- #1
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$$w = \frac{(ab - d) }{c - a - b}$$
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
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