Hi, unfortunately math isn't my biggest strength so hopefully someone can help me out.

I have the following formula.

4.78 = ((1000 / 3001) * 86) / 6.0

Works good.

Now lets say I have a number of 4.54, how can I calculate a number to go where 6.0 is that would give me 4.54 using the same numbers from the formula above such as.

4.54 = ((1000 / 3001) * 86) / 6.32

So the last number is what needs to be calculated using 4.54 and the rest.

Thanks

 

hi rob,
Work out the value within the brackets first to give a Constant value, then transposing becomes easy.
E

 

Thanks Eric, you're always there for me. Appreciate it.

6.31 = ((1000 / 3001) * 86) / 4.54

 

hi,
The Constant C =28.657

x= C/y
E

 

Yes makes sense now. Thanks again!

 

Hi,

When you have a formula with a number on the left and another number on the right that is in the denominator, because you can multiply by one and divide by the other you essentially 'swap' the two numbers.

For example:
F=1/T

To calculate T instead of F, swap the places of the F and then T:
T=1/F

Note this is the same as dividing both sides by F and multiplying both sides by T.

 

Hi, unfortunately math isn't my biggest strength so hopefully someone can help me out.

You've actually identified the root cause and the solution right here. Your math skills aren't where you need them to be for the kind of things you are doing. So the correct solution is to strengthen your math skills so that they are.

It looks like you are most in need of basic algebra skills. There are tons of books out there, both physical and online. There's also courses you can take, both in-seat and online. You might search out online math placement tests to evaluate your current skills and figure out what level would be most appropriate for you to start your journey at.

 

Hi, unfortunately math isn't my biggest strength so hopefully someone can help me out.

I have the following formula.

4.78 = ((1000 / 3001) * 86) / 6.0

Works good.

Now lets say I have a number of 4.54, how can I calculate a number to go where 6.0 is that would give me 4.54 using the same numbers from the formula above such as.

4.54 = ((1000 / 3001) * 86) / 6.32

So the last number is what needs to be calculated using 4.54 and the rest.

Thanks

What you have is not a formula.
It is a mathematical equation with values of numbers.

You are confusing two areas of mathematics: arithmetic and algebra.

We will tackle the second problem first. Algebra makes use of unknown quantities in an equation. Since the quantities are not yet known, we give them letters of the alphabet. Let us take the first equation:

4.78 = ((1000 / 3001) * 86) / 6.0

Replace this with an Algebra equation:

Y = ((A / B) * C ) / X

If A = 1000, B = 3001, C = 86, and these values never change, we call them constants.
In this specific example, both X and Y are not constant. We call them variables. X depends on Y. Y depends on X.

In the way it is written, Y is called the dependent variable, because Y depends on X.
X is called the independent variable, because we can independently choose X, and Y must follow along.

What you refer to as "reverse formula" is known as the "inverse relationship". In other words, how do we make Y the independent variable and X the dependent variable.

We want to create a new formula (we call it a function) so that instead of Y being a function of X, written as

Y = f(X)

we want X as a function of Y, written as

X = f(Y)

In order to create this new formula, we go back to the rules of basic arithmetic.
The initial equation is:

Y = ((A / B) * C ) / X

We want to move Y to the right hand side of the equal sign, and X to the left hand side.
This is an easy example. Another equation can be much more complex. Some are very difficult to solve. Some are impossible to solve.

Multiply both sides of the equation by X.
X * Y = X * ((A / B) * C ) / X

Can you see how this reduces to:
X * Y = ((A / B) * C )

Now, divide both sides of the equation by Y
X * Y / Y = ((A / B) * C ) / Y

Can you see how this reduces to :
X = ((A / B) * C ) / Y

And there you have your desired inverse function.

Now as an exercise, do this common conversion. Convert temperature in °C to °F assuming that you are given this formula:

C = ( F - 32 ) * 5 / 9

In other words, if the temperature in Canada is now 25°C. What is the same temperature in °F?

 

What you have is not a formula.
It is a mathematical equation with values of numbers.

You are confusing two areas of mathematics: arithmetic and algebra.

We will tackle the second problem first. Algebra makes use of unknown quantities in an equation. Since the quantities are not yet known, we give them letters of the alphabet. Let us take the first equation:

4.78 = ((1000 / 3001) * 86) / 6.0

Replace this with an Algebra equation:

Y = ((A / B) * C ) / X

If A = 1000, B = 3001, C = 86, and these values never change, we call them constants.
In this specific example, both X and Y are not constant. We call them variables. X depends on Y. Y depends on X.

In the way it is written, Y is called the dependent variable, because Y depends on X.
X is called the independent variable, because we can independently choose X, and Y must follow along.

What you refer to as "reverse formula" is known as the "inverse relationship". In other words, how do we make Y the independent variable and X the dependent variable.

We want to create a new formula (we call it a function) so that instead of Y being a function of X, written as

Y = f(X)

we want X as a function of Y, written as

X = f(Y)

In order to create this new formula, we go back to the rules of basic arithmetic.
The initial equation is:

Y = ((A / B) * C ) / X

We want to move Y to the right hand side of the equal sign, and X to the left hand side.
This is an easy example. Another equation can be much more complex. Some are very difficult to solve. Some are impossible to solve.

Multiply both sides of the equation by X.
X * Y = X * ((A / B) * C ) / X

Can you see how this reduces to:
X * Y = ((A / B) * C )

Now, divide both sides of the equation by Y
X * Y / Y = ((A / B) * C ) / Y

Can you see how this reduces to :
X = ((A / B) * C ) / Y

And there you have your desired inverse function.

Now as an exercise, do this common conversion. Convert temperature in °C to °F assuming that you are given this formula:

C = ( F - 32 ) * 5 / 9

In other words, if the temperature in Canada is now 25°C. What is the same temperature in °F?

Excellent description, but a little daunting.

Proficiency in algebra depends first on knowing what’s important and what’s not.

Immediately, one should notice that the individual values of A, B, and C are not relevant by themselves to obtaining an answer.

The first step is to recognize this and calculate a new value. Let’s call this K. K is equal to the following:

K=((A / B) * C )​

Now, the algebraic problem becomes tractable. Consider the two equations;

1. Y = ((A / B) * C ) / X
2. Y = K / X

Which is easier to solve?

My Dad taught me as a young boy, that learning what is happening is more important than learning what to do. Starting by condensing a problem into what is relevant (variable quantities) versus what (in some sense) is less relevant (values that do not change in the problem space) is as important as the nitty picky rules. His advice was well taken. My math SATs were perfect. I got an BS in Applief Math in college. He was a teacher (professionally) and taught me well.

 

Excellent description, but a little daunting.

Proficiency in algebra depends first on knowing what’s important and what’s not.

Immediately, one should notice that the individual values of A, B, and C are not relevant by themselves to obtaining an answer.

The first step is to recognize this and calculate a new value. Let’s call this K. K is equal to the following:

K=((A / B) * C )​

Now, the algebraic problem becomes tractable. Consider the two equations;

1. Y = ((A / B) * C ) / X
2. Y = K / X

Which is easier to solve?

My Dad taught me as a young boy, that learning what is happening is more important than learning what to do. Starting by condensing a problem into what is relevant (variable quantities) versus what (in some sense) is less relevant (values that do not change in the problem space) is as important as the nitty picky rules. His advice was well taken. My math SATs were perfect. I got an BS in Applief Math in college. He was a teacher (professionally) and taught me well.

The TS may not realize that that only happened because he presented a "trivial" example where the formula is reduced to

Y = K / X

This is not the case with a common temperature conversion example of

C = ( F - 32 ) * 5 / 9

One cannot reduce this in a similar fashion. It requires some degree of proficiency at arithmetic.

 

Exactly! The second most important skill s to know when the other skills don’t apply.

My point being is that the goal is to be able to analyze problems and know almost intuitively which general rules apply and which don’t.

This is the issue I have with the way math (arithmetic) is taught at the elementary level. Kids get so caught up in grouping, a graphical representation of arithmetic, that the problem 2+2=4 becomes an arduous exercise.

First, learn that 2+2=4, then and only then, progress to a graphical representation. While substitute teaching, I ran into students who where overwhelmed by 2+2+4! And IMHO that totally should never happen.

So, I presented my example, because in my opinion, that’s the proper approach. Your response is a great follow up, but if anyone doesn’t ‘get’ Y=K/X, there is no way in hell they are going to understand your response.

 

Using the principal of combining constants to simplify the equation and recognizing that it might take multiple steps, solving your example problem is trivialized.

Yes it takes a few steps, but deals with a smaller set of terms, because the constants are recursively simplified to... other constants.

You and I know we are dealing with multiple numeric abstractions - but most importantly, each step below results in a simpler form. We go from complex to simple in a heartbeat.

C = ( F - 32 ) * 5 / 9
C = ( F - 32 ) * 0.556
C = 0.556 * F - 0.556 * 32
C = 0.556 * F - 17.778
C + 17.778 = 0.556 * F
F = ( C + 17.778 ) / 0.556
F = C / 0.556 + 17.778 / 0.556
F = C / 0.556 + 32
F = 1.78 * C + 32​

Ok, a litter longer than a heartbeat. But the process reduced three parameters to two. In a simple standard linear form.

I was taught techniques that worked. The why and the complicated explanstions came later. And as a matter of fact, I wasn’t taught the why, but rather taught the teachers why.

 

Division is simply multiplication with an inverse. Use the notation X' = 1/X to denote the inverse of X. The inverse of the inverse of X is of course X itself. So

C = (F - 32) * 5 / 9

becomes

C = (F - 32) * 5 * 9'

Rearranging we get:

(F - 32)' = C' * 5 * 9'

We don't want the left side to be in terms of an inverse so we just go through and "flip" all of the multiplicative terms in the equation.

(F - 32) = C * 5' * 9

Finally add 32 to both sides.

F = (C * 5' * 9) + 32

In other words:

*** EDIT ***
F = 9/5C + 32
*** EDIT ***

 

C = ( F - 32 ) * 5 / 9
C = ( F - 32 ) * 0.556
​

I would never teach mathematics this way.
I would not reduce the numbers (unless the results are whole numbers).
Also I would hold off on using a calculator until you get to the very final equation.

A simpler route is
C = ( F - 32 ) * 5 / 9
C * 9 / 5 = ( F - 32 )
( C * 9 / 5 ) + 32 = F

( I would have preferred to see the TS work through this, but then again... who knows?)
​

 

Division is simply multiplication with an inverse. Use the notation X' = 1/X to denote the inverse of X. The inverse of the inverse of X is of course X itself. So

C = (F - 32) * 5 / 9

becomes

C = (F - 32) * 5 * 9'

Rearranging we get:

(F - 32)' = C' * 5 * 9'

We don't want the left side to be in terms of an inverse so we just go through and "flip" all of the multiplicative terms in the equation.

(F - 32) = C * 5' * 9

Finally add 32 to both sides.

F = (C * 5' * 9) + 32

In other words:

F = 5/9C + 32

Tada!

Just a simple misstep at the end there.
No Tada!

 

Also, if one could not recall if it is supposed to be 5/9 or 9/5, perhaps it would help to use the following:

C = ( F - 32 ) * 100 / 180

( C * 180 / 100 ) + 32 = F

if it helps to remember that there are 100 degrees Celsius (previously known as Centigrade for a good reason) between water freezing and boiling versus 180 degrees Fahrenheit.

Hence,

F = 1.80 * C + 32

(and not 1.78 * C + 32 )

 

Just a simple misstep at the end there.
No Tada!

Fair enough, but also case in point!

 

Oh no, I was hoping this thread would disappear.

As soon as ericgibbs gave me the answer I felt pretty stupid as I should have figured that out, one of those nights.

 

Oh no, I was hoping this thread would disappear.

As soon as ericgibbs gave me the answer I felt pretty stupid as I should have figured that out, one of those nights.

That's OK. You started off the thread by saying that mathematics is not one of your strengths.
That's OK here too. Our role here on AAC is to help you improve on wherever you need help.
Glad to be able to provide assistance.

 

Hello again,

I forgot to mention that the 'reverse' formula here is actually referred to as the "inverse" function.
There is a bunch written on calculating inverse functions for more complicated functions but the main idea goes like this.
If you have a function like:
y=x^2

then you want to solve for x rather than y.
For this function it happens to be easy but there are two possibilities:
x=sqrt(y)
x=-sqrt(y)

Some other functions are more difficult such as:
y=x^4+x^3+x^2+x+1

so you'd have to read up on how to handle the more complicated ones.
Sometimes only a numerical solution is possible.