1. **State the problem:** We need to solve the simultaneous linear equations using the graph method: $$3x + y = 2$$ $$2x - y = 3$$ 2. **Rewrite each equation in slope-intercept form ($y = mx + c$) to graph easily:** For the first equation: $$3x + y = 2 \implies y = 2 - 3x$$ For the second equation: $$2x - y = 3 \implies -y = 3 - 2x \implies y = 2x - 3$$ 3. **Interpretation:** - The first line has slope $-3$ and y-intercept $2$. - The second line has slope $2$ and y-intercept $-3$. 4. **Graphing:** - Plot the first line by starting at $(0,2)$ on the y-axis and using the slope $-3$ (down 3 units, right 1 unit). - Plot the second line by starting at $(0,-3)$ and using the slope $2$ (up 2 units, right 1 unit). 5. **Find the intersection point:** The solution to the system is the point where the two lines intersect. 6. **Algebraic verification of intersection:** Set the two expressions for $y$ equal: $$2 - 3x = 2x - 3$$ Solve for $x$: $$2 + 3 = 2x + 3x$$ $$5 = 5x$$ $$x = 1$$ Substitute $x=1$ into one of the equations to find $y$: $$y = 2 - 3(1) = 2 - 3 = -1$$ 7. **Final answer:** The solution to the simultaneous equations is: $$\boxed{(1, -1)}$$ This means the lines intersect at the point $(1, -1)$, which satisfies both equations.