# IMAT 2013 Q59 [Set of Values]

What is the set of values for which 12 - x^2 > 8 and 2x + 3 ≥ 5?

A. 1 ≤ x
B. 1 < x ≤ 2
C. 1 ≤ x < 2
D. 2 < x
E. ­-1 ≤ x < 2

To solve this question we will have two parts.

1. part:
12-x^2>8 \to 12>8 +x^2 \to 12-8 > x^2 \to 4>x^2
Or better typed: x^2<4. This means that -2<x<2.

2. part:
2x +3 \ge 5 \to 2x \ge 5-3 \to 2x \ge 2. Dividing the whole equation with 2 gives us the final form: x\ge 1

Now, when both parts are finished, we need to see which areas collide. The best way to do this is to draw a numeric line:

The area that collides is the solution: (C)1\le x<2

TIP: Put the circles on the numeric line to indicate that the x can take that value, so you don’t exclude it.