# Subtracting two equations

1. Aug 26, 2012

### 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: Aug 26, 2012
2. Aug 26, 2012

### HallsofIvy

Staff Emeritus
I'm not sure what your difficulty is- just go ahead and subtract the second from the first:
$$\frac{4\pi r^3}{3}(\rho_1- \rho_2)g- 6\pi\eta v_0- \left(\frac{4\pi r^3}{3}(\rho_1-\rho_2)g- 6\pi\eta v_1- qE\right)= 0- 0= 0$$
Rearranging (technically, using the "associative" and "commutative" laws) we have
$$\left(\frac{4\pi r^3}{3}(\rho_1- \rho_2)g- \frac{4\pi r^3}{3}(\rho_1- \rho_2)g\right)- \left(6\pi\eta v_0- 6\pi\eta v_1\right)+ qE= 0$$

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.

(Strictly speaking we do not define arithmetic operations for "equations" as we do for numbers, variables, and expressions. "Subracting equations" is shorthand for "subtracting each side of the equations".)