|3-x|= x-3, Its solutions are
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Answered by
1
x= 3
|3-3|=3-3
|0|=0
0=0hence proved
Answered by
2
As the LHS of the equation contains multiple modulus fuctions, we would have to define it one by one.
Case 1 :- x>=3.
Then all the modulus opens with positive sign.
So, |x - 3 - x - 2| = 5
|-3 - 2| = 5
5 = 5.
Hence all the values of x >= 3are the solutions.
Case 2 :- - 2 =< x < 3.
Now, |x - 3| opens as 3 - x (because for x less than 3, x - 3 is negative and its modulus will open with negative sign) and |x + 2| opens as x + 2.
So, |3 - x -x -2| = 5.
|1 - 2x| = 5.
If x >1/2, then 2x - 1 = 5.
2x = 4
X = 2.
It can be an answer as it is between 1/2 and 3.
If x <= 1/2, then 1 - 2x = 5.
2x = -4
X = -2
Case 3 :- x<-2.
|x - 3| opens as 3 - x and |x + 2| opens as -(x + 2).
So, |3 - x - (-(x+2))| = 5
Or, |3 - x + x + 2| = 5
Hence, 5 = 5.
So, all the values of x less than -2 are the solutions.
Therefore, the solution is
x <= -2
x = 2
And x >= 3.
Case 1 :- x>=3.
Then all the modulus opens with positive sign.
So, |x - 3 - x - 2| = 5
|-3 - 2| = 5
5 = 5.
Hence all the values of x >= 3are the solutions.
Case 2 :- - 2 =< x < 3.
Now, |x - 3| opens as 3 - x (because for x less than 3, x - 3 is negative and its modulus will open with negative sign) and |x + 2| opens as x + 2.
So, |3 - x -x -2| = 5.
|1 - 2x| = 5.
If x >1/2, then 2x - 1 = 5.
2x = 4
X = 2.
It can be an answer as it is between 1/2 and 3.
If x <= 1/2, then 1 - 2x = 5.
2x = -4
X = -2
Case 3 :- x<-2.
|x - 3| opens as 3 - x and |x + 2| opens as -(x + 2).
So, |3 - x - (-(x+2))| = 5
Or, |3 - x + x + 2| = 5
Hence, 5 = 5.
So, all the values of x less than -2 are the solutions.
Therefore, the solution is
x <= -2
x = 2
And x >= 3.
Digraskarsushant12:
Ooo
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