# Check my arithmetic for this formula.

by .:Endeavour:.
Tags: arithmetic, check, formula
 P: 80 I have this formula: F = qVe + (Pe - Pa) * Ae; I want to get q by its self. This what I did to get q by its self. F = qVe + (Pe - Pa) * Ae $$\frac{F - (P_e - P_a)}{(A_e)} = \frac{(V_e * q)(A_e)}{(A_e)}$$ [( F - (Pe - Pa)) ÷ Ae] ÷ Ve = q This is how I got q by its self in order to solve for q. I'm not sure if its correct, so please look over it to see if its correct. Thank you for your time.
 Sci Advisor P: 2,193 Check my arithmetic for this formula. I think what you did, correct me if I'm wrong, is subtract (Pe-Pa) from both sides and then divide everything by Ae. The problem with this is that subtracting from the right hand side doesn't eliminate the (Pe-Pa) because it has an Ae attached to it. What Office_Shredder is saying is you have to move the entire term, Ae(Pe-P), over to the other side: $$F=qV_{e}+(P_{e}-P_{a})A_{e}$$ Subtract $$(P_{e}-P_{a})A_{e}$$ from both sides: $$F-(P_{e}-P_{a})A_{e}=qV_{e}+(P_{e}-P_{a})A_{e}-(P_{e}-P_{a})A_{e}$$ $$F-(P_{e}-P_{a})A_{e}=qV_{e}$$ $$\frac{F-(P_{e}-P_{a})A_{e}}{V_{e}}=q$$ To make explicit what you did: $$F=qV_{e}+(P_{e}-P_{a})A_{e}$$ $$F-(P_{e}-P_{a})=qV_{e}+(P_{e}-P_{a})A_{e}-(P_{e}-P_{a})$$ And you see the terms don't drop out on the right hand side. Hope this clarifies.