# Subtracting two equations

Looh
Hello!

I'm having trouble subtracting two equations, I'm not really sure how to go about it.

$\\ \frac{4\pi r^3}{3}(\rho_1 - \rho_2)g - 6\pi r\eta v_0 = 0\\ \frac{4\pi r^3}{3}(\rho_1 - \rho_2)g - 6\pi r\eta v_1 - qE = 0$

For clarification, there are three different equations used.
$\\ F_g = \frac{4\pi r^3}{3}(\rho_1 - \rho_2)g \\ F_f = 6\pi r\eta v_0\\ F_e = qE$

Which gives:
$\\ F_g - F_f = 0 \\ F_g - F_f - F_e = 0$

This, in turn, results in $F_e$, or $qE$, but I'm sure that's wrong.

How should I go about solving this?

Last edited:

$$\frac{4\pi r^3}{3}(\rho_1- \rho_2)g- 6\pi\eta v_0- \left(\frac{4\pi r^3}{3}(\rho_1-\rho_2)g- 6\pi\eta v_1- qE\right)= 0- 0= 0$$
$$\left(\frac{4\pi r^3}{3}(\rho_1- \rho_2)g- \frac{4\pi r^3}{3}(\rho_1- \rho_2)g\right)- \left(6\pi\eta v_0- 6\pi\eta v_1\right)+ qE= 0$$
The the terms in the first parentheses are identical and so cancel. The terms in the second parentheses are almost identical but one has "$v_1$" and the other "$v_0$". We can factor out the other terms (distributive law), leaving $6\pi\eta(v_0- v_1)$. That is, "subtracting the two equations" gives
$$6\pi\eta(v_0- v_1)+ qE= 0$$
I'm not sure how your other equations are involved but certainly if you have $F_g- F_f= 0$ and $F_g- F_f- F_e= 0$ and you subtract the second from the first you have $F_g- F_x- (F_g- F_x- F_e)= (F_g- F_g)- (F_x- F_x)+ F_e= F_e= 0$ because the first two pairs cancel.