Systems of Equations in 4 or more unknowns

Question

I am doing problems that involve partial decomposition. I can't for the life of me remember how to solve some of the systems of equations that I end up with after multiplying out the partial fractions. For example...

$$A+B+C=0$$ $$A+C+D=0$$ $$A+B+D=2$$ $$A=-4$$

As far as I can see, elimination doesn't work. Maybe it does, and I'm just not seeing it. I've just been punching them into the solve function of my graphing calculator which works great, but I would like to solve them with pencil and paper.

Answer 1

Substitute the 4th equation in all the others, you get: $$B+C=4$$ $$C+D=4$$ $$B+D=6$$

The third equation is: $B=6-D$ and substitute this in the first equation, you get: $$6-D+C=4$$ $$\implies D-C=2$$ So now you have: $$D-C=2$$ $$D+C=4$$

Again, the second equation is: $D=4-C$ and substitute in the first one: $$4-C-C=2$$ $$\implies2C=2\implies C=1$$Since: $$D=4-C=4-1=3$$ and $$B=4-C=4-1=3$$ So: $$A=-4, B=3,C=1,D=3$$

In general (for simple linear system of equations such as this) put one equation in terms of a variable, and substitute this in another one, each time "reducing" a variable. Repeat this until only one variable is left.

For larger system of linear equations, matrices are the way to go.

Answer 2

Elimination does work. In fact, it is guaranteed to do so, though it may turn out to become less systematic, the larger the system grows.

First, you'll want to eliminate $A$ from the first three equations because you already know its value (so here, elimination consists of just plugging in the known value). This way, you obtain $B+C=4$, $C+D=4$, $B+D=6$ as three equations in three unknowns. Next eliminate ... why not $B$ using the first equation? So you plug $B=4-C$ into $B+D=6$ to arrive at $D-C=2$. You could now eliminate $C$ and find $D$ from the two equations $C+D=4$, $D-C=2$ - but at this point (if not earlier), I find it easier to simply add ($2D=6$) and subtract ($2C=2$) the two equations. Now equipped with $D=3$ and $C=1$, we find $B=4-C=3$, and already know $A=-4$.