absolute value of(x + 3) + absolute value of (3 - x) < 9.
I am not sure how to go about this question. Although, we generally use three cases.
let's rewrite it as
|x+3| < 9 - |3-x|
x+3 < 9 - |3-x| or -x-3 < 9 - |3-x|
|3-x| < 6-x , #1)
or
|3-x| < 12 + x , (#2)
from #1
3-x < 6-x
no solution
or
-3 + x < 6-x
2x < 9
x < 9/2
from #2
3-x < 12+x
-2x < 9
x > -9/2
or
-3+x < 12 +x
no solution
so we have x< 9/2 and x > -9/2
our critical values of -9/2 and +9/2 divide our number line into 3 parts
1. x < -9/s
2. x between -9/2 and 9/2
3. x > 9/2
I usually pick an arbitrary number in each region and test it in the original
1. let x = -10 --> |-7|< 9 -|13| false
2. let x = 0 ---> |3| < 9 - |3| true
3. let x = 10 --> |13| < 9 - |-7| false
so -9/2 < x < 9/2
Thank You!
To solve this inequality, you are correct that we can use three cases. Let's break down the steps to determine the values of x that satisfy the inequality:
1. Case 1: When both expressions inside the absolute value signs are non-negative (greater than or equal to zero):
- (x + 3) ≥ 0 and (3 - x) ≥ 0
- Solving both inequalities:
- x + 3 ≥ 0 --> x ≥ -3
- 3 - x ≥ 0 --> x ≤ 3
- Combining the two solutions, we have: -3 ≤ x ≤ 3
2. Case 2: When (x + 3) is negative and (3 - x) is non-negative:
- (x + 3) < 0 and (3 - x) ≥ 0
- Solving both inequalities:
- x + 3 < 0 --> x < -3
- 3 - x ≥ 0 --> x ≤ 3
- Combining the two solutions, we have: x < -3
3. Case 3: When (x + 3) is non-negative and (3 - x) is negative:
- (x + 3) ≥ 0 and (3 - x) < 0
- Solving both inequalities:
- x + 3 ≥ 0 --> x ≥ -3
- 3 - x < 0 --> x > 3
- Combining the two solutions, we have: x > 3
Now, we need to consider the overlapping regions of these cases to find the values of x that satisfy the original inequality.
Based on the first case, -3 ≤ x ≤ 3 satisfies the inequality.
Based on the second case, x < -3 is a possible solution.
Based on the third case, x > 3 is a possible solution.
Combining all the cases, the values of x that satisfy the inequality are:
- ∞ < x < -3, -3 ≤ x ≤ 3, 3 < x < ∞
Therefore, the solution to the inequality |x + 3| + |3 - x| < 9 is the set of real numbers that fall into these intervals.