Find the set of values of x for which a) 3(x−2) < 8−2x b) (2x−7)(1+x) < 0 c) both 3(x−2) < 8−2x and (2x−7)(1+x) < 0 - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 1

To solve this inequality, we first identify the critical values where the expression equals zero:

1. Setting each factor to zero:

   • From $2x - 7 = 0$, we find x = rac{7}{2}
   • From $1 + x = 0$, we find $x = -1$

2. The critical points are $x = -1$ and x = rac{7}{2}. To determine the intervals:

   • Test regions: Choose points from the intervals: (-∞, -1), (-1, 3.5), (3.5, ∞).
   • Checking these:
     • For $x = -2$ (in (-∞, -1)), $(2(-2)-7)(1-2) = (-11)(-1) > 0$
     • For $x = 0$ (in (-1, 3.5)), $(2(0)-7)(1+0) = (-7)(1) < 0$
     • For $x = 4$ (in (3.5, ∞)), $(2(4)-7)(1+4) = (1)(5) > 0$

3. The solution is where the product is negative:

Thus, the solution for part (b) is:

Answer: -1 < x < rac{7}{2}

We need to find the intersection of the solutions to both parts:

1. From part (a), we have $x < 2.8$.
2. From part (b), we have -1 < x < rac{7}{2}.
3. The intersection of these inequalities is:
   • The lower bound is $-1$ and the upper bound is $2.8$, but since rac{7}{2} = 3.5 which is greater than $2.8$, it does not affect the upper limit. Thus, the combined solution based on the constraints is:

Answer: $-1 < x < 2.8$