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Comments
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@Root how much for you to do my homework. The current math teacher isn't hardcore enough. I need a professional to torture me with numbers.
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let a = x*j = y/k
let b = k/j
ab = xk = y/j
since a and b are known, if you find x, you will get k = ab/x and j = a/x, y = ak = a²b/x
That's all the variables
If you know y, j = y/ab, x = a/j = a²b/y, k = y/a
In short, you need one more variable, this is not enough, but this can help you find patterns. -
@theabbie thank you abbie. Figured it was going to be something like that.
Also voice-to-text insisted your handle is "applepie" and not "abbie." -
iiii90854y@Wisecrack not any, but one that gives additional information, so that two if them cannot be reduced to just one.
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Helpful answer, and I'm starting to grasp what you're explaining.
Suppose we have another unknown variable d.
We know d*y and we know d*k, but we don't obviously know d, y or k.
Additionally you know the
value of x/d
But obviously not x or d alone.
Related Rants
If I have four unknown variables, x, y, j, and k, but know the values of x*j, y/k, and k/j, and x*j == y/k
How do I go about getting the values of the individual unknown variables?
random
math