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

To find the values of x, first rearrange the inequality:

1. Subtract 36 from both sides:

   $x^2 - 9x - 36 ≼ 0$

2. Factor the quadratic:

   $(x - 12)(x + 3) ≼ 0$

3. Identify the critical points: x = 12 and x = -3.

4. Test intervals between these points:

   • For x < -3, choose x = -4: $(x - 12)(x + 3) = (-16)(-1) > 0$
   • For -3 < x < 12, choose x = 0: $(x - 12)(x + 3) = (-12)(3) < 0$
   • For x > 12, choose x = 13: $(x - 12)(x + 3) = (1)(16) > 0$

5. The valid intervals are from -3 to 12 (including endpoints where the product is 0):

   $-3 ≼ x ≼ 12$

We already found the solutions to parts (a) and (b).

1. From part (a): $x > 2.5$

2. From part (b): $-3 ≼ x ≼ 12$

3. The final solution combines these intervals:

   • Since x must be greater than 2.5 and within -3 to 12:

   Therefore, the solution is:

   $2.5 < x ≼ 12$