1. **State the problem:** We have two sets of inequalities and equations to analyze. **Problem 8:** Find the region satisfying $$-2 < x \leq 1,$$ $$y > -2,$$ $$y \leq x + 1.$$ **Problem 9(a):** Solve the simultaneous equations $$3y + 2x = 12,$$ $$y = x - 1.$$ **Problem 9(b):** Mark points satisfying $$3y + 2x > 12,$$ $$x < y - 1,$$ $$x \text{ and } y \text{ are integers},$$ $$x < 6.$$ --- 2. **Problem 8: Find the region satisfying all inequalities** - The inequality $$-2 < x \leq 1$$ restricts $$x$$ between $$-2$$ and $$1$$, not including $$-2$$ but including $$1$$. - The inequality $$y > -2$$ means $$y$$ is above the horizontal line $$y = -2$$. - The inequality $$y \leq x + 1$$ means $$y$$ is below or on the line with slope 1 and y-intercept 1. The region $$R$$ is the intersection of these three conditions. --- 3. **Problem 9(a): Solve simultaneous equations** Given: $$3y + 2x = 12$$ $$y = x - 1$$ Substitute $$y = x - 1$$ into the first equation: $$3(x - 1) + 2x = 12$$ $$3x - 3 + 2x = 12$$ $$5x - 3 = 12$$ Add 3 to both sides: $$5x - \cancel{3} + \cancel{3} = 12 + 3$$ $$5x = 15$$ Divide both sides by 5: $$\frac{5x}{\cancel{5}} = \frac{15}{\cancel{5}}$$ $$x = 3$$ Now find $$y$$: $$y = x - 1 = 3 - 1 = 2$$ So the solution is: $$x = 3,$$ $$y = 2.$$ --- 4. **Problem 9(b): Mark points satisfying inequalities** Inequalities: $$3y + 2x > 12,$$ $$x < y - 1,$$ $$x < 6,$$ $$x, y \in \mathbb{Z}$$ (integers). We look for integer points where these inequalities hold. --- **Final answers:** - For problem 8, the region $$R$$ is bounded by $$-2 < x \leq 1$$, above $$y = -2$$, and below or on $$y = x + 1$$. - For problem 9(a), the solution to the simultaneous equations is: $$x = 3,$$ $$y = 2.$$ - For problem 9(b), points with integer coordinates satisfying all inequalities should be marked with a cross on the grid.