An algebra problem by Kenneth Gravamen

Assuming it is x+2y=100, range of y is 0-50 when range of x is 100-0 (only even numbers for x, say when y=1, x=98, skipping every odd number) hence the total amount of solution for this situation is 50 seeing y can only be 50 numbers and x is complementary to it. However, with modulus, x and y can be negatives, as such the possibilities are: (+x, +y), (-x, +y), (+x,-y), (-x,-y) we have 4 sets and the total solutions become 4(50)=200

• If $|y|= 50$ $=>$ $2 |y| = 100$ , this implies that , $|x| = 0$ ; which makes one solution for $y = -50$ and $y = 50$

• But from the range of $y = -49$ to $y =49$ there will be two values of $x$ for each value of $y$ ;

$Example --->$ for the value : $y = -20$ $=>$ $2|y| = 40$ $=>$ $|x| = 60$ . Hence $x = 60$ or $x = -60$

• Therefore; from $y = -49$ to $y = 49$ there are,

  $49 ( - ve ) + 1 (zero) + 49 (+ ve ) = 99$ values of y, each of which has two solutions.

Hence , there are $\boxed{[2 \times 99] + 2 = 200}$ , distinct solutions to the given equation.