jklops686 said:
Homework Statement
I have a picture attached with this static equilibrium problem.
(Solve system for Fac and Fab)
Homework Equations
Sum of x and y both=0
The Attempt at a Solution
I just don't have a systematic easy approach to solving these systems. Does anyone have one? From the picture, they used the system of equations and add them together cancelling variables, but I don't know how exactly they did that. This class would be so much easier if I could do this well!
There is a very standard way to solve such problems. It is not always the fastest way, but it is one that works every time. You just use one of the equations to solve for one variable in terms of the other; then you put that expression into the other equation.
In your case you have two equations of the form a*x + b*y = 0, c*x + d*y = e. The unknowns are x and y, and the given parameters are a, b, c, d, e. You can solve for x in terms of y from the first equation: a*x + b*y = 0 ==> x = -(b/a)*y (since a ≠ 0). Now, where you see x in the second equation, just substitute in that expression, so the second equation now becomes c*(-b/a)*y + d*y = e, or [d - (b*c/a)]*y = e. If d - (b*c/a) ≠ 0 we can solve for y by division: y = e/[d - (b*c/a)]. Once we know y, we can substitute that into our x-expression to find x.
Note that if d - (b*c/a) = 0 we will have an equation of the form 0*y = e. If e ≠ 0 this is an impossible equation, which indicates that we must have made an error in setting up the problem (or else were handed an inconsistent system by somebody evil). On the other hand, if e = 0 we have the equation 0*y = 0, which allows any value whatsoever for y.
Later, if and when you take linear algebra you will learn such methods as "row reduction", etc., but they are really nothing more than the above method made more streamlined.