Linear and Quadratic Equations

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Counting question with modulus also thrown in. Worry about all those negative possibilities as well.

Counting - Linear Equations

    (|x| - 3) (|y| + 4) = 12
    1. 4
    2. 6
    3. 10
    4. 8

 

  • Correct Answer
    Choice C. 10

Explanatory Answer

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Detailed Solution

If 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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More Questions from Linear & Quad. Equations

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