## Algebra problem with two variables

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
tbmac61
Posts: 1
Joined: Tue Oct 29, 2013 7:31 pm

### Algebra problem with two variables

Hi, I'm a new member and working on a project at work and not coming up with an answer.
We mix two products together with different known densities.
An example is the 2 individual ingredient densities are 48 and 70.
After they are mixed the known total density equals an example of 52.
My question is:
I need a formula to figure out how many pounds of each density material do I need to get to the total mixed density of 52.
The total pounds of both ingredients must equal 2000 lbs.
I have the answer by way of trial and error.
The answer is 1636 lbs of 48 lb material and 364 lbs of 70 lb material.
This gives me
1636x48=78528
364x70=25480
The 1636 + 364 = 2000 (constant)
Add 78528 + 25480 = 104008 /2000 = 52.004 total density
I need a formula where the (2) individual and mixed densities are known and the 2000 lb limit is constant and will tell me the ingredient totals.
Thank you very much for any help you can provide!

FWT
Posts: 153
Joined: Sat Feb 28, 2009 8:53 pm

### Re: Algebra problem with two variables

It looks like you're doing a mixture thing, like they show here. The mixes don't usually work like this in real life, but if they're working for you then that's okay. Just do the set-up like you did: mixing a lbs of x and b lbs of y to get 2000 lbs of z, then its a + b = 2000 and ax + by = 2000z. Then solve the system for the answer. You could substitute: (2000-b)x + by = 2000z, because you know x, y, and z. Then solve for b and back-solve for a.

jg.allinsymbols
Posts: 72
Joined: Sat Dec 29, 2012 2:42 am
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### Re: Algebra problem with two variables

Difficult to do this using pure text, but easier on paper.

Batch size, B=2000 pounds.
x pounds of r pound per volume material
y pounds of t pound per volume material
You want density of mixture to be d pounds per volume.

The pounds of x+y divided by volume of x and y must be d. Assume you unknown variables are only x and y. In fact, B could be some value other than 2000; just that we KNOW B in our example.

$(x+y)/(x/r+y/t)=d$ AND $x+y=B$

Solve this system of equations for x and y.
Working through the complex rational equation, you can obtain it in a form as $(rt-dt)x=(rd-rt)y$
You may then choose a substitution using the B equation and solve for one of the variables. The advantage of keeping all this in symbols is that when the values for the symbols change for a different example, you would not need to re-solve the whole problem again.

In real life, volumes are not always uninteractable when different density materials are mixed. If solid granular materials, smaller particles may fit between fewer larger particles and the resulting density might be higher than predicted from the model given above.

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