1. ## solving for proportions

Seasons greetings!

I apologize for the format of this post. The FAQ link on how to format equations seems to be broken.

I have a "compound" (C1) that is to be made from two other compounds and two pure elements (C2, C3, L, T)

Compound 2 is composed of elements in this proportion: C2 = 30X + 70L
Compound 3 is composed of elements in this proportion: C3 = 4X + Z + 95L

The final compound must have the following proportion of elements:
C1 = 8X + 2T + .2Z +89.8L

So, to find the proportions needed to make compound C1, I would need to solve the following for a, b, c, and d?
C1 = aC2 + bC3 + cL + dT

I this makes no sense, I'm sorry. My head is already hurting...

2. Hello, scunning!

I have a compound $\displaystyle (C_1)$ to be made from two other compounds
. . and two pure elements: .$\displaystyle C_2,\:C_3,\:L,\: T.$

Compound 2 has elements in this proportion: .$\displaystyle C_2 \:=\: 30X + 70L$
Compound 3 has elements in this proportion: .$\displaystyle C_3 \:=\: 4X + Z + 95L$

The final compound must have the following proportion of elements:
. . $\displaystyle C_1 \:=\: 8X + 2T + 0.2Z +89.8L$

So, to find the proportions needed to make compound $\displaystyle C_1$
. . . I would need to solve the following for $\displaystyle a, b, c,\text{ and }d\,?$
. . . . . $\displaystyle C_1 \:=\: aC_2 + bC_3 + cL + dT$ . Yes!

We have: . . . . . . . . . . . $\displaystyle aC_2 + bC_3 + cL + dT \;=\;8X + 2T + 0.2Z + 89.8L$

. $\displaystyle a(30X + 70L) + b(4X + Z + 95L) + cL + dT \;=\;8X + 2T + 0.2Z + 89.8L$

. .$\displaystyle 30aX + 70aL + 4bX + bZ + 95bL + cL + dT \;=\;8X + 2T + 0.2Z + 89.8L$

. . . .$\displaystyle (30a + 4b)X + dT + bZ + (70a + 95b + c)L \;=\;8X + 2T + 0.2Z + 89.8L$

Equate coefficients: . $\displaystyle \begin{bmatrix}30a + 4b \:=\: 8 \\ d \:=\: 2 \\ b \:=\: 0.2 \\ 70a + 95b + c \:=\: 89.8 \end{bmatrix}$

And solve the system of equations: . $\displaystyle \begin{bmatrix} a &=& \frac{6}{25} \\ b &=& 0.2 \\ c &=& 54 \\ d &=& 2 \end{bmatrix}$

Therefore, we will use: .$\displaystyle \begin{Bmatrix} \frac{6}{25}\text{ units} & \text{compoind 2} \\ 0.2\text{ units} & \text{compound 3} \\ 54\text{ units} & \text{element L} \\ 2\text{ units} & \text{element T} \end{Bmatrix}$

3. Soroban, you are a lifesaver! Thank you so much for the explanation. This forum is a wonderful resource.

Have a great Christmas!

Edit: It didn't seem like those proportions worked, but then I realized that since I expressed the compounds in terms of 100 parts, the elements needed to be expressed the same way. I change it to the following, and your logic worked perfectly! Thanks again.

C1 = aC2 + bC3 + c100L + d100T