Solve an Algebraic Equation: A+B=C

  • Thread starter seasnake
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It does.In summary, the conversation is about finding a non-time-series, algebraic equation for a set of data where A + B always equals 10 and C is a dependent variable. The proposed equations include (A - B) / (A + B) = C, (A - B) / (A times B) = C, and (A - B)^2 / (A + B) = C. One approach suggested is to use
  • #1
seasnake
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0
Can anyone figure this out... searching for an algebraic equation (not a student)

I'm trying to unravel some equations in a spreadsheet and as such I need to find a non-time-series, algebraic equation for the following data expressed for finding C:

A + B always equals 10 (i.e. A1 + B1 = 10; A2 + B2 = 10; ...)

when:

A0 = 0... B0 = 10 ... C0 = 10

A1 = 1 ... B1 = 9 ... C1 = 4

A2 = 2 ... B2 = 8 ... C2 = 2

A3 = 3 ... B3 = 7 ... C3 = 0.83 repeating 3

A4 = 4 ... B4 = 6 ... C4 = 0.2

A5 = 5 ... B5 = 5 ... C5 = 0

A6 = 6 ... B6 = 4 ... C6 = 0.2

A7 = 7 ... B7 = 3 ... C7 = 0.83 repeating 3

A8 = 8 ... B8 = 2 ... C8 = 2

A9 = 9 ... B9 = 1 ... C9 = 4

A10 = 10 ... B10 = 0 ... C10 = 10

an eqational attempts would be something along the lines of:

(A - B) / (A + B) = C

(A - B) / (A times B) = C

(A - B)^2 / (A + B) = C

Note: C3 = C7 = 0.83 = 5 / 6 (serves as a decent clue as it displays both a numerator and a denominator)

once again, I am not seeking a time series equation...
 
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  • #2


What's a 'not time series equation'? Do you want an exact polynomial through those points or what?
 
  • #3


seasnake said:
I'm trying to unravel some equations in a spreadsheet and as such I need to find a non-time-series, algebraic equation for the following data expressed for finding C:

A + B always equals 10 (i.e. A1 + B1 = 10; A2 + B2 = 10; ...)

when:

A0 = 0... B0 = 10 ... C0 = 10

A1 = 1 ... B1 = 9 ... C1 = 4

A2 = 2 ... B2 = 8 ... C2 = 2

A3 = 3 ... B3 = 7 ... C3 = 0.83 repeating 3

A4 = 4 ... B4 = 6 ... C4 = 0.2

A5 = 5 ... B5 = 5 ... C5 = 0

A6 = 6 ... B6 = 4 ... C6 = 0.2

A7 = 7 ... B7 = 3 ... C7 = 0.83 repeating 3

A8 = 8 ... B8 = 2 ... C8 = 2

A9 = 9 ... B9 = 1 ... C9 = 4

A10 = 10 ... B10 = 0 ... C10 = 10

an eqational attempts would be something along the lines of:

(A - B) / (A + B) = C

(A - B) / (A times B) = C

(A - B)^2 / (A + B) = C

Note: C3 = C7 = 0.83 = 5 / 6 (serves as a decent clue as it displays both a numerator and a denominator)

once again, I am not seeking a time series equation...
You are seeking an equation. "Time" is not a mathematical concept so "time series" is not relevant here.

Since x+ y= 10, y is fixed once you know x so you can just look for a function of x giving those values. I notice this is symmertric about x= 5 so I would look for f(x)= a(x-5)2n Taking f(0)= a(5)2n= 10 and f(1)= a(4)2n= 4 we can divide the first equation by the second and get
[tex]\frac{5^{2n}}{4^{2n}}= \left(\frac{5}{4}\right)^{2n}= \frac{10}{4}= \frac{5}{2}[/itex]
Taking log of both sides, 2n log(5/4)= log(5/2) or 2n(0.096910)=0.39794 so 2n= 4.106. Let's just call that "4". Then, from f(0)= a(5)2n= a(54)= 125a= 10, a= 10/125= 0.08.

See if f(x,y)= 0.08 x4 fits the other values.
 

Related to Solve an Algebraic Equation: A+B=C

What is an algebraic equation?

An algebraic equation is a mathematical expression that contains variables, numbers, and mathematical operations. The goal is to find the value of the variable that makes the equation true.

What is the purpose of solving an algebraic equation?

Solving an algebraic equation allows us to find the value of the variable in the equation. This helps us to solve real-world problems and make predictions based on given information.

What are the basic steps for solving an algebraic equation?

The basic steps for solving an algebraic equation are:
1. Simplify both sides of the equation by combining like terms.
2. Isolate the variable by performing inverse operations.
3. Check your solution by plugging it back into the original equation.

What are inverse operations?

Inverse operations are mathematical operations that undo each other. For example, addition and subtraction are inverse operations, as well as multiplication and division. In solving an algebraic equation, we use inverse operations to isolate the variable.

How do you check your solution for an algebraic equation?

To check your solution for an algebraic equation, simply plug the value you obtained for the variable back into the original equation. If the equation is true, then your solution is correct. If the equation is false, then you need to recheck your work.

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