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: \(\displaystyle |x+1|+|x-2| = 7\).
3 Cases:
a) \(\displaystyle x < -1\).
b) \(\displaystyle x > 2\).
c) \(\displaystyle -1 \le x \ge 2\).
a) \(\displaystyle x < -1\). In this case \(\displaystyle x+1 < 0\) which means that \(\displaystyle |x+1| = -(x+1)\) and \(\displaystyle x-2 < 0\) which means that \(\displaystyle |x-2| = -(x-2)\).
solve: \(\displaystyle -(x+1)-(x-2) = 7 \rightarrowtail \)(because i wanted to)\(\displaystyle \rightarrowtail -x+1-x+2 = 7 \rightarrowtail -2x+3 = 7 \rightarrowtail -2x = 6 \rightarrowtail x = -3\).
check: \(\displaystyle |x+1|+|x-2| = 7 \rightarrowtail |-3+1|+|-3-2| = 7 \rightarrowtail |-2|+|-5| = 7 \rightarrowtail 2+5 = 7 \rightarrowtail 7 = 7\).
Thus, x<-1.
b) \(\displaystyle x > 2\). In this case \(\displaystyle x+1 > 0\) which means that \(\displaystyle |x+1| = (x+1)\) and \(\displaystyle x-2 > 0\) which means that \(\displaystyle |x-2| = (x-2)\).
solve: \(\displaystyle (x+1)+(x-2) = 7 \rightarrowtail x+1+x-2 = 7 \rightarrowtail 2x-1 = 7 \rightarrowtail 2x = 8 \rightarrowtail x = 4\).
check: \(\displaystyle |x+1|+|x-2| = 7 \rightarrowtail |4+1|+|4-2| = 7 \rightarrowtail |5|+|2| = 7 \rightarrowtail 7 = 7\).
Thus, x >2.
c) \(\displaystyle -1 \le x \ge 2\). In this case \(\displaystyle x+1 > 0\) which means that \(\displaystyle |x+1| = (x+1)\) and \(\displaystyle x-2 < 0\) which means that \(\displaystyle |x-2| = -(x-2)\).
solve: \(\displaystyle (x+1)-(x-2) = 7 \rightarrowtail x+1-x+2 = 7 \rightarrowtail 3 \neq 7\). Not a Solution.
Thus the set of numbers x such that \(\displaystyle |x+1|+|x-2| = 7\) is \(\displaystyle \{-3,4\}\).
The equation is: \(\displaystyle |x+1|+|x-2| = 7\).
3 Cases:
a) \(\displaystyle x < -1\).
b) \(\displaystyle x > 2\).
c) \(\displaystyle -1 \le x \ge 2\).
a) \(\displaystyle x < -1\). In this case \(\displaystyle x+1 < 0\) which means that \(\displaystyle |x+1| = -(x+1)\) and \(\displaystyle x-2 < 0\) which means that \(\displaystyle |x-2| = -(x-2)\).
solve: \(\displaystyle -(x+1)-(x-2) = 7 \rightarrowtail \)(because i wanted to)\(\displaystyle \rightarrowtail -x+1-x+2 = 7 \rightarrowtail -2x+3 = 7 \rightarrowtail -2x = 6 \rightarrowtail x = -3\).
check: \(\displaystyle |x+1|+|x-2| = 7 \rightarrowtail |-3+1|+|-3-2| = 7 \rightarrowtail |-2|+|-5| = 7 \rightarrowtail 2+5 = 7 \rightarrowtail 7 = 7\).
Thus, x<-1.
b) \(\displaystyle x > 2\). In this case \(\displaystyle x+1 > 0\) which means that \(\displaystyle |x+1| = (x+1)\) and \(\displaystyle x-2 > 0\) which means that \(\displaystyle |x-2| = (x-2)\).
solve: \(\displaystyle (x+1)+(x-2) = 7 \rightarrowtail x+1+x-2 = 7 \rightarrowtail 2x-1 = 7 \rightarrowtail 2x = 8 \rightarrowtail x = 4\).
check: \(\displaystyle |x+1|+|x-2| = 7 \rightarrowtail |4+1|+|4-2| = 7 \rightarrowtail |5|+|2| = 7 \rightarrowtail 7 = 7\).
Thus, x >2.
c) \(\displaystyle -1 \le x \ge 2\). In this case \(\displaystyle x+1 > 0\) which means that \(\displaystyle |x+1| = (x+1)\) and \(\displaystyle x-2 < 0\) which means that \(\displaystyle |x-2| = -(x-2)\).
solve: \(\displaystyle (x+1)-(x-2) = 7 \rightarrowtail x+1-x+2 = 7 \rightarrowtail 3 \neq 7\). Not a Solution.
Thus the set of numbers x such that \(\displaystyle |x+1|+|x-2| = 7\) is \(\displaystyle \{-3,4\}\).
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