations by graphing and find the number of solutions. Given system: $$5x + 5y = 15$$ $$-8x + 2y = 16$$ 2. **Convert each equation to slope-intercept form $y = mx + b$:** For the first equation: $$5x + 5y = 15$$ Subtract $5x$ from both sides: $$\cancel{5x} + 5y = 15 - \cancel{5x}$$ $$5y = -5x + 15$$ Divide both sides by 5: $$\frac{5y}{\cancel{5}} = \frac{-5x}{\cancel{5}} + \frac{15}{5}$$ $$y = -x + 3$$ For the second equation: $$-8x + 2y = 16$$ Add $8x$ to both sides: $$\cancel{-8x} + 2y = 16 + \cancel{8x}$$ $$2y = 8x + 16$$ Divide both sides by 2: $$\frac{2y}{\cancel{2}} = \frac{8x}{\cancel{2}} + \frac{16}{2}$$ $$y = 4x + 8$$ 3. **Analyze the slopes and intercepts:** - First line: slope $m_1 = -1$, intercept $b_1 = 3$ - Second line: slope $m_2 = 4$, intercept $b_2 = 8$ Since the slopes are different ($-1 \neq 4$), the lines intersect at exactly one point. 4. **Find the point of intersection by solving the system algebraically:** Set the two expressions for $y$ equal: $$-x + 3 = 4x + 8$$ Add $x$ to both sides: $$3 = 5x + 8$$ Subtract 8 from both sides: $$3 - 8 = 5x$$ $$-5 = 5x$$ Divide both sides by 5: $$\frac{-5}{\cancel{5}} = \frac{5x}{\cancel{5}}$$ $$x = -1$$ Substitute $x = -1$ into $y = -x + 3$: $$y = -(-1) + 3 = 1 + 3 = 4$$ 5. **Final answer:** - The system has one solution. - The solution point is $(-1, 4)$. --- "content" summary: - Equations in slope-intercept form: $$y = -x + 3$$ $$y = 4x + 8$$ - One solution at point $(-1, 4)$.