1. **State the problem:** We need to plot the lines $y=3x-11$ and $y=-2x+4$ by completing their tables of values, then use the graph to solve the simultaneous equations. 2. **Complete the tables:** For $y=3x-11$: - When $x=0$, $y=3(0)-11=0-11=-11$ - When $x=1$, $y=3(1)-11=3-11=-8$ - When $x=2$, $y=3(2)-11=6-11=-5$ Table: \begin{array}{c|ccc} x & 0 & 1 & 2 \\ y & -11 & -8 & -5 \\ \end{array} For $y=-2x+4$: - When $x=0$, $y=-2(0)+4=0+4=4$ - When $x=1$, $y=-2(1)+4=-2+4=2$ - When $x=2$, $y=-2(2)+4=-4+4=0$ Table: \begin{array}{c|ccc} x & 0 & 1 & 2 \\ y & 4 & 2 & 0 \\ \end{array} 3. **Plot the points:** - For $y=3x-11$: points $(0,-11)$, $(1,-8)$, $(2,-5)$ - For $y=-2x+4$: points $(0,4)$, $(1,2)$, $(2,0)$ 4. **Solve the simultaneous equations algebraically:** Set $3x-11 = -2x+4$ $$3x - 11 = -2x + 4$$ Add $2x$ to both sides: $$3x + 2x - 11 = 4$$ $$5x - 11 = 4$$ Add $11$ to both sides: $$5x = 4 + 11$$ $$5x = 15$$ Divide both sides by $5$: $$\frac{\cancel{5}x}{\cancel{5}} = \frac{15}{5}$$ $$x = 3$$ Substitute $x=3$ into $y=3x-11$: $$y = 3(3) - 11 = 9 - 11 = -2$$ 5. **Final answer:** The solution to the simultaneous equations is $\boxed{(3, -2)}$. This means the two lines intersect at the point $(3, -2)$ on the graph.