Challenge problems - Set solutions 06/23/03

[Analistul]

Hi,

Shouldn't x=2, x=3 and x=5 be tested also (either included in the intervals or tested separately?)

Regards and thanks,

[dbernst]

Analistul,

So all students can benefit from the posts, please include the entire problem and its answer choices. Once posted, an instructor will be glad to address your concerns.

Thanks!

[Analistul]

For the following problem I believe that x=2, x=3 and x=5 should also be considered in the intervals or separately.


[b]"Which of the following sets includes ALL of the solutions of x that will satisfy the equation: ?

|x-2|-|x-3|=|x-5|



Answer
One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change.

In the equation , there are 4 intervals of interest:

x < 2: In this interval, the value inside each of the three absolute value expressions is negative.

2 < x < 3: In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative.

3 < x < 5: In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative.

5 < x: In this interval, the value inside each of the three absolute value expressions is positive.

Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs.

For the first interval, x < 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs):



Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x < 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest).

For the second interval 2 < x < 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative:



Notice, again, that the solution is NOT a valid solution since it lies outside the interval 2 < x < 3.

For the third interval 3 < x < 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative:



The solution x = 4 is a valid solution since it lies within the interval 3 < x < 5.

Finally, for the fourth interval 5 < x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive:



The solution x = 6 is a valid solution since it lies within the interval 5 < x.

We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set. Therefore, C is the correct answer."[/b][/b]

[StaceyKoprince]

If x=2, then |x-2|-|x-3|=|x-5| becomes (2-2) - (2-3) = (2-5) or -1 = -3. So x can't be 2
If x=3, then |x-2|-|x-3|=|x-5| becomes (3-2) - (3-3) = (3-5) or 1 = -2. So x can't be 3
If x=5, then |x-2|-|x-3|=|x-5| becomes (5-2) - (5-3) = (5-5) or 1 = 0. So x can't be 5

[pranjali.deshpande]

what about x=0 ...

If x = 0 then the equation becomes :-

|-2| - |-3| = |-5|

2 - 3 = 5

-1 = 5 => which is clearly wrong....

Is this question even correct????

[jnelson0612]

pranjali, this is a very old thread--can you please post the question? Thank you.