3 variables as addendds and proportions. I know process but not rationale.

[thecoast]

Hi, folks. I've figured out how to do something but not how/why it works.
Here's the problem:

x+y+z=871
The proportion of x:y = 4:5 and the proportion of y:z = 3:8.
What is the value of y?

I've attached an image I made on the free iOS app MathMagic. I'm sure I've NOT written the thing out in technically correct terms (would welcome help), but the example aids in illustration. The algorithm works, but I just want to know how/why mathematically. Any help would be appreciated to make it look better and say what I intend it to say.

The process with plain text is as follows:
4:5 = 4/5 and 3:8 = 3/8.
1) Multiply the numerator of 4/5 by the numerator of 3/8 to get 12.
2) Multiply the numerator of 3/8 by the denominator of 4/5 to get 15.
3) Multiply the denominator of 4/5 by the denominator of 3/8 to get 40.
4) Add the three products to get 12+15+40=67.
5) 12/67 + 15/67 + 40/67 = x+y+z = 871
6) 15/67 = y/871 =195/871
7) Answer: y = 195

I'd like to be able to explain why it works and figure out how to extend this to problems with more variables and more proportions.




Thank you.

 

[tkhunny]

Build Your Own Version and See. Personally, an "algorithm" to solve a specific problem, tends to annoy me. If you have a specific application and wish to code it on a specific machine, I can see it, but generally, no! Just learn to use your best algebra.

x + y + z = 871
Rewrite the ratios

x:y :: 4:5 MEANS $\displaystyle \dfrac{x}{y} = \dfrac{4}{y} \implies x = \dfrac{4}{5}\cdot y$

y:z :: 3:8 MEANS $\displaystyle \dfrac{y}{z} = \dfrac{3}{8} \implies z = \dfrac{8}{3}\cdot y$

Substitute these new expressions for x an z an you have an arithmetic problem - just finding a common denominator and adding. You can drag the magic algorithm out of the arithmetic - but why?

 

[HallsofIvy]

Hi, folks. I've figured out how to do something but not how/why it works.
Here's the problem:

x+y+z=871
The proportion of x:y = 4:5 and the proportion of y:z = 3:8.
What is the value of y?

Here's how I would do it: x:y::4:5 so x/y= 4/5 from which x= (4/5)y. y:z::3:8 so y/z= 3/8 so z= (8/3)y. From that x+ y+ z= (4/5)y+ y+ (8/3)y= (12/15)y+ (15/15)y+ (40/15)y= (67/15)y= 871. y= (871)(15)/67= 13(15)= 195.

I've attached an image I made on the free iOS app MathMagic. I'm sure I've NOT written the thing out in technically correct terms (would welcome help), but the example aids in illustration. The algorithm works, but I just want to know how/why mathematically. Any help would be appreciated to make it look better and say what I intend it to say.

The process with plain text is as follows:
4:5 = 4/5 and 3:8 = 3/8.
1) Multiply the numerator of 4/5 by the numerator of 3/8 to get 12.
2) Multiply the numerator of 3/8 by the denominator of 4/5 to get 15.
3) Multiply the denominator of 4/5 by the denominator of 3/8 to get 40.
4) Add the three products to get 12+15+40=67.
5) 12/67 + 15/67 + 40/67 = x+y+z = 871
6) 15/67 = y/871 =195/871
7) Answer: y = 195

I'd like to be able to explain why it works and figure out how to extend this to problems with more variables and more proportions.


View attachment 8026

Thank you.

 

[thecoast]

Build Your Own Version and See. Personally, an "algorithm" to solve a specific problem, tends to annoy me. If you have a specific application and wish to code it on a specific machine, I can see it, but generally, no! Just learn to use your best algebra.

x + y + z = 871
Rewrite the ratios

x:y :: 4:5 MEANS $\displaystyle \dfrac{x}{y} = \dfrac{4}{y} \implies x = \dfrac{4}{5}\cdot y$

y:z :: 3:8 MEANS $\displaystyle \dfrac{y}{z} = \dfrac{3}{8} \implies z = \dfrac{8}{3}\cdot y$

Substitute these new expressions for x an z an you have an arithmetic problem - just finding a common denominator and adding. You can drag the magic algorithm out of the arithmetic - but why?

Thanks for the reply. I'm brand-spankin' new here so I'm not sure what the "Build Your Own Version and See" refers to.

Would I be correct in assuming that you meant to write $\displaystyle \dfrac{4}{5}$ in the first line instead of $\displaystyle \dfrac{4}{y}$?

In reference to your annoyance (for which I apologize), isn't the best algebra following general sets of steps applicable to particular problems? Don't mathematicians come up with general formulas or expressions into which one can plug in particular values and the formulas/expressions are general because they are always true given implied or explicit criteria? Maybe I shouldn't have used the term "algorithm" or I used it in a contextually inappropriate way. Thank you for your helpfulness.