Modulus Equality Solving Concept -II

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SOME MORE CONCEPTS,

if we have to solve,
|x - 1| = 2

means find the points where y = 2 and y = |x - 1| are equal

|x - 1| consists of two lines

for x \le 1,  \Rightarrow |x - 1| = - (x - 1) = 1 - x

for, x \ge 1, |x - 1| = x - 1


Therefore, solution of the equation have those values of x where y = x - 1 and y = 2 are equal  & y = 1 - x and y = 2 are equal.

therefore, 

x - 1 = 2,
or, x = 3

and 1 - x = 2
or, x = -1
 Therefore solution of |x - 1| = 2 is x = 3 and -1

SECOND METHOD,


Now,

what is |x - a|,
 it is the distance between 'x' and 'a'.

consider, |x - 1| = 2


Straight away from figure we get the solution, 
i.e. x = -1 and 3,


Consider the previous post question,

 |x - 1| + |x + 1| = 3

Means, we have find those value of 'x' from which sum of distance to -1 and 1 equals to 3

We can see that if we take x =2 the sum of distance from -1 and 1 is 4 , 
or, d' denotes distance

the distance of 2 from -1 = 3 , ...................( |x + 1|, put x = 2)
d( - 1,2) = 3
the distance of 2 from 1 = 1,..................... ( |x - 1|, put x = 2)
d(1,2) = 1

 SUM = 4,................................................. (|x + 1| + |x - 1|, put x = -2)
d( - 1,2) + d(1,2) = 3 + 1 = 4


 similarly, 

the distance of -2 from -1 = 1, ......................( |x + 1|, put x = -2)
d( - 1, - 2) = 1

the distance of -2 from 1 = 3, ....................... ( |x - 1|, put x = -2 )
d(1, - 2) = 3

SUM = 4, .................................................. (|x + 1| + |x - 1|, put x = -2)
d( - 1, - 2) + d(1, - 2) = 1 + 3 = 4

and distance between -1 and 1 = 2
d( - 1,1) = 2

if we take x =1
the distance from -1 to 1 = 2, ......................  ( |x + 1|, put x = 1)
d( - 1,1) = 2

and the distance from 1 to 1 is 0, ....................( |x - 1|, put x = 1 )
d(1,1) = 0

SUM = 2, ......................................................(|x + 1| + |x - 1|)
d( - 1,1) + d(1,1) = 2 + 0 = 2

  Therefore, solution lies between 1 and 2 & -2 and -1 ,because we have to find the sum which equals to 3,

By, Hit and Trial Method,

We take x = 3/2,

the distance of 3/2 from -1, = 2 + 1/2 = 2.5, ( |x + 1|)
d\left( { - 1,\frac{3}{2}} \right) = 2.5

the distance of 3/2 from 1, = 0.5,  ( |x - 1| )
d\left( {1,\frac{3}{2}} \right) = 0.5

SUM = 3,  (|x + 1| + |x - 1|)
d\left( { - 1,\frac{3}{2}} \right) + d\left( {1,\frac{3}{2}} \right) = 2.5 + 0.5 = 3

 Therefore, x = 3/2 is the solution of the equation

 Similarly , x = -3/2 is also the solution of the equation ( similar arguments)


if we have the question,
|x - 1| + |x + 1| = 2

how we find the solution,
Since, distance between -1 and 1 = 2 

We take any point between -1 and 1 including -1 and 1 we get the equation satisfied,
How????




 If we take any point x = a between -1 and 1 , clearly sum of distance from -1 to 'a' and 1 to 'a' is equal to 2 (from figure).

and for x = 1 , distance of 1 from 1 is zero
and distance from -1 to 1 is 2
therefore, sum is equal to 2

Similarly for -1.

  therefore solution of |x - 1| + |x + 1| = 2 is,

  x \in [ - 1,1] 

 and if question is,

|x - 1| + |x + 1| = 1

Since, distance between -1 and 1 is 2
d( - 1,1) = 2

therefore, you cant find any value for x for which sum of distance from -1 and 1 will be 1,

Therefore, No Solution




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