## Simultaneous Equations

Simultaneous equations represent two or more functions in a Cartesian plane and solving them is their point of intersection.

There are three types of simultaneous equations we can have where there is one solution, no solution or infinite solutions.

• There is one solution when there is a single intersection point.
• No solution when there are no intersections and the lines are parallel different lines.
• Infinite solutions when they represent the same line.

There are two main methods used to solve simultaneous equations one is by elimination and the other is through matrices.

### Elimination Method

Elimination is done by making the coefficients of a variable the same across two equations and then minusing the two equations to eliminate the variable for example: $4y+5x=6\newline 8y+2x=4$

Equation 1 ×2 $8y+10x=12$

Equation 3-Equation 2 $8y+10x -(8y+2x)=12-4 \newline 8x=8 \newline x=1$

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Sub in x=1 into equation 1 $4y+5(1)=6 \newline y=\frac{1}{4}$

This gets more complicated with more variables but a few more steps of elimination will allow you to break the equation down to one variable. As long as you have as many equations as variables you can find the solution.

### Matrix

The other method is the matrix method. Matrices can be used to solve simultaneous equations and are the easiest methods if you are able to use a calculator.

The basic form of simultaneous equations is:

AX=B

Where:

• A is the coefficients of the variables in a n by n matrix
• X is the n number of variables to be solved for in a n by 1 matrix
• B is the set of numbers the equations equal in a n by 1 matrix

To solve the matrix we need to A inverse. Once we find A inverse we multiply both sides: $A^{-1}AX=A^{-1}B$

Remember $A^{-1}A = I$ $IX=A^{-1}B \newline X==A^{-1}B$

Since IX multiplies out to just X

### The Inverse Matrix

Determining the inverse matrix gets very complicated for any matrices larger than a 2 by 2 matrix, however it can be easily calculated by using your calculator.

Lets look at the same equations as before: $4y+5x=6\newline 8y+2x=4$

But now in the matrix form: $\begin{bmatrix} 4 & 5 \\ 8 & 2 \end{bmatrix}\left|\begin{array}{cc}xy\\ x\end{array}\right|=\left|\begin{array}{cc}6\\ 4\end{array}\right|$

How do we find the inverse? First we need the determinant.

#### Determinant $\left|\begin{array}{cc}a&b\\ c&d\end{array}\right|$ $det(A)=ad-bc \newline det(A)=4 \times 2 - 5 \times 8=-32$

#### Inverse $A^{-1}=\frac{1}{det(A)}\left|\begin{array}{cc}d&-b\\ -c&a\end{array}\right| \newline=-\frac{1}{32}\left|\begin{array}{cc}2&-5\\ -8&2\end{array}\right|$ $X=A^{-1}B$ $\left|\begin{array}{cc}y\\ x\end{array}\right|=-\frac{1}{32}\left|\begin{array}{cc}2&-5\\ -8&2\end{array}\right| \left|\begin{array}{cc}6\\ 4\end{array}\right|$ $\left|\begin{array}{cc}y\\ x\end{array}\right|=-\frac{1}{32}\left|\begin{array}{cc}2\times 6+-5\times 4\\ -8\times 6+4\times4\end{array}\right|$ $\left|\begin{array}{cc}y\\ x\end{array}\right|=\left|\begin{array}{cc}\frac{8}{32}\\ \frac{32}{32}\end{array}\right|=\left|\begin{array}{cc}\frac{1}{4}\\ 1\end{array}\right|$

This is the same as our result earlier. This time the method took a lot longer however if we had a calculator you can use that to calculate the inverse of A and to do the matrix multiplication. It is also essential to use a calculator to solve any set of simultaneous equations with more than three equations if you wish to use the matrix method.