# How to solve simultaneous equations using Delta?

I can solve simultaneous equations using multiple methods, but came across this new procedure while revising for my exam. I've never seen anything like this before and can't find any explanations of it on the net, so I was wondering if someone could please give me a hand to understand it?

Put equations $(1)$ and $(2)$ into matrix form.

$$\begin{bmatrix}1/s\\-4/(s+2)\end{bmatrix}=\begin{bmatrix}2+4s&-2\\-2&s+2\end{bmatrix}\begin{bmatrix}I_1\\I_2\end{bmatrix}$$

$$\boxed{\Delta=\frac2s\left(s^2+2s+4\right),\quad\Delta_1=\frac{s^2-4s+4}{s(s+2)},\quad\Delta_2=\frac{-6}s}$$

$$\boxed{I_1=\frac{\Delta_1}\Delta=\frac{1/2\cdot\left(s^2-4s+4\right)}{(s+2)\left(s^2+2s+4\right)}=\frac A{s+2}+\frac{Bs+C}{s^2+2s+4}}$$

$$1/2\cdot\left(s^2-4s+4\right)=A\left(s^2-4s+4\right)+B\left(s^2+2s\right)+C(s+2)$$

Equating coefficients:

\begin{align} s^2:1/2&=A+B\\ s^1:-2&=2A+2B+C\\ s^0:2&=4A+2C \end{align}

Solving these equations leads to $\quad A=2,\quad B=-3/2,\quad C=-3$

It's just getting the delta equations that is unknown to me.

The solution of any $2\times2$ linear system $$ax+by=c\\a'x+b'y=c'$$ can be established by eliminating both unknowns in turn.

1. eliminate $y$ by multiplying by $b'$ and $-b$ and adding: $$(ab'-a'b)x=cb'-c'b.$$i.e. $$\Delta\cdot x=\Delta_x.$$

2. eliminate $x$ by multiplying by $-a'$ and $a$ and adding: $$(ab'-a'b)y=c'a-ca'.$$i.e. $$\Delta\cdot y=\Delta_y.$$

Those "deltas" are called determinants. The main one, $\Delta$ (sometimes called the discriminant of the system) tells you if there is a unique solution when it is nonzero.

This rule generalizes to larger systems of equations.

In the given case, as the coefficients are parametric, the solution is expressed as the quotient of polynomials (in the variable $s$). Another technique, decomposition in simple fractions, was used to "simplify" it.

Look up determinant, and Cramer's Rule. It is an efficient way to solve a $2\times 2$ system, and even a $3\times 3$, but gets inefficient after a while.