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Modulus of a number cannot be negative. A simple idea that somehow escapes our mind every now and then.

Counting - Linear Equations

x + |y| = 8, |x| + y = 6. How many pairs of x, y satisfy these two equations?
1. 2
2. 4
3. 0
4. 1

Choice D. 1

Detailed Solution

We start with the knowledge that the modulus of a number can never be negative, though the number itself may be negative.

The first equation is a pair of lines defined by the equations
y = 8 – x        ------- (i) {when y is positive}
y = x – 8        ------- (ii) {when y is negative}

With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be negative, which is not allowed)

The second equation is a pair of lines defined by the equations:
y = 6 – x        ------- (iii) {when x is positive}
y = 6 + x        ------- (iv) {when x is negative}

with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be negative.

On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).

Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative.

Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This satisfies both equations.

Only one solution is possible for this system of equations.

Correct Answer: Only one value is possible.

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