# How can I find a solution for c and d for all real integer values?

by x86
Tags: integer, real, solution, values
 P: 94 $$w = \frac{(ab - d) }{c - a - b}$$ I have to solve the above equation for variables c and d if w can be any number from $$w \in (-\infty, +\infty)$$ If we set w = 0, then w = 1 we can solve for c and d $$0 = ab - d$$ $$d = ab$$ $$c = a + b$$ Now if I can substitute the values to check the solution for w = 1 $$c - a - b = ab - d$$ Substituting c, $$a + b - a - b = ab - d$$ $$0 = ab - d$$ $$d = ab$$ I know that my solution is true for both w = 0 and w = 1 but how can I prove that my solution is true for $$w \in (-\infty, +\infty)$$ I've tried this: $$w(c - a - b) = (ab - d)$$ $$w(a + b - a -b) = ab - d$$ $$0 = ab - d$$ $$ab = d$$ But is this really an acceptable way of solving the solution? I am very confused. I've proved that the equations I found earlier (when I set w = 1 and w = 0) are true when w = w by putting it into the mother equation