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4 Answers
- cidyahLv 75 years ago
2x-1 > 0 and x+1 > 0
|2x-1| = 2x-1 and |x+1| = x+1
2x-1 + 3(x+1) > 5
2x-1+3x+3 > 5
5x+2 > 5
5x > 3
x > 3/5 -----(1)
2x-1 > 0 and x+1 < 0
|2x-1| = 2x-1 and |x+1| = -(x+1)
2x-1 + 3(-x-1) > 5
2x-1-3x-3 > 5
-x -4 > 5
-(x+4) > 5
(x+4) < -5
x < - 9 -------(2)
2x-1 < 0 and x+1 > 0
|2x-1| = -(2x-1) and |x+1| = x+1
-(2x-1)+3(x+1) > 5
-2x+1+3x+3 > 5
x + 4 > 5
x > 1 -----(3)
2x-1 < 0 and x+1 < 0
|2x-1| = -(2x-1) and |x+1| = -(x+1)
-2x+1 +3(-x-1) > 5
-2x+1-3x-3 > 5
-5x-2 > 5
-5x > 7
5x < -7
x < -7/5 ------(4)
from (1) & (3) x > 3/5
from (2) & (4) x > -7/5
x < -7/5 & x > 3/5
- PinkgreenLv 75 years ago
|2x-1|=2x-1, if x>=1/2
.........=1-2x, if x<=1/2
3|x+1|=3x+3, if x>=-1
..........=-3x-3, if x<=-1
=>
1-2x-3x-3>5=>x<-7/5
3x+3+2x-1>5=>x>3/5
=>
|2x-1|+3|x+1|>5 has the
solution as
x<-7/5 or x>3/5
- L. E. GantLv 75 years ago
Think of the situations you have to cover...
|2x -1| ==> 2x -1 and -(2x-1)
|x+1| ==> (x+1) and -(x+1)
So you could break it into four sets
(1)...(2x-1) + 3(x+1) > 5
==> 2x -1 + 3x + 3 > 5
==> 5x > 3
==> x > 3/5
(2)... (2x-1) -3(x+1) > 5
==> 2x -1 -3x -3 > 5
==> -x > 9
==> x > -9
(3)... -(2x-1) + 3(x+1) >5
==> -2x + 1 + 3x + 3 > 5
==> x > 1
(4)... -(2x-1) -3(x+1) > 5
etc.
