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

We first rearrange the inequality to:

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

Next, we can factor the quadratic:

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

The critical points are $x = 12$ and $x = -3$. We can analyze the intervals based on these points to determine where the inequality holds:

1. Test interval $(-\infty, -3)$: Choose $x = -4$,

   • Result: $(-4 - 12)(-4 + 3) > 0$.

2. Test interval $(-3, 12)$: Choose $x = 0$,

   • Result: $(0 - 12)(0 + 3) < 0$.

3. Test interval $(12, \infty)$: Choose $x = 13$,

   • Result: $(13 - 12)(13 + 3) > 0$.

The solution to part (b) is:

$$-3 \leq x \leq 12.$$

From part (a), we have:

$$x > 2.5$$

From part (b), we found:

$$-3 \leq x \leq 12.$$

To find the intersection of these two results, we combine them:

1. The value of $x$ must be greater than 2.5.
2. The value of $x$ must also be at most 12.

Thus, the solution for part (c) is:

$$2.5 < x \leq 12.$$