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Updated on: 13 Sep 2018, 08:16
Originally posted by EgmatQuantExpert on 13 Sep 2018, 03:04.
Last edited by EgmatQuantExpert on 13 Sep 2018, 08:16, edited 2 times in total.
Last edited by EgmatQuantExpert on 13 Sep 2018, 08:16, edited 2 times in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Different methods to solve absolute value equations and inequalities- Exercise Question #1
If x is an integer, then how many values of x will satisfy the equation ||x - 2| + 7| = 6?
Options
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To read the article: Different methods to solve absolute value equations and inequalities
If x is an integer, then how many values of x will satisfy the equation ||x - 2| + 7| = 6?
Options
- a) 0
b) 1
c) 2
d) 3
e) 4
Next Question
To read the article: Different methods to solve absolute value equations and inequalities
12 Oct 2018, 17:25
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Hey Jancekmichal --
The trick here is the idea that once you take the absolute value of something, it must be positive.
So based on that |x + 2| must be positive. This is important because of what follows.
Because an absolute value sign inside an absolute value sign is confusing, let's pretend that |x+2| = y instead. (This is just to simplify the equation. You could leave it as is)
That's going to give you | y + 7| = 6. Now solve that as an actual equation. Remember that with an absolute value sign you need to solve this twice, once as if what's inside the signs is negative and once as if it's positive. That will give you two equations:
y + 7 = 6, so y = -1 and y + 7 = -6, so y = - 13.
BUT WAIT. We said that y = |x + 2|, which we said must always be positive. That means that |x + 2| can't ever equal -6 or - 13, which means that there are no valid values for x that would make this equation true.
The trick here is the idea that once you take the absolute value of something, it must be positive.
So based on that |x + 2| must be positive. This is important because of what follows.
Because an absolute value sign inside an absolute value sign is confusing, let's pretend that |x+2| = y instead. (This is just to simplify the equation. You could leave it as is)
That's going to give you | y + 7| = 6. Now solve that as an actual equation. Remember that with an absolute value sign you need to solve this twice, once as if what's inside the signs is negative and once as if it's positive. That will give you two equations:
y + 7 = 6, so y = -1 and y + 7 = -6, so y = - 13.
BUT WAIT. We said that y = |x + 2|, which we said must always be positive. That means that |x + 2| can't ever equal -6 or - 13, which means that there are no valid values for x that would make this equation true.
General Discussion
Updated on: 17 Sep 2018, 03:47
Originally posted by EgmatQuantExpert on 13 Sep 2018, 03:09.
Last edited by EgmatQuantExpert on 17 Sep 2018, 03:47, edited 2 times in total.
Last edited by EgmatQuantExpert on 17 Sep 2018, 03:47, edited 2 times in total.
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Solution
Given:
- • We are given that x is an integer, and
• We are also given an absolute value equation, ||x - 2| + 7| = 6
To find:
- • We need to find the number of values of x, that satisfy the given equation
Approach and Working:
- • If we observe, we can see that there are two absolute value functions in the equation, ||x - 2| + 7| = 6
• So, let us first solve the outer modulus by substituting |x - 2| as t, which gives
- o |t + 7| = 6
- o t + 7 = 6, if t ≥ -7
- Implies, t = -1
Can we consider this as a possible value of t?
Obviously, No.
Since, the value of t = |x - 2| is always greater than or equal to 0.
Thus, t = -1 is not a possible value
- This is again not possible, since, t cannot be a negative number
• Therefore, the number of possible values of x is 0
Now, let’s see another approach to solve this equation.
- • In the equation, |t + 7| = 6, the minimum value of t = 0, since t = |x - 2|
• So, the minimum value of |t + 7| = |0 + 7| = 7
• Thus, the value of |t + 7| is always greater than or equal to 7
• Therefore, the number of possible values of t and x is 0
Hence, the correct answer is option A.
Answer: A
