how to solve the following equation
Please explain in details how to solve the following equation
How many real numbers x are solutions to the following equation?
|x - 1| = |x - 2| + |x - 3|
Answer
CaseI:
If x < 1, the equation becomes
(1 - x) = (2 - x) + (3 - x)
=> x = 4,
CaseII:
If 1 ≤ x ≤ 2,
we get (x - 1) = (2 - x) + (3 - x)
=> x = 2.
CaseIII:
If 2 ≤ x ≤ 3,
we get (x - 1) = (x - 2) + (3 - x)
=>x = 2.
CaseIV:
If x ≥ 3,
we get (x - 1) = (x - 2) + (x - 3),
=> x = 4. So 2 and 4 are the only
solutions, and the answer is 2.
Please note that x=4
Please note that x=4 solution is valid from CaseIII only since
for CaseI it is arrived at by assuming x<1.
Therefore,for Case I: Assumption : X<1
Outcome : X=4
which is contradictory.
Yes Divya your are right,
Yes Divya your are right, Sorry I Think I have written the above the answer in a confusing way
Let me clarify I think I missed to write the complete solution. Adding to the above comment.
For case I : if we assume x < 1 and solve the eqn we get x = 4 so its not a valid solution as both are contradicting to each other.
(
To verify lets take x = -2 then left side will be 3 and right side 9 so its not a valid solun.
for x = -3, LHS is 4 and RHS 11
It’s clear that there will be difference of 7 between LHS and RHS.)
No solution for this condition.
For case II: I think above comment is not confusing.
and we have one solution i.e 2.
For case III: we have also one solution 2 and its same as case II.
No. Of solutions 1
For case IV: 2 and 4 are the possible solutions but as we have considered x≥3 , 2 cant be the solution. So one possible solution is 4.
With all the conditions we have two solutions - 2 & 4.
So the answer is 2 - we have two possible solutions.