1. **Problem statement:** We are given two lines on a coordinate plane and need to find their point of intersection, which is the solution to the system of linear equations represented by these lines. 2. **Identify the equations of the lines:** - First line passes through points (-2, 0) and (4, 7). - Second line passes through points (-1, 7) and (3, 1). 3. **Find the slope of each line using the formula** $m=\frac{y_2 - y_1}{x_2 - x_1}$: - For the first line: $m_1=\frac{7 - 0}{4 - (-2)}=\frac{7}{6}$. - For the second line: $m_2=\frac{1 - 7}{3 - (-1)}=\frac{-6}{4}=-\frac{3}{2}$. 4. **Write the equation of each line in slope-intercept form $y=mx+b$ by finding $b$ (y-intercept):** - First line: Using point (-2, 0), $0=\frac{7}{6}(-2)+b \Rightarrow 0= -\frac{14}{6}+b \Rightarrow b=\frac{14}{6}=\frac{7}{3}$. - Equation: $$y=\frac{7}{6}x + \frac{7}{3}$$ - Second line: Using point (-1, 7), $7= -\frac{3}{2}(-1)+b \Rightarrow 7= \frac{3}{2}+b \Rightarrow b=7 - \frac{3}{2}=\frac{14}{2} - \frac{3}{2}=\frac{11}{2}$. - Equation: $$y= -\frac{3}{2}x + \frac{11}{2}$$ 5. **Find the intersection by setting the two equations equal:** $$\frac{7}{6}x + \frac{7}{3} = -\frac{3}{2}x + \frac{11}{2}$$ 6. **Solve for $x$:** Multiply both sides by 6 to clear denominators: $$6 \times \left(\frac{7}{6}x + \frac{7}{3}\right) = 6 \times \left(-\frac{3}{2}x + \frac{11}{2}\right)$$ $$7x + 14 = -9x + 33$$ Add $9x$ to both sides: $$7x + 9x + 14 = 33$$ $$16x + 14 = 33$$ Subtract 14 from both sides: $$16x = 19$$ Divide both sides by 16: $$x = \frac{19}{16}$$ 7. **Find $y$ by substituting $x$ back into one of the line equations, e.g., first line:** $$y = \frac{7}{6} \times \frac{19}{16} + \frac{7}{3} = \frac{133}{96} + \frac{224}{96} = \frac{357}{96} = \frac{119}{32}$$ 8. **Final answer:** The lines intersect at $$\left(\frac{19}{16}, \frac{119}{32}\right)$$. This point is the solution to the system of equations represented by the two lines.