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Question: (|x| – 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy these two equations?

(a) 4
(b) 6
(c) 10
(d) 8

DETAILED SOLUTION

f x and y are integers, so are |x| –3 and |y| + 4. So, we start by finding out in how many ways 12 can be written as the product of two integers.
12 can be written as 12 × 1, or 6 × 2, or 3 × 4. To start with, we can eliminate the possibilities where the two terms are negative as |y| + 4 cannot be negative.
Further, we can see that |y| + 4 cannot be less than 4. So, among the values, we can have |y| +4 take values 4, 6 or 12 only, or |y| can take values 0, 2 and 8 only.
When |y| = 0, |x| – 3 = 3, |x| = 6, x can be +6 or -6. Two pairs of values are possible: (6, 0) and (-6, 0)
When |y| = 2, |x| – 3 = 2, |x| = 5, x can be +5 or –5. There are four possible pairs here: (5, 2) , (-5, 2), (5, -2), (-5, -2)
When |y| = 8, |x| – 3 = 1, |x| = 4, x can be +4 or –4. There are four possible pairs here: (4, 8) , (-4, 8), (4, -8), (-4, -8)

Correct Answer: 10 Pairs

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3 Comments

Comments are closed.

  1. Ravikant Yadav 3 years ago

    thanks

  2. Thank u sir❤️

  3. Sir a small doubt at last we have two equations mod x = 6 and mod y =0 so we will be having 3 solutions at last but u said only 2, please explain me this

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