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# Equation with Mod

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Equation with Mod [#permalink]  28 Dec 2006, 14:33
Solve :

| |Câ€“1|+|C| | = |C+3|â€“C
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[#permalink]  28 Dec 2006, 14:47
| |Câ€“1|+|C| | = |C+3|â€“C

As |c+3| - c is always positive, the equation becomes:

|Câ€“1|+|C| = |C+3|â€“C

o If c < -3, the equation is simplified to:

-(c-1) - c = -(c+3) - c
<=> -2*c + 1 = -2*c -3
<=> 1 = -3

No solution.

o If -3 < c < 0, the equation is simplified to:

-(c-1) - c = (c+3) - c
<=> -2*c + 1 = 3
<=> c = 2/(-2) = -1

One solution is c = -1

o If 0 < c < 1, the equation is simplified to:

-(c-1) + c = (c+3) - c
<=> 1 = 3

No solution.

o If c > 1, the equation is simplified to:

(c-1) + c = (c+3) - c
<=> 2*c - 1 = 3
<=> c = 2

One solution is c = 2

To conclude, there are 2 solutions for the equation : c = -1 and c = 2.
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Re: Equation with Mod [#permalink]  28 Dec 2006, 14:58
Swagatalakshmi wrote:
Solve :

| |Câ€“1|+|C| | = |C+3|â€“C

Let C<0, then the equation becomes

||-(c-1)+-c|=-(c+3)-c
|-2c+1|=2c-3
2c-1=-2c-3
4c=-2
c=-1/2

Let C>0, using |a+b|=|a|+|b|, the equation becomes

||c-1||+||c||+c=|c+3|
c-1+c+c=c+3
c=2

Solutions for C are 2,-1/2.

I plugged these values into the original equation and they satisfy.
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Re: Equation with Mod [#permalink]  28 Dec 2006, 15:06
Swagatalakshmi wrote:
Solve :

| |Câ€“1|+|C| | = |C+3|â€“C

Let C<0, then the equation becomes

||-(c-1)+-c|=-(c+3)-c
|-2c+1|=2c-3
2c-1=-2c-3
4c=-2
c=-1/2

Let C>0, using |a+b|=|a|+|b|, the equation becomes

||c-1||+||c||+c=|c+3|
c-1+c+c=c+3
c=2

Solutions for C are 2,-1/2.

I plugged these values into the original equation and they satisfy.

Sorry... but ||-1/2-1| + |-1/2|| = |1,5 + 0,5| = 2 and |-1/2 + 3| - (-1/2) = 2,5 + 0,5 = 3. Thus, the solution -1/2 is not correct.
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[#permalink]  28 Dec 2006, 15:09
You are right.

I discounted the scenario where |C+3| will be C+3, when C<0 & C>-4.
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[#permalink] 28 Dec 2006, 15:09
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# Equation with Mod

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