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Date: 2017-06-03
Last Updated: 2017-06-03
NB: In school when we say Simultaneous Equations, we generally mean to say a system of linear equations or a linear system.
Well it's a fair question! They were always one thing I struggled with in school, but I could usually wrangle the answers for the 2 dimensional set of equations but I could never get the correct answers for a 3 dimensional set of equations. We probably all learnt the same method for solving Simultaneous Equations, or symbos as one teacher in my school referred to them as. I now know the method to be called "Elimination of Variables" which makes perfect sense when you think about it, because that's what it is.
Well initially I learnt them through algebra, so they were just arbitrary stupid problems that I had to do for homework to avoid getting in trouble. But when we did more complicated Coordinate Geometry, I learned that the equations are lines on the x-y plane and the point we find is the point of intersection. This knowledge enables us to apply the mathematical tools of variable elimination and Gauss-Jordan elimination (discussed below), to a wide range of real world problems.
The simplest method for solving a system of equations.
\[
\begin{aligned}
2x + 3y &= 6 \newline
4x + 9y &= 15 \newline
\hline
-4x - 6y &= -12 \newline
4x + 9y &= 15 \newline
\hline
0x + 3y &= 3 \newline
\end{aligned}
\]
\[
\therefore y = 1
\]
\[
\begin{aligned}
2x + 3(1) &= 6 \newline
2x &= 3
\end{aligned}
\]
\[
\therefore x = \frac{3}{2}
\]
The above figures show that this elimination method works fine for 2D systems but as you can imagine get's tricky for higher order systems.
This method I learnt in my Linear Algebra module in university, this makes the process seem much easier (at least to me). You convert the system of equations into whats called an "Augmented Matrix", and you perform basic row (usually arithmetic) operations on each row until the matrix reaches whats known as "reduced row echelon form". There are three primary row operations, namely;
1. Swap the positions of two rows.
2. Multiply a row by a nonzero scalar.
3. Add to one row a scalar multiple of another.
\[
\begin{aligned}
x + 3y - 2z &= 5 \newline
3x + 5y + 6z &= 7 \newline
2x + 4y + 3z &= 8
\end{aligned}
\]
$$
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
3 & 5 & 6 & 7 \newline
2 & 4 & 3 & 8
\end{array}\right]
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
0 & -4 & 12 & -8 \newline
2 & 4 & 3 & 8
\end{array}\right]
$$
$$
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
0 & -4 & 12 & -8 \newline
0 & -2 & 7 & -2
\end{array}\right]
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
0 & 1 & -3 & 2 \newline
0 & -2 & 7 & -2
\end{array}\right]
$$
$$
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
0 & 1 & -3 & 2 \newline
0 & 0 & 1 & 2
\end{array}
\right]
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 3 & -2 & 5 \newline
0 & 1 & 0 & 8 \newline
0 & 0 & 1 & 2
\end{array}
\right]
$$
$$
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 0 & -2 & -19 \newline
0 & 1 & 0 & 8 \newline
0 & 0 & 1 & 2
\end{array}\right]
\rightarrow
\left[
\begin{array}{rrr|r}
1 & 0 & 0 & -15 \newline
0 & 1 & 0 & 8 \newline
0 & 0 & 1 & 2
\end{array}\right]
$$
\[
\begin{aligned}
x &= -15 \newline
y &= 8 \newline
z &= 2
\end{aligned}
\]
In fact both methods are the exact same, and variable elimination could be used for \>3 dimensional systems. It is just how the mathematics is represented on paper.
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