doughishere
Junior Member
- Joined
- Dec 18, 2015
- Messages
- 59
Hey all. Im back. Quick question to test my understanding. The problem is, Find all numbers x satisfying the given equation.
The equation is: ∣x+1∣+∣x−2∣=7.
3 Cases:
a) x<−1.
b) x>2.
c) −1≤x≥2.
a) x<−1. In this case x+1<0 which means that ∣x+1∣=−(x+1) and x−2<0 which means that ∣x−2∣=−(x−2).
solve: −(x+1)−(x−2)=7↣(because i wanted to)↣−x+1−x+2=7↣−2x+3=7↣−2x=6↣x=−3.
check: ∣x+1∣+∣x−2∣=7↣∣−3+1∣+∣−3−2∣=7↣∣−2∣+∣−5∣=7↣2+5=7↣7=7.
Thus, x<-1.
b) x>2. In this case x+1>0 which means that ∣x+1∣=(x+1) and x−2>0 which means that ∣x−2∣=(x−2).
solve: (x+1)+(x−2)=7↣x+1+x−2=7↣2x−1=7↣2x=8↣x=4.
check: ∣x+1∣+∣x−2∣=7↣∣4+1∣+∣4−2∣=7↣∣5∣+∣2∣=7↣7=7.
Thus, x >2.
c) −1≤x≥2. In this case x+1>0 which means that ∣x+1∣=(x+1) and x−2<0 which means that ∣x−2∣=−(x−2).
solve: (x+1)−(x−2)=7↣x+1−x+2=7↣3=7. Not a Solution.
Thus the set of numbers x such that ∣x+1∣+∣x−2∣=7 is {−3,4}.
The equation is: ∣x+1∣+∣x−2∣=7.
3 Cases:
a) x<−1.
b) x>2.
c) −1≤x≥2.
a) x<−1. In this case x+1<0 which means that ∣x+1∣=−(x+1) and x−2<0 which means that ∣x−2∣=−(x−2).
solve: −(x+1)−(x−2)=7↣(because i wanted to)↣−x+1−x+2=7↣−2x+3=7↣−2x=6↣x=−3.
check: ∣x+1∣+∣x−2∣=7↣∣−3+1∣+∣−3−2∣=7↣∣−2∣+∣−5∣=7↣2+5=7↣7=7.
Thus, x<-1.
b) x>2. In this case x+1>0 which means that ∣x+1∣=(x+1) and x−2>0 which means that ∣x−2∣=(x−2).
solve: (x+1)+(x−2)=7↣x+1+x−2=7↣2x−1=7↣2x=8↣x=4.
check: ∣x+1∣+∣x−2∣=7↣∣4+1∣+∣4−2∣=7↣∣5∣+∣2∣=7↣7=7.
Thus, x >2.
c) −1≤x≥2. In this case x+1>0 which means that ∣x+1∣=(x+1) and x−2<0 which means that ∣x−2∣=−(x−2).
solve: (x+1)−(x−2)=7↣x+1−x+2=7↣3=7. Not a Solution.
Thus the set of numbers x such that ∣x+1∣+∣x−2∣=7 is {−3,4}.
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