# Equation solving: L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT

## Homework Statement

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT

How can we solve for dT? I should get those dTs to one side but by dividing they'd just cancel out.

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT -- Subtract both sides by L[0,Fe]

L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] -- Divide both sides by L[O,Fe]a[Fe]

dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] / L[O,Fe]a[Fe] -- What now? I don't know how to get that dT from the right side to left side. If we divide, it just cancels out.

Thank you!

## Homework Equations

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT

## The Attempt at a Solution

Above ^^

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DrClaude
Mentor
By addition/subtraction on both sides, get an equation where all terms including dT are on the left-hand side and all terms without dT are on the right-hand side. Then use the distributive property of multiplication (in reverse) to get a single dT, then divide both sides by the factor by which dT is multiplied.

By addition/subtraction on both sides, get an equation where all terms including dT are on the left-hand side and all terms without dT are on the right-hand side. Then use the distributive property of multiplication (in reverse) to get a single dT, then divide both sides by the factor by which dT is multiplied.

I understand the basic idea that you described in your response, but I'm not entirely sure on how to do that. I can't get all dTs to the left-hand side because

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT --- Subtract by L[0,Fe]

L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] --- Multiply by dT

L[0,Fe]α[Fe]*dT*dT = L[0,A])+L[0,Al]α[Al]*dT*dT - L[0,Fe] -- dTs are on the left and right. I don't know how to get them to one side because if we divide, they just cancel out.

dT(L[0,Fe]α[Fe]) = dT(L[0,A])+L[0,Al]α[Al]) - L[0,Fe] -- We can take dT as a common factor like this <--

I don't think this solution works because dTs are not on only one side of the equation. How can I do so?

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DrClaude
Mentor
L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] --- Multiply by dT
Don't multiply by dT. You need to move L[0,Al]α[Al]*dT to the LHS, so subtract this term on both sides.

L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe]

L[0,Fe]α[Fe]*dT - L[0,Al]α[Al]*dT = L[0,Al] - L[0,Fe]

dT(L[0,Fe]α[Fe]-L[0,Al]α[Al]) = L[0,Al] - L[0,Fe]

dT = L[0,Al] - L[0,Fe] / (L[0,Fe]α[Fe]-L[0,Al]α[Al])

Is this right?

Don't multiply by dT. You need to move L[0,Al]α[Al]*dT to the LHS, so subtract this term on both sides.

DrClaude
Mentor
dT = L[0,Al] - L[0,Fe] / (L[0,Fe]α[Fe]-L[0,Al]α[Al])
You are missing some parentheses. Otherwise it is correct.

dT = (L[0,Al] - L[0,Fe]) / (L[0,Fe]α[Fe]-L[0,Al]α[Al])

• JesseK
Mark44
Mentor
L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT
What do Fe, Al, and A represent? The first two look like the symbols for iron and aluminum, but I don't know what A represents.

What does L[0, Fe] and the other expressions like this represent?

• LCKurtz and epenguin
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT

How can we solve for dT? I should get those dTs to one side but by dividing they'd just cancel out.

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT -- Subtract both sides by L[0,Fe]

L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] -- Divide both sides by L[O,Fe]a[Fe]

dT = L[0,A])+L[0,Al]α[Al]*dT - L[0,Fe] / L[O,Fe]a[Fe] -- What now? I don't know how to get that dT from the right side to left side. If we divide, it just cancels out.

Thank you!

## Homework Equations

L[0,Fe]+L[0,Fe]α[Fe]*dT = L[0,A])+L[0,Al]α[Al]*dT

## The Attempt at a Solution

Above ^^

You are making more work for yourself by not simplifying the notation first, before proceeding. Many problem solvers (myself included) would first re-write the equation before solving it.

So, let A= L[0,Fe], B = L[0,Fe]*α(Fe) and C = L[0,Al], D= L[0,Al]*α(Al). Then the equation reads as
A+B * dT = C + D * dT.
Therefore, if B≠D we have dT = (C-A)/(B-D). Now you can put back the values for A,B,C,D.

Doing it that way sometimes makes for less work (although not by much in this simple example) and reduces the chance of making errors---always a serious concern when you have complicated-looking equations that are simple in reality.

What do Fe, Al, and A represent? The first two look like the symbols for iron and aluminum, but I don't know what A represents.

What does L[0, Fe] and the other expressions like this represent?

I'm afraid this is not really relevant to solving the equation for something. However, Fe and Al are the symbols for iron and aluminum. :)

Mark44
Mentor
I'm afraid this is not really relevant to solving the equation for something.
That's true, but if you're solving an equation, it's helpful to know what the symbols represent in an equation.

If you follow Ray's suggestion, you will end up with this:
$$dT = \frac{L[0, Al] - L[0, Fe]}{L[0,Fe]*α(Fe) - L[0,Al]*α(Al)}$$
If I needed to simplify this further I would need to know how to do the subtractions and multiplications in this expression, which would require understanding what the symbols mean.

It has often been that case that a member presents an equation to be solved, but it later turns out that the equation is meaningless, and is incorrect or unrelated to the problem the member actually is trying to solve. If we solve the equation, it's a case of "garbage in, garbage out." Being able to understand what the symbols in an equation mean is important in cases like this.

However, Fe and Al are the symbols for iron and aluminum. :)
As I guessed, but I'm still at a loss to know what A represents, since there is no element with A as its symbol. The elements whose names start with A or whose symbols start with A are Silver (Ag - from Argentum, the Latin name), Argon (Ag), Aluminum (Al), Arsenic (As), Americium (Am), and Actinium (Ac).

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