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

by x86
Tags: integer, real, solution, values
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x86
#1
Jan12-14, 05:08 AM
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
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mfb
#2
Jan12-14, 08:17 AM
Mentor
P: 11,819
There are no fixed c,d, such that the equation is true for more than one w. If you know a,b,c,d, you can calculate w, it cannot be more than one value.

c=a+b is impossible, this would make the denominator zero.

Either you try something impossible, or it is unclear what you want to do.


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