In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
re 1 shows a line $l$ with equation $x+y=6$ and a curve $C$ with tion $y=6 x-2 x^{2}+1$
\[
y=6-x
\]
line $l$ intersects the curve $C$ at the points $P$ and $Q$ as shown in Figure 1 Find, using algebra, the coordinates of $P$ and the coordinates of $Q$. region $R$, shown shaded in Figure 1 , is bounded by $C, l$ and the $x$-axis. Use inequalities to define the region $R$.

Step 1 ：Set the equations of the line and the curve equal to each other and solve for x: \(x + y = 6\) and \(y = -2x^{2} + 6x + 1\).

Step 2 ：Solve the equation to find the x-values: \(x = 1\) and \(x = \frac{5}{2}\).

Step 3 ：Substitute the x-values back into either the equation of the line or the curve to find the corresponding y-values: \(y = 5\) and \(y = \frac{7}{2}\).

Step 4 ：The coordinates of points P and Q are \((1, 5)\) and \(\left(\frac{5}{2}, \frac{7}{2}\right)\) respectively.

Step 5 ：Define the region R using inequalities: \(0 \leq y \leq 6-x\) for \(0 \leq x \leq 1\) and \(0 \leq y \leq 6x - 2x^{2} + 1\) for \(1 \leq x \leq \frac{5}{2}\).

Step 6 ：\(\boxed{\text{The coordinates of P are }(1, 5)\text{ and the coordinates of Q are }\left(\frac{5}{2}, \frac{7}{2}\right)\text{. The region R is defined by the inequalities }0 \leq y \leq 6-x\text{ for }0 \leq x \leq 1\text{ and }0 \leq y \leq 6x - 2x^{2} + 1\text{ for }1 \leq x \leq \frac{5}{2}\text{.}}\)