johnnymbikes wrote:
What is wrong with the following in the negative scenario B below?
|2x+3| > |7x-2|
(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1
(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?
This doesn't match ans choice C
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?A. x < -1/9 or x > 5
B. -1 < x < 1/9
C. -1/9 < x < 1
D. -1/9 < x < 5
E. x < -1/9 or x > 1
The critical points (also known as transition points or key points) for |2x + 3| and |7x - 2| are x = -3/2 and x = 2/7, respectively. These critical points are where the expressions inside the modulus equal zero, signifying a transition from negative to positive values (or vice-versa).
Considering the two critical points, we analyze three ranges:
1. If x < -3/2, then 2x + 3 < 0 and 7x - 2 < 0. Thus, |2x + 3| = -(2x + 3) and |7x - 2| = -(7x - 2). Hence, for this range we'd get: -(2x + 3) > -(7x - 2), which gives x > 1. Discard this range because it contradicts with the range we consider: x < -3/2.
2. If -3/2 ≤ x ≤ 2/7, then 2x + 3 ≥ 0 and 7x - 2 ≤ 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = -(7x - 2). Hence, for this range we'd get: 2x + 3 > -(7x - 2), which gives x > -1/9. Combining with the range we consider, we'd get -1/9 < x ≤ 2/7.
3. If x > 2/7, then 2x + 3 > 0 and 7x - 2 > 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = 7x - 2. Hence, for this range we'd get: 2x + 3 > 7x - 2, which gives x < 1. Combining with the range we consider, we'd get 2/7 ≤ x < 1.
So, we get two ranges: -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1, which give: -1/9 < x < 1.
Answer: C.
10. Absolute Value
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Hope it helps.