l 4x-2 l < l x-2 l   find x values.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

l 4x-2 l < l x-2 l

We have 4 cases:

case(1):

4x -2 < x-2

==> 3x < 0

==> x < 0

==> x= ( -inf , 0)

 

case(2):

-(4x-2) < x-2

-4x + 2 < x-2

-5x < -4

==> x < 4/5

==> x= ( -inf , 4/5)

 

case(3) :

-(4x-2) < -(x-2)

-4x + 2 < -x + 2

-3x < 0

==> x < 0

==> x= ( -inf, 0)

 

case(4):

4x-2 <-(x-2)

4x-2 < -x + 2

==> 5x < 4

==> x < 4/5

==> x = ( -inf , 4/5)

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

|4x-2| < |x-2|.

When x < 1/2 ,  |4x-2| < |x-2|  implies 2-4x < 2-x.

0 <  3x

x > 0 when  x < 1/2. So 0 < x < 1/2 is a solution.

When x = 1/2,  0 < |x-2| ==> 0 < 2-x .  So x < 2. Therefore , x = 1/2 is  a solution. (Please note that the symbol ==> stands for implies, as the symbol , => looks like equal or greater than.)

When 1/2 < x < 2, |4x-2| < |x-2| ==>  4x-2 < 2-x , or  5x < 4 . So x  < 4/5 is a solution. Therefore 1/2 < x < 4/5 is a solution.

When x = 2. |4x-2| < |x-2| ==> 4x-2 < 0, or x < 2/4. So  x = 1/2. But x = 1/2 is a solution already covered. But x = 1/2 when x = 2 is contradiction.

When x > 2, |4x-2| < |x-2| ==> 4x-2 > x-2 , Or 4x-x = 0, or x= 0 does not go together with  when x > 2.

So the solution is    0 < x 1/2  or  1/2 < x < 4/5.

Combining  we get:  0 < x < 4/5 is a solution.

Therefore  solution  set is   ] 0  , 4/5 [ which is both left open and right open interval .

Tally :

We take a value x < 0, say  x = -2,  |4x-2| = |4*-2-2| = |-10| = 10. The RHS = |-4-2| = 6. So 10 > 6 the inequality does not hold.

At x = 0, we get LHS |4*0-2|  = 2 and RHS |0-2 | = 2. So it becomes an equality.

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