Find the set of values of $x$ for which (a) $3x - 7 > 3 - x$ (b) $x^2 - 9x \leq 36$ (c) both $3x - 7 > 3 - x$ and $x^2 - 9x \leq 36$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

Start by rewriting the inequality:

$x^2 - 9x - 36 \leq 0$

Next, factor the quadratic:

$(x - 12)(x + 3) \leq 0$

To find the critical points, set each factor to zero:

1. $x - 12 = 0 \implies x = 12$
2. $x + 3 = 0 \implies x = -3$

Using a sign chart or interval test, we determine the intervals:

• Test $x < -3$ (say $x = -4$): $(-)(-)$ = $+$ (positive)
• Test $-3 < x < 12$ (say $x = 0$): $(-)(+)$ = $-$ (negative)
• Test $x > 12$ (say $x = 13$): $(+)(+)$ = $+$ (positive)

The solution comes from the interval where the product is non-positive:

$-3 \leq x \leq 12$

From part (a), we found that:

$x > 2.5$

From part (b), we found that:

$-3 \leq x \leq 12$

To find the set of values for $x$ that satisfy both conditions, we take the intersection of the two results:

$2.5 < x \leq 12$

Thus, the final answer for part (c) is:

$2.5 < x \leq 12$