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# Silly doubt about Modulus questions

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01 Feb 2012, 05:13
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Silly doubt

When we solve the below equation , how many possible solutions we have?
|x – 3| = |2x – 3|

Which of the below cases we need to consider when we face such a equation?
Case 1
+LHS = +RHS
Case 2
-LHS = -LHS
Case 3
-LHS = +RHS
Case 4
LHS = -RHS
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01 Feb 2012, 05:20
rohitgoel15 wrote:
Silly doubt

When we solve the below equation , how many possible solutions we have?
|x – 3| = |2x – 3|

Which of the below cases we need to consider when we face such a equation?
Case 1
+LHS = +RHS
Case 2
-LHS = -LHS
Case 3
-LHS = +RHS
Case 4
LHS = -RHS

|x – 3| = |2x – 3| can take only two forms: when they expand with the same sign (notice that ++ and -- will give you the same exact expression) and when they expand with the different signs (notice that +- and -+ will give you the same exact expression).

Case 1: x-3=2x-3 --> x=0;
Case 1: x-3=-(2x-3) --> x=2.

Check Tough and Tricky Absolute values and Inequalities Questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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02 Feb 2012, 01:39
1
KUDOS
Expert's post
rohitgoel15 wrote:
Silly doubt

When we solve the below equation , how many possible solutions we have?
|x – 3| = |2x – 3|

Which of the below cases we need to consider when we face such a equation?
Case 1
+LHS = +RHS
Case 2
-LHS = -LHS
Case 3
-LHS = +RHS
Case 4
LHS = -RHS

Or you can forget all cases and do it logically!
|x – 3| means distance from point 3 on the number line.
|2x – 3| = 2|x – 3/2| means twice of distance from 1.5 on number line.

Where will these distances be equal?

It will be at a third of the distance between 1.5 and 3 and also at 1.5 to the left of 1.5 as shown in the two diagrams below:
Attachment:

Ques3.jpg [ 6.93 KiB | Viewed 1271 times ]

i.e. x can take two values 2 and 0.

For more on the theory of mods, check:
http://www.veritasprep.com/blog/2011/01 ... edore-did/
http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
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05 Feb 2012, 21:33
Thanks Karishma and Bunuel. Kudos to both !
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