How would one solve a system of 4 equations with 4 variables, where the variables multiply each other

Question

The equations are as follows: \begin{align*} 0.3 A &= 0.1 B \\ B C &= 0.37 (A + B) \\ 0.5 A + D B &= 0.43 (A + B) \\ C + D &= 0.9 B \end{align*}

I'm aware I won't get a numerical answer for any single variable, but I need ratios anyways

EDIT for clarity full task reads as: the pot contains 43% liquid 1, 20% liquid 2 and 37% liquid 3 if we move all of liquid 3 into a second pot, while leaving the first at a 50/50 blend of liquids 1 and 2, what is the exact composition of pot 2 if it has 10% liquid 2

I had V0 = V1 + V2, then I made the other equations from the rest of the data I see no other way to solve this, but I also can't solve it this way, so I'm open to suggestions

EDIT 2: the method of moving is distillation, assuming no losses, ALL of liquid 3 is moved, while the others are distributed to leave pot 1 at a 50/50 blend I also forgot to note that pot 2 has 10% liquid 2

Answer 1

The first equation lets you eliminate $A$. There will be a common factor of $B$ in the second that you can divide out if $B \neq 0$ giving you a value for $C$. The third equation will give you a value for $D$ after dividing out $B$. Once you have $C,D$ the last will give a value for $B$.

Answer 2

I would solve the original problem this way. You have two states: the first being one container $V$ with compositions $\alpha_1 = 0.43$ liquid 1, $\alpha_2 = 0.20$ liquid 2, and the remaining $\alpha_3 = 0.37$ liquid 3. The second state has two containers: one container $\beta V$ with compositions $\delta_1 = 0.5$ liquid 1 and $\delta_2 = 0.5$ liquid 2; and the other container $(1 - \beta) V$ with composition $\gamma_1$ liquid 1, $\gamma_2 = 0.1$ liquid 2, and $\gamma_3$ liquid 3.

The only unknowns here are $\beta$ (which is the fraction of the original volume still left in the first container), and $\gamma_1, \gamma_3$ (the fractions of liquids 1 and 3 in the second container).

By conservation of volume of liquid 2, we have

$$\alpha_2 = \delta_2 \beta + \gamma_2 (1 - \beta)$$

Everything is known here, except $\beta$! So, we can solve for it:

$$\beta = \frac{\alpha_2 - \gamma_2}{\delta_2 - \gamma_2}$$

and

$$1 - \beta = \frac{\delta_2 - \alpha_2}{\delta_2 - \gamma_2}$$

By conservation of volume of liquid 3, we have the relation (after dividing by $V \ne 0$)

$$\alpha_3 = \gamma_3 (1 - \beta)$$

giving

$$\gamma_3 = \frac{\alpha_3}{1 - \beta} = \boxed{\frac{\alpha_3 (\delta_2 - \gamma_2)}{\delta_2 - \alpha_2}}$$

By conservation of volume of liquid 1, we have

$$\alpha_1 = \delta_1 \beta + \gamma_1 (1 - \beta)$$

giving

$$\gamma_1 = \frac{\alpha_1 - \delta_1 \beta}{1 - \beta}$$

$$= \boxed{\frac{\alpha_1 \delta_2 + \gamma_2 \delta_1 - \alpha_1 \gamma_2 - \alpha_2 \delta_1}{\delta_2 - \alpha_2}}$$

I'll leave it to you to check, but when I substitute the given values of the known variables, I get $\gamma_3 = \boxed{\frac{37}{75}}$ and $\gamma_1 = \boxed{\frac{61}{150}}$. Although not asked for, the proportion of $V$ that remains in the first container is $\beta = \boxed{\frac{1}{4}}$.

This answer looks like a solution to the system you suggested as well, based on one of the comments.

Answer 3

Let's try to be smart, instead of solving equations with many variables, try to minimise their number.

Assume volume of Pot 1 to be $100$ L. So, we have $43$ L of A, $20$L of B and $37$L of C.

Now, we have transfer full liquid C to Pot $2$ along with some Liquid A and B, and in the end, we are supposed to have equal volumes of A and B in Pot $1$. So, if 'x' L liquid B is removed then we need to remove $(23+x)$ L of Liquid B to ensure $50$-$50$ in pot $1$.

This gives us a single equation, $\frac{x}{23+x+x+37} \cdot 100 = 10 \%\Rightarrow x=7.5 L$

Therefore, pot 2 has $30.5$ L of liquid A, $7.5$ L of liquid B and $37$ L of liquid of C.

Now, I hope you can compute the composition percentage.