#### 16_hope_dylan

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- Thread starter 16_hope_dylan
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What are "r" & "i"? Please describe.

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If you have

Now you say "the two ingredients also need to be in a variable ratio (r/k = i/s, where k and s can be any two numbers that have a sum of 100)". Leaving k as a parameter, s= 100- k so that second equation becomes r/k= i/(100- k). Multiplying both sides by 100- k, i= [(100- k)/k]r= (100/k- 1)r= (100/k)r- r. So now the first equation becomes 0.8r+ 2.1[(100/k)r- r= 0.8r- 2.1r+ (210/k)r= (210/k- 1.3)r= 1.68v. r= 1.68v/(210/k- 1.3)= 1.68kv/(210- 1.3k).

Once you have that value of r, i= (1.68v- 0.8r)/2.1. Of course both r and i will depend on the two parameters, v and k.

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So I kinda get your work but towards the ending it gets a little confusing. What of those equations should I be using?If you haveweightr of ingredient 1 (I presume in grams) and its density is 2.1 grams per cubic cm, then its volume is r/2.1 cubic cm. Similarly if the weight of ingredient 2 is i grams and its density is 0.8 grams per cubic cm then its volume is i/0.8 cubic cm. You want that to fit into volume v cubic cm. so r/2.1+ i/0.8= v. I don't like fractions so I would multiply both sides of the equation by 2.1(0.8)= 1.68. The equation becomes 0.8r+ 2.1i= 1.68v.

Now you say "the two ingredients also need to be in a variable ratio (r/k = i/s, where k and s can be any two numbers that have a sum of 100)". Leaving k as a parameter, s= 100- k so that second equation becomes r/k= i/(100- k). Multiplying both sides by 100- k, i= [(100- k)/k]r= (100/k- 1)r= (100/k)r- r. So now the first equation becomes 0.8r+ 2.1[(100/k)r- r= 0.8r- 2.1r+ (210/k)r= (210/k- 1.3)r= 1.68v. r= 1.68v/(210/k- 1.3)= 1.68kv/(210- 1.3k).

Once you have that value of r, i= (1.68v- 0.8r)/2.1. Of course both r and i will depend on the two parameters, v and k.