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I'm not really a maths guy, apologies if this is either trivial or impossible

If I have 2 formulas, say (ax + cxy) / (ax + b + cxy + d) = r and (cxy + dy) / (ax + b + cxy + d) = s

and I need to find both x and y, how would I go about this, can it be done?

I'm trying to find the adjustments needed to make 2 sets of properties fit desired ratios The first property would put it in either a or c if true, b or d if not The second property would put it in either a or b if true, c or d if not

Maybe this would work as an analogy Say you had 2 teams of people and wanted to work out wage adjustments as ratios needed so that Team 1 earned 60% of the total wages and Team2 40%, and the split between men and women was equal.

In this case A and C are Team 1, and C and D are women (I'm not sure if I'm confusing things by splitting into 4 groups?)

x would then be the multiplier for Team 1s wages and y would be the multiplier to womens wages

gubitza
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1 Answers1

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Notice that each ratio has the common denominator $ax + b + cxy + d.$

Therefore, its quite easy to subtract s from r giving

$$(r - s) = \frac{(ax + cxy) - (cxy + dy)}{ax + b + cxy + d} = \frac{ax - dy}{ax + b + cxy + d}$$

Then you can multiply and expand out the terms as follows:

$$(r-s)ax + (r-s)b + (r-s)cxy + (r-s)d = ax - dy$$

And solving for y looks something like this,

$$((r - s)cx + d)y = (r - s + 1)ax - (r - s)b$$

$$y = \frac{(r - s + 1)ax - (r - s)b}{(r - s)cx + d}$$

The variable x can be chosen arbitrarily, and y is given by the formula above. I'm not quite sure what you meant by the properties, but you'd certainly need something else to find exact values for x and y. Hope this helps

Luca Armstrong
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  • Thanks for replying, sorry if my question was poorly worded I'm trying to figure out if it works now, am I right in thinking that this is the same as ((r-s+1)(ax))-((r-s)b) / ((r-s)((c*x)+d)) (I'm not sure how to format my formulas properly here, apologies) – gubitza Feb 12 '23 at 15:13
  • Not quite, $(r-s)cx + d$ isn't the same as $(r-s)(cx+d).$ If you're trying to cancel the (r-s) factors, I'm afraid you can't because of the denominator. You should also try using MathJax notation when writing questions or comments. – Luca Armstrong Feb 12 '23 at 17:44
  • I think the trouble I'm having is where the brackets would go. e.g. (r-s) + 1 is different to r - (s+1), so just r-s+1 could be either. Is it possible to put in idiot proof brackets? – gubitza Feb 24 '23 at 21:40
  • The expression $r-s+1$ is the same thing as $(r-s)+1$, however, $r-(s+1)$ is equal to $r-s-1$ which is different. – Luca Armstrong Feb 26 '23 at 10:51