1. **State the problem:** Solve the system of linear equations by graphing: $$2x + 5y = 1$$ $$3x - y = -7$$ 2. **Rewrite each equation in slope-intercept form $y = mx + b$ to graph easily:** For the first equation: $$2x + 5y = 1$$ Subtract $2x$ from both sides: $$5y = 1 - 2x$$ Divide both sides by 5: $$y = \frac{1}{5} - \frac{2}{5}x$$ For the second equation: $$3x - y = -7$$ Subtract $3x$ from both sides: $$-y = -7 - 3x$$ Multiply both sides by $-1$: $$y = 7 + 3x$$ 3. **Graph the lines:** - First line: $y = -\frac{2}{5}x + \frac{1}{5}$ - Second line: $y = 3x + 7$ 4. **Find the intersection point algebraically to confirm the graph:** Set the two expressions for $y$ equal: $$-\frac{2}{5}x + \frac{1}{5} = 3x + 7$$ Multiply both sides by 5 to clear denominators: $$-2x + 1 = 15x + 35$$ Bring all terms to one side: $$-2x - 15x = 35 - 1$$ $$-17x = 34$$ Divide both sides by $-17$: $$x = \frac{34}{-17} = -2$$ 5. **Substitute $x = -2$ into one of the equations to find $y$:** Using $y = 3x + 7$: $$y = 3(-2) + 7 = -6 + 7 = 1$$ 6. **Solution:** The lines intersect at the point $$\boxed{(-2, 1)}$$ which is the solution to the system. This means the system has one unique solution where both equations are satisfied.