QUESTION IMAGE
Question
solve and check each inequality. you cant check all the solutions, so pick a few samples.
|x| + 2 < 11
|x| < 9
check: x < 9 and x > - 9
6: |6| + 2 = 6 + 2 = 8 < 11
-4: |-4| + 2 = 4 + 2 = 6 < 11
|x| > 12
|x - 2| > 3
x - 2 > 3 or x - 2 < - 3
x > 5 or x < - 1
check:
9: |9 - 2| = 7 > 3
-4: |-4 - 2| = |-6| = 6 > 3
|x| ≤ 4
check:
|x| + 4 < 20
check:
|x| - 6 > 1
check:
|x - 1| < 5
check:
|x + 10| > 5
check:
|12x| ≥ 24
check:
|\frac{x}{3}| < 8
check:
Step1: Recall absolute - value inequality rules
For \(|a|0\)), it is equivalent to \(-b < a < b\); for \(|a|>b\) (\(b>0\)), it is equivalent to \(a>b\) or \(a < - b\).
Step2: Solve each inequality
Apply the above rules to isolate \(x\) in each absolute - value inequality.
Step3: Check solutions
Pick sample values from the solution sets and substitute them into the original inequalities to verify.
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- For \(|x|+2 < 11\):
- Solution: \(|x|<9\), which means \(- 9 < x < 9\).
- Check: Let \(x = 6\), then \(|6|+2=6 + 2=8<11\); let \(x=-4\), then \(|-4|+2=4 + 2=6<11\).
- For \(|x|>12\):
- Solution: \(x>12\) or \(x < - 12\).
- Check: Let \(x = 13\), \(|13|=13>12\); let \(x=-13\), \(|-13|=13>12\).
- For \(|x| + 4<20\):
- Solution: \(|x|<16\), which means \(-16 < x < 16\).
- Check: Let \(x = 10\), \(|10|+4=10 + 4=14<20\); let \(x=-10\), \(|-10|+4=10 + 4=14<20\).
- For \(|x - 1|<5\):
- Solution: \(-5
- Check: Let \(x = 0\), \(|0 - 1|=1<5\); let \(x = 5\), \(|5 - 1|=4<5\).
- For \(|12x|\geq24\):
- Solution: \(|x|\geq2\), which means \(x\geq2\) or \(x\leq - 2\).
- Check: Let \(x = 2\), \(|12\times2|=24\geq24\); let \(x=-2\), \(|12\times(-2)|=24\geq24\).
- For \(|x - 2|>3\):
- Solution: \(x-2>3\) or \(x - 2<-3\), so \(x>5\) or \(x < - 1\).
- Check: Let \(x = 9\), \(|9 - 2|=7>3\); let \(x=-4\), \(|-4 - 2|=6>3\).
- For \(|x|\leq4\):
- Solution: \(-4\leq x\leq4\).
- Check: Let \(x = 0\), \(|0|=0\leq4\); let \(x = 4\), \(|4|=4\leq4\).
- For \(|x|-6>1\):
- Solution: \(|x|>7\), so \(x>7\) or \(x < - 7\).
- Check: Let \(x = 8\), \(|8|-6=2>1\); let \(x=-8\), \(|-8|-6=2>1\).
- For \(|x + 10|>5\):
- Solution: \(x+10>5\) or \(x + 10<-5\), so \(x>-5\) or \(x < - 15\).
- Check: Let \(x = 0\), \(|0 + 10|=10>5\); let \(x=-20\), \(|-20 + 10|=10>5\).
- For \(|\frac{x}{3}|<8\):
- Solution: \(-8<\frac{x}{3}<8\), multiplying all parts by 3 gives \(-24 < x < 24\).
- Check: Let \(x = 10\), \(|\frac{10}{3}|\approx3.33<8\); let \(x=-10\), \(|\frac{-10}{3}|\approx3.33<8\).