Q13/052) x,y are negative integers given : y=10x/(100−x) find all the possible solutions of y

y(100-x) = 10 x
or xy + 10x - 100y = 0
or x(y+10) - 100(y+10) = - 1000
or (x-100)( y+ 10) = - 1000 ..1

y < 0 so y + 10 < 10
x < 0 so x - 100 < -100

- 1000 = 1 * (- 1000) => y = - 9 , x = -900
similarly you can find other solutions
= 2 * (-500) => y = - 8, x = - 400
= 5 * (-200) => y = -5
= 4 * (-250) => y = - 6
= 8 * (-125) => y = -2

y + 10 has to be positive(as x-100) is –ve  and < 10 so these are only solutions