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14 Jan 2021, 04:56
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BrentGMATPrepNow
Bunuel
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
A. 5
B. 10
C. 11
D. 20
E. 21
For this question, it's useful to know that |a - b| represents the DISTANCE from point a to point b on the number line.
For example, |3 - 10| = the DISTANCE from 3 to 10 on the number line.
Since |3 - 10| = |-7| = 7, we know that 7 is the distance from 3 to 10 on the number line
So, in this case, |x - 5| = the distance from x to 5
Likewise, since x + 5 = x - (-5), we know that |x + 5| = |x - (-5)|
So, |x + 5| = the distance from x to -5
The equation y = |x + 5| - |x - 5| has two critical points. These are x-values that MINIMIZE the value of |x + 5| and MINIMIZE the value of |x - 5|
First, |x + 5| is minimized when x = -5. That is, when x = -5, |x + 5| = |(-5) + 5| = |0| = 0
Next, |x - 5| is minimized when x = 5. That is, when x = 5, |x - 5| = |5 - 5| = |0| = 0
Let's add these critical points ( x = 5 and x = -5) to the number line.
Notice that these critical points divide the number line into 3 regions.
Let's see what happens to the value of y when x lies in each region.
Let's start with region 1
When x lies in region 1 (e.g., x = -7), notice that the blue bar represents the value of |x - 5|
When x lies in region 1 (e.g., x = -7), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the blue bar must be 10 units LONGER than the red bar
So, it must be true that |x + 5| - |x - 5| = -10
Now let's generalize.
For ANY value of x in region 1, the length of the blue bar will be 10 units LONGER than the red bar.
So, for ANY value of x in region 1, |x + 5| - |x - 5| = -10
---------------------------------------------------------------------
Now let's see what's going on in region 3
When x lies in region 3 (e.g., x = 8), notice that the blue bar represents the value of |x - 5|
When x lies in region 3 (e.g., x = 8), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the red bar must be 10 units LONGER than the blue bar
So, it must be true that |x + 5| - |x - 5| = 10
To generalize, we can say that, for ANY value of x in region 3, |x + 5| - |x - 5| = 10
IMPORTANT ASIDE: At this point, we've shown that |x + 5| - |x - 5| can equal 10 and -10
---------------------------------------------------------------------
Now onto region 2
Let's see what happens when x = -4
If x = -4, then |x + 5| (the red bar) = 1 and |x - 5| = 9
So, |x + 5| - |x - 5| = 1 - 9 = -8
Now let's see what happens when x = 3.5
[img]https://i.imgur.com/1SyK8Ii.png[/img
If x = 3.5, then |x + 5| = 8.5 and |x - 5| = 1.5
So, |x + 5| - |x - 5| = 8.5 - 1.5 = 7
---------------------------------------------------------------------
CONCLUSION: At this point, we've shown that |x + 5| - |x - 5| can equal 10, -10, and all values BETWEEN 10 and -10
There are 21 INTEGERS, from -10 to 10 inclusive.
So, y can have 21 different integer values
Answer: E
Cheers,
Brent
Hi BrentGMATPrepNow
great post just one question why are you using only strict inequality sign, for second region i used -5 ≤ x <5
in this case -10<=x <10 can can take 20 values in second region
for third region i used non strict inequality 5 ≤ x
thank you
btw in what cases to use strict vs non strict inequality sign ?
14 Jan 2021, 12:14
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dave13
Hi BrentGMATPrepNow great post just one question
why are you using only strict inequality sign, for second region i used -5 ≤ x <5
in this case -10<=x <10 can can take 20 values in second region
for third region i used non strict inequality 5 ≤ x
thank you
btw in what cases to use strict vs non strict inequality sign ?
great post just one question why are you using only strict inequality sign, for second region i used -5 ≤ x <5
in this case -10<=x <10 can can take 20 values in second region
for third region i used non strict inequality 5 ≤ x
thank you
btw in what cases to use strict vs non strict inequality sign ?
Good question.
Once I tested x = 5 and x = -5 (resulting in y = 10 and y = -10), I already had two possible values of y.
So it was unnecessary to include x = 5 and x = -5 in my range of values.
30 Dec 2021, 22:17
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Given that \(y=|x+5|-|x-5|\) and we need to find how many integer values can y take
Now, whenever we have more than one Absolute value in the problem then we need to try to reduce the number of cases we need to solve.
We do that by putting the values inside the absolute value = 0 and then try to find the points on the number line where the value inside the absolute value will change sign
|x+5| -> x+5 = 0 => x = -5
|x-5| -> x-5 = 0 => x = 5
So, we can divide the number line into three parts
x < - 5, -5 <= x < 5, x > = 5
-5 to 5.JPG [ 20.07 KiB | Viewed 2457 times ]
Case 1: x < - 5
=> |x+5| = -(x+5) (as x+5 will be negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = -(x+5) - (-(x-5)) = -x -5 +x - 5 = -10
So, for any value of x < -5, y = -10
Case 2: -5 <= x < 5
=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (-(x-5)) = x + 5 + x - 5 = 2x
So, for any value of x between -5 and +5, y = 2x
So, we can take integer and decimal values with .5 in the end of x from -5 to 5 (as y=2x and 0.5*2 will give us integer)
=> x = -5, -4.5, -4, -3.5,...0, 0.5, 1..., 4.5 and corresponding values of y will be
=> y = -10, -9, -8, ....0, 1, 2, ... 9
=> 9-(-10) = 1 = 20 values
Case 3: x >=5
=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = x-5 (as x-5 will be non negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (x-5) = x + 5 - x + 5 = 10
So, y can take 21 integer values from -10 to 10
So, Answer will be E
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
Now, whenever we have more than one Absolute value in the problem then we need to try to reduce the number of cases we need to solve.
We do that by putting the values inside the absolute value = 0 and then try to find the points on the number line where the value inside the absolute value will change sign
|x+5| -> x+5 = 0 => x = -5
|x-5| -> x-5 = 0 => x = 5
So, we can divide the number line into three parts
x < - 5, -5 <= x < 5, x > = 5
Attachment:
-5 to 5.JPG [ 20.07 KiB | Viewed 2457 times ]
Case 1: x < - 5
=> |x+5| = -(x+5) (as x+5 will be negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = -(x+5) - (-(x-5)) = -x -5 +x - 5 = -10
So, for any value of x < -5, y = -10
Case 2: -5 <= x < 5
=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (-(x-5)) = x + 5 + x - 5 = 2x
So, for any value of x between -5 and +5, y = 2x
So, we can take integer and decimal values with .5 in the end of x from -5 to 5 (as y=2x and 0.5*2 will give us integer)
=> x = -5, -4.5, -4, -3.5,...0, 0.5, 1..., 4.5 and corresponding values of y will be
=> y = -10, -9, -8, ....0, 1, 2, ... 9
=> 9-(-10) = 1 = 20 values
Case 3: x >=5
=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = x-5 (as x-5 will be non negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (x-5) = x + 5 - x + 5 = 10
So, y can take 21 integer values from -10 to 10
So, Answer will be E
Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
