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03 Aug 2019, 22:40
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A
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Difficulty:
85%
(hard)
Question Stats:
51% (02:18) correct
49%
(02:23)
wrong
based on 437
sessions
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3y + 2|x| = 33. How many positive integral values of (x, y) are possible?
A. 5
B. 6
C. 10
D. 11
E. 12
A. 5
B. 6
C. 10
D. 11
E. 12
04 Aug 2019, 11:09
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How many positive integral values of (x, y) are possible?
Solution:
x,y must be positive and has to be an integer
\(3y + 2|x| = 33\)
\(2|x| = 33-3y\)
in the equation Right hand side must be an even number to balance LHS, for which y must be an odd number
=>since, odd-odd=even => 33 is odd... 3y must be odd
y cannot be >10 because |x| has to be >0 ( positive and integer)
y can be [1,3,5,7,9]
Answer A
To explain further \(2|x| = 33-3y\) => \(|x| = \frac{{33-3y}}{{2}}\) (To make |x| an even integer, \(\frac{{33-3y}}{{2}}\) should be equal to an even integer
\(2|x| = 33-3y\)
2|x| = 33-3=30=2*15
2|x| = 33-9=24=2*12
2|x| = 33-15=18=2*9
2|x| = 33-21=12=2*6
2|x| = 33-27=6=2*3
Solution:
x,y must be positive and has to be an integer
\(3y + 2|x| = 33\)
\(2|x| = 33-3y\)
in the equation Right hand side must be an even number to balance LHS, for which y must be an odd number
=>since, odd-odd=even => 33 is odd... 3y must be odd
y cannot be >10 because |x| has to be >0 ( positive and integer)
y can be [1,3,5,7,9]
Answer A
To explain further \(2|x| = 33-3y\) => \(|x| = \frac{{33-3y}}{{2}}\) (To make |x| an even integer, \(\frac{{33-3y}}{{2}}\) should be equal to an even integer
\(2|x| = 33-3y\)
2|x| = 33-3=30=2*15
2|x| = 33-9=24=2*12
2|x| = 33-15=18=2*9
2|x| = 33-21=12=2*6
2|x| = 33-27=6=2*3
General Discussion
17 Jul 2021, 10:27
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3y + 2|x| = 33 and we need to find positive integral values of (x,y)
Since, x is positive so |x| = x
=> 3y + 2|x| = 33 will become
3y + 2x = 33
=> 2x = 33-3y
=> x = \(\frac{33-3y}{2}\)
Now, for x to be an integer 33-3y should be divisible by 2
33 is odd so 3y also has to be odd to make 33-3y as even.
=> y will be odd number and 0 < 3y < 33
=> 0 < y < 11
So there are 5 possible values of y (1,3,5,7,9) (so there will be 5 positive value of x which are possible too)
So, answer will be A
Hope it helps!
Watch the following video to learn Basics of Absolute Values
Since, x is positive so |x| = x
=> 3y + 2|x| = 33 will become
3y + 2x = 33
=> 2x = 33-3y
=> x = \(\frac{33-3y}{2}\)
Now, for x to be an integer 33-3y should be divisible by 2
33 is odd so 3y also has to be odd to make 33-3y as even.
=> y will be odd number and 0 < 3y < 33
=> 0 < y < 11
So there are 5 possible values of y (1,3,5,7,9) (so there will be 5 positive value of x which are possible too)
So, answer will be A
Hope it helps!
Watch the following video to learn Basics of Absolute Values
