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How do you solve this algebraically?

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by shankar.ashwin » Wed Sep 14, 2011 8:42 pm
Whenever you remove a modulus, you need to consider both the positive and negative signs,
you always have 2 cases in such problems.

Case 1:

x-4=4-x
2x=8
x=4.

Case 2:

-x+4=4-x
x=0
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
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by tttrn333 » Wed Sep 14, 2011 8:47 pm
in case two x can be anything.
shankar.ashwin wrote:Whenever you remove a modulus, you need to consider both the positive and negative signs,
you always have 2 cases in such

problems.

Case 1:

x-4=4-x
2x=8
x=4.

Case 2:

-x+4=4-x
x=0
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
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by shankar.ashwin » Wed Sep 14, 2011 8:52 pm
Yes indeed. X could take any value. Agree..
Could you post the actual problem.
tttrn333 wrote:in case two x can be anything.
shankar.ashwin wrote:Whenever you remove a modulus, you need to consider both the positive and negative signs,
you always have 2 cases in such

problems.

Case 1:

x-4=4-x
2x=8
x=4.

Case 2:

-x+4=4-x
x=0
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
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by Anurag@Gurome » Wed Sep 14, 2011 8:55 pm
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.

Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
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by tttrn333 » Wed Sep 14, 2011 8:58 pm
Anurag,

Do we use this "method" when there appears to be a infinite solution?

How does your way relate to the what we are normally taught. Case1, case2 then test values.
Anurag@Gurome wrote:
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.

Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
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by Anurag@Gurome » Wed Sep 14, 2011 8:59 pm
shankar.ashwin wrote:Whenever you remove a modulus, you need to consider both the positive and negative signs,
you always have 2 cases in such problems.

Case 1:

x-4=4-x
2x=8
x=4.

Case 2:

-x+4=4-x
x=0
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
'Case 2:' maths is not correct.
-x+4 = 4-x does not mean x = 0. It only means LHS = RHS.
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by Anurag@Gurome » Wed Sep 14, 2011 9:02 pm
tttrn333 wrote:Anurag,

Do we use this "method" when there appears to be a infinite solution?

How does your way relate to the what we are normally taught. Case1, case2 then test values.
Anurag@Gurome wrote:
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.



Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
Use this method when there is a modulus sign.
You can easily verify that the equation will not hold for x > 4.
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by shankar.ashwin » Wed Sep 14, 2011 9:02 pm
Ya, completely agree.
But the catch here is to consider values only less than 4,
Anurag@Gurome wrote:
shankar.ashwin wrote:Whenever you remove a modulus, you need to consider both the positive and negative signs,
you always have 2 cases in such problems.

Case 1:

x-4=4-x
2x=8
x=4.

Case 2:

-x+4=4-x
x=0
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.
'Case 2:' maths is not correct.
-x+4 = 4-x does not mean x = 0. It only means LHS = RHS.
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by tttrn333 » Fri Sep 16, 2011 12:42 am
Anurag,

I just came across this problem that doesn't quite work with your method, can you explain if I'm doing it correctly?

|X^2-16|=x-4

If X^2-16>0 then X>4 or X<4 and X^2-16=x-4 Solve=> you get X=4 or X=-3. Neither fit the constraint either X>4 or X<-4

Similarly, X^2-16<0 X>-4 X<4 Solve=> X=4 or -5. Again the same issue.


Anurag@Gurome wrote:
tttrn333 wrote:Anurag,

Do we use this "method" when there appears to be a infinite solution?

How does your way relate to the what we are normally taught. Case1, case2 then test values.
Anurag@Gurome wrote:
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.



Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
Use this method when there is a modulus sign.
You can easily verify that the equation will not hold for x > 4.
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by Abhishek009 » Fri Sep 16, 2011 6:53 am
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.

Let's plugin in some arbitrary values for X ....


CASE - I

Plugin an value of x less than 4...

Say x = 2

Now put it the equation


|X-4|=4-x

|2 - 4 | = 4 -2

2 = 2 { Makes sense }

CASE - II

Again take a value more than 4

Say x = 5


|X-4|=4-x

|5 - 4|= 4-5

1 = -1 { Doesn't make any sense }


CASE III

Now take value of x as 4...

|X-4|=4-x

|4-4| = 4-4

0 = 0 { Makes sense }


CASE IV

Take -ve values of x

Say x = -2

|X-4|=4-x

|-2-4| = 4 + 2

6 = 6 { Makes sense }


So from the above conclusions we can safely say that , we can find the value of x only when x is either equal to 4 or less than 4......


Though the method I hav used seems a bit time consuming , but one can solve such problems in 20 - 30 seconds mentally....
Abhishek
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by Anurag@Gurome » Fri Sep 16, 2011 10:11 pm
tttrn333 wrote:Anurag,

I just came across this problem that doesn't quite work with your method, can you explain if I'm doing it correctly?

|X^2-16|=x-4

If X^2-16>0 then X>4 or X<4 and X^2-16=x-4 Solve=> you get X=4 or X=-3. Neither fit the constraint either X>4 or X<-4

Similarly, X^2-16<0 X>-4 X<4 Solve=> X=4 or -5. Again the same issue.


Anurag@Gurome wrote:
tttrn333 wrote:Anurag,

Do we use this "method" when there appears to be a infinite solution?

How does your way relate to the what we are normally taught. Case1, case2 then test values.
Anurag@Gurome wrote:
tttrn333 wrote:|X-4|=4-x?

I did the two case thing but when X-4=-4+X, there seems to be an infinite possibility.



Solution:
lxl = +x if x >= 0 and = -x if x < 0
So, lx-4l = x-4 if x-4 >= 0 or x >= 4.
This implies x-4 = 4-x if x >= 4 or x = 4 is one solution.
Next, lx-4l = 4-x if x-4 < 0 or if x < 4.
So, given equation becomes 4-x = 4-x if x < 4.
This implies x < 4.
So, the solution is x <= 4.
Use this method when there is a modulus sign.
You can easily verify that the equation will not hold for x > 4.
The method definitely works.
The problem given by you is lx^2-16l = x-4.
This implies that the equation becomes
case 1) x^2 - 16 = x - 4 if x^2 - 16 >= 0 or if either x >= 4 or x <= -4. (Note the equality sign which you have missed)
=> (x-4)(x+3) = 0 if x >= 4 or x <= -4.
Now, x = 4 or -3. So, x = 4 is a valid solution because this satisfies the inequality x >= 4.

Case 2) -x^2 + 16 = x - 4 if x^2 - 16 < 0 or if -4 < x < 4.
=> x^2 + x - 20 = 0 if -4 < x < 4
=> (x+5)(x-4) = 0 if -4 < x < 4
=> x = -5 or 4 if -4 < x < 4.
Now this is not possible. So, case 2) is rejected.

Only solution is x = 4.
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