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Absolute fundamentals

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Absolute fundamentals [#permalink]

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New post 10 Mar 2011, 09:06
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?
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Re: Absolute fundamentals [#permalink]

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New post 11 Mar 2011, 11:01
Only two cases have to be considered.

case 1
------
3 + 2x > 4 - x
3x > 1
x > 1/3

case 2
------
3 + 2x > x - 4
x > -7

Taking the most restricted value x > 1/3

rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?

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Re: Absolute fundamentals [#permalink]

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New post 11 Mar 2011, 12:53
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rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?


Above solution is not correct.

\(|3+2x|> |4-x|\)"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are \(-\frac{3}{2}\) and \(4\). Hence we'll have three ranges to check:

A. \(x<-\frac{3}{2}\) --> \(-(3+2x)>{4-x}\) --> \(x<{-7}\);

B. \(-\frac{3}{2}\leq{x}\leq{4}\) --> \(3+2x>4-x\) --> \(x>\frac{1}{3}\), as we consider the range \(-\frac{3}{2}\leq{x}\leq{4}\), then \(\frac{1}{3}<{x}\leq{4}\);

C. \(x>4\) --> \(3+2x>-(4-x)\) --> \(x>{-7}\), as we consider the range \(x>4\), then \(x>4\);

Ranges from A, B and C give us the solution as: \(x<{-7}\) or \(x>\frac{1}{3}\) (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html
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Re: Absolute fundamentals [#permalink]

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New post 11 Mar 2011, 18:22
Bunuel

I got this thoroughly. Please enlighten me on these statements -

a) For a quadratic equation - the value is positive beyond the roots and negative between the roots. In other words the value of the quadratic alternates +/- in between the root intervals.
b) For an inequality - you must examine the root intervals. "sign" of the inequality MAY flip between the root intervals. To determine the exact sign - use the numberline.

I hope I have absorbed the information flawlessly.

thanks

Bunuel wrote:
rxs0005 wrote:
Hi


if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?


Above solution is not correct.

\(|3+2x|> |4-x|\)"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are \(-\frac{3}{2}\) and \(4\). Hence we'll have three ranges to check:

A. \(x<-\frac{3}{2}\) --> \(-(3+2x)>{4-x}\) --> \(x<{-7}\);

B. \(-\frac{3}{2}\leq{x}\leq{4}\) --> \(3+2x>4-x\) --> \(x>\frac{1}{3}\), as we consider the range \(-\frac{3}{2}\leq{x}\leq{4}\), then \(\frac{1}{3}<{x}\leq{4}\);

C. \(x>4\) --> \(3+2x>-(4-x)\) --> \(x>{-7}\), as we consider the range \(x>4\), then \(x>4\);

Ranges from A, B and C give us the solution as: \(x<{-7}\) or \(x>\frac{1}{3}\) (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html

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Re: Absolute fundamentals [#permalink]

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New post 12 Mar 2011, 21:30
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rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?


Bunuel has provided you the algebraic solution. Let me add the graphical approach. It may seem daunting at first, but if you take the effort of understanding it, it will seem very straight forward and easy.
Draw the graphs of both the mods
y = | 3 + 2x |
and y = |4 - x |
as shown below.

Attachment:
Ques2.jpg
Ques2.jpg [ 14.42 KiB | Viewed 4399 times ]

Now where is the graph of | 3 + 2x | > the graph of |4 - x | ?
Where is the Red line above the Purple line?
Can I say it's so for the values of x as depicted by the green arrows?
Now all we need to do is find these points.
Point A: 2x+3 = 4-x which implies x = 1/3
Point B: -2x - 3 = 4-x which implies x = -7
So the given condition is satisfied when x > 1/3 or x < -7

Note 1: For more on how to draw graphs of mods, check:
http://www.veritasprep.com/blog/2011/01 ... h-to-mods/

Note 2: If you are wondering how do we decide whether we need to take (2x+3) or (-2x-3), (4-x) or (x-4) while finding points A and B, notice that we need to find the intersection of 2 lines to find these points.
When we make the graph of | 3 + 2x |, one line is (3+2x), and the other line is (-3-2x). When we make graph of |4 - x |, one line is (4-x) and the other is (x-4).
While finding point A, the Red line is going up from left to right so co-efficient of x must be positive hence we use (2x+3). The purple line is going down from left to right so the co-efficient of x must be negative so we use (4-x).
Similarly for point B, the red line is going down from left to right so co-efficient of x must be negative so we use (-2x-3). The purple line is going down from left to right so the co-efficient of x must be negative so we use (4-x).
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Re: Absolute fundamentals [#permalink]

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New post 12 Mar 2011, 23:30
+1 This is epic. I wish you taught me inequalities back in school. :-D

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Re: Absolute fundamentals [#permalink]

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New post 13 Mar 2011, 08:15
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gmat1220 wrote:
+1 This is epic. I wish you taught me inequalities back in school. :-D


If you like alternative solutions, I suggest you check out:

http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

An even shorter and more intuitive approach to these questions.
Rephrase the question as:

| 2x + 3| - |4 - x | > 0

2| x + 3/2| - |x - 4 | > 0

The '2' outside the mod means 'twice the distance from -3/2'. Also, |4 - x | is the same as |x - 4 |.

Then see if you can figure out the answer from the number line. (I must tell you that it takes some effort to figure out the first time... A few people in my batches have done it though so its not terribly difficult...)
I could give an explanation later if you want to verify...
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Re: Absolute fundamentals [#permalink]

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New post 13 Mar 2011, 12:50
Karishma, as usual - u are great. I would be more than happy to learn more about it.

thanks! +1
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Re: Absolute fundamentals [#permalink]

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New post 14 Mar 2011, 18:36
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144144 wrote:
Karishma, as usual - u are great. I would be more than happy to learn more about it.

thanks! +1


Check out the inequality posts on my blog. I have discussed these methods there in posts titled 'Bagging the Graphs' and 'Holistic approach to mods'
They will help you get comfortable with the basics... Get back in case you have any doubts... Then try and solve this question.... I will give the solution using the method mentioned in the post if you like then....
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Re: Absolute fundamentals [#permalink]

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New post 14 Mar 2011, 23:09
Karishma,
I have read your blog. Pls verify this - I am doing it intuitively.
If a - b > 0 this means at some point a = b. Lets determine that point. a = |2x+3| and b = |4-x|

Solving -
|2x+3| = |4-x|

Case 1
------
2x + 3 = 4 - x
3x = 1
x = 1/3
Therefore to make a > b, x > 1/3. i.e. move x further right of zero.

Case 2
-------
2x + 3 = x - 4
x + 7 = 0
x = -7

Therefore to make a > b, x < -7. i.e. move x further left of zero.

VeritasPrepKarishma wrote:
gmat1220 wrote:
+1 This is epic. I wish you taught me inequalities back in school. :-D


If you like alternative solutions, I suggest you check out:

http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

An even shorter and more intuitive approach to these questions.
Rephrase the question as:

| 2x + 3| - |4 - x | > 0

2| x + 3/2| - |x - 4 | > 0

The '2' outside the mod means 'twice the distance from -3/2'. Also, |4 - x | is the same as |x - 4 |.

Then see if you can figure out the answer from the number line. (I must tell you that it takes some effort to figure out the first time... A few people in my batches have done it though so its not terribly difficult...)
I could give an explanation later if you want to verify...

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Re: Absolute fundamentals [#permalink]

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New post 15 Mar 2011, 19:10
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gmat1220 wrote:
Karishma,
I have read your blog. Pls verify this - I am doing it intuitively.
If a - b > 0 this means at some point a = b. Lets determine that point. a = |2x+3| and b = |4-x|

Solving -
|2x+3| = |4-x|

Case 1
------
2x + 3 = 4 - x
3x = 1
x = 1/3
Therefore to make a > b, x > 1/3. i.e. move x further right of zero.

Case 2
-------
2x + 3 = x - 4
x + 7 = 0
x = -7

Therefore to make a > b, x < -7. i.e. move x further left of zero.


This is absolutely fine but I would not worry about equating... I would let common sense guide me... Let me tell you what I have in mind

Let's focus on just the x axis i.e. the number line

Since you have gone through the post, you know that mod is nothing but distance from 0 on the number line...
|x| = 2 means 'x is at a distance of 2 from point 0 on the number line'
|x-4| = 6 means 'the distance of x from 4 on the number line is 6' so x must be 10 or -2
These are the basics.

Let's take an easier example first. What does the following mean?
|x+2| > |x-4|
It means the value of x is such that distance from -2 is greater than distance from 4.
Attachment:
Ques3.jpg
Ques3.jpg [ 3.01 KiB | Viewed 4157 times ]

Where is the distance from -2 equal to distance from 4? At point 1 on the number line, right? Red and green arrows will be equal.
So if x > 1, red arrow will be longer than green i.e. the distance of the point from -2 will be more than the distance from 4. So x > 1 satisfies this inequality.
What about points on the left of -2? Will there be any point such that its distance from -2 is equal to the distance from 4? Obviously not. All points will be closer to -2 than to 4.
Hence the only region is x > 1.


Now let's take the question at hand:
|2x+3| > |4-x|
2|x+3/2| > |x-4|
It means the value for x is such that twice the distance from -3/2 is more than the distance from 4.
Where will twice the distance from -1.5 be exactly equal to distance from 4?

Attachment:
Ques2.jpg
Ques2.jpg [ 7.97 KiB | Viewed 4156 times ]


We should divide the distance of 5.5 between them in 3 equal parts to get 5.5/3
Now lets go 5.5/3 ahead of -1.5 to get -1.5+5.5/3 = 1/3. This is the point where twice the distance from -3/2 is equal to distance from 4. So you go to the right to make twice the distance from -3/2 greater than the distance from 4. So one solution is x > 1/3
Is there some other point where the same thing will happen?
Yes, at x = -7. How do I get it? because distance between -1.5 and 4 is 5.5
When I add this to the left of -1.5, I get point 7 which is where double the distance from -1.5 will be equal to distance from point 4. To the left of -7, twice the distance from -1.5 will be greater than the distance from 4.
So another solution is x < -7

It is much more intuitive and all you need to do is draw a number line and then reason it out. Let me warn you, it's not everyday that I come across people who are interested in and appreciate alternative strategies. So when I do get an audience, I tend to get a little out of hand... If it makes sense to you, go ahead and try it out.. let me know if you get stuck with anything.. if it doesn't make sense, ignore it...
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Re: Absolute fundamentals [#permalink]

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New post 18 Mar 2011, 00:47
Bunuel wrote:
rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x


3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?


Above solution is not correct.

\(|3+2x|> |4-x|\)"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are \(-\frac{3}{2}\) and \(4\). Hence we'll have three ranges to check:

A. \(x<-\frac{3}{2}\) --> \(-(3+2x)>{4-x}\) --> \(x<{-7}\);

B. \(-\frac{3}{2}\leq{x}\leq{4}\) --> \(3+2x>4-x\) --> \(x>\frac{1}{3}\), as we consider the range \(-\frac{3}{2}\leq{x}\leq{4}\), then \(\frac{1}{3}<{x}\leq{4}\);

C. \(x>4\) --> \(3+2x>-(4-x)\) --> \(x>{-7}\), as we consider the range \(x>4\), then \(x>4\);

Ranges from A, B and C give us the solution as: \(x<{-7}\) or \(x>\frac{1}{3}\) (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html


Hello, here is my solution: testing all possibilities with signs

1) 3+2x>4-x
x>1/3
2) -3-2x>-4+x
x<1/3
3) -3-2x>4-x
-7>x
4)3+2x>4+x
x>-7

but i have read somewhere that we shall only test both positive and positive neg.
so we have solution x>1/3 x>-7 is this correct ?
thx

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Re: Absolute fundamentals [#permalink]

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New post 18 Mar 2011, 01:20
My solution:
Because both \(|3+2x|\) and \(|4-x| >= 0\), I can square the two sides:
\((3+2x)^2 > (4-x)^2\)
\(9+12x+4x^2 > 16-8x+x^2\)
\(3x^2+20x-7 > 0\)
\(x1=\frac{1}{3}\)
\(x2={-7}\)
=> we have two ranges: \(x<{-7}\) or \(x>\frac{1}{3}\)
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Re: Absolute fundamentals [#permalink]

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New post 18 Mar 2011, 02:16
I think this table will help illustrate the solution of Bunuel.
1. \(x<-\frac{3}{2}\)
\(|3+2x|= -(3+2x)\) and \(|4-x|={4-x}\)

2. \(-\frac{3}{2}\leq{x}\leq{4}\)
\(|3+2x|={3+2x}\) and \(|4-x|={4-x}\)

3. \(x>4\)
\(|3+2x|={3+2x}\) and \(|4-x|=-(4-x)\)
Attachments

Table.jpg
Table.jpg [ 6.54 KiB | Viewed 3735 times ]


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