Help, Math-Pa, how to solve |2x-1|+3|x+1|>5?

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

|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

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.

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