1. Stating the problem: We need to plot the graph of the simultaneous linear equations: $$y - 2x = 5$$ $$2y + x = 0$$ 2. Find the intercepts for the first equation $y - 2x = 5$: - For the y-intercept, set $x=0$: $$y - 2(0) = 5 \\ y = 5$$ - So, the y-intercept is $(0, 5)$. - For the x-intercept, set $y=0$: $$0 - 2x = 5 \\ -2x = 5 \\ x = -\frac{5}{2} = -2.5$$ - So, the x-intercept is $(-2.5, 0)$. 3. Find the intercepts for the second equation $2y + x = 0$: - For the y-intercept, set $x=0$: $$2y + 0 = 0 \\ 2y = 0 \\ y = 0$$ - So, the y-intercept is $(0, 0)$. - For the x-intercept, set $y=0$: $$2(0) + x = 0 \\ x = 0$$ - So, the x-intercept is also $(0, 0)$. 4. Interpretation: - The first line intercepts the y-axis at $(0, 5)$ and the x-axis at $(-2.5, 0)$. - The second line intercepts both axes at the origin $(0,0)$. 5. Using these points, you can plot the two lines and find their point of intersection graphically. 6. To find the exact solution algebraically: - From the second equation: $$2y + x = 0 \Rightarrow x = -2y$$ - Substitute into the first: $$y - 2(-2y) = 5 \Rightarrow y + 4y = 5 \Rightarrow 5y = 5 \Rightarrow y = 1$$ - Then $$x = -2(1) = -2$$ 7. So, the solution where the two lines intersect is $(-2, 1)$. Final answer: - First line intercepts at $(0,5)$ and $(-2.5,0)$. - Second line intercepts at $(0,0)$. - Intersection point is $(-2,1)$.