Factorise 12, and think about what integer values the terms on the Left Hand Side can take that satisfy the equation.

1. Linear Equation

2. Linear Equation
(|x| – 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy these
two equations?
(a) 4 (b) 6
(c) 10 (d) 8

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

4. Linear Equation
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)
Totally, 10 pairs of values are possible. Answer choice (c)
(|x| – 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy these
two equations?

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