Simultaneous Equations

Simultaneous equations act as an entry point for understanding problems in linear algebra.

These equations have multiple unknown values that we want to solve. Solving a linear equation means finding a point at which a system of equations intersect. For example, the two lines below \( y = 2x - 2 \) and \( y = -\frac{1}{2}x + 5 \) intersect at point \( (2, 2) \) meaning that this is where the solution for this system of equations is.

Importantly, it is the co-efficients of the unknowns in different equations that describe how the equations behave. For example, in the well known form \( y = mx \) the \( m \) describes the actual character of the equation. Modifying \( m \) modifies the gradient.

To solve a system of equations with \( n \) unknowns requires at least \( n \) number of equation. So to solve a system of three dimensional equations \( x + y + z = m \) requires at least three equations.

Two Unknowns

A system of linear equations with two unknowns (namely \( x \) and \( y \)) can be stated as

\[ ax + by = m \]

\[ cx + dy = n \]

With unique co-efficients indicated by \( a \), \( b \), \( c \) and \( d \).

Parallel Lines

You can show that a system of two dimensional linear equations has a solution if the gradient of the two equations is different. For the system of equations

\[ ax + by = m \]

\[ cx + dy = n \]

A solution exists if

\[ ad - bc \neq 0 \]

If the gradient of the lines is the same it would mean that the lines are parallel and would never meet. Therefore there would be no solution.

To show this convert the standard form of the equations into y intercept form by dividing by the associated co-efficient of \( y \) and subtracting the \( x \) term.

\[ \frac{ax + by = m}{b} \]

\[ \frac{cx + dy = n}{d} \]

\[ \frac{a}{b}x + y = \frac{m}{b} \]

\[ \frac{c}{d}x + y = \frac{n}{d} \]

In standard form as

\[ y = - \frac{a}{b}x + \frac{m}{b} \]

\[ y = - \frac{c}{d}x + \frac{n}{d} \]

Now that the gradient is isolated as \( \frac{a}{b}x \) and \( \frac{c}{d}x \) respectively it is clear that if \( \frac{a}{b} = \frac{c}{d} \) then the lines are parallel and this system of equations does not have a solution.

This can be stated more elegantly by cross multiplying each side of the equation by their denominators to get

\[ ad = cb \]

If this case then the lines are parallel. This is just a restating of the above fractional form. This where the rule that if

\[ ad - cb \neq 0 \]

Then the lines are parallel and the system has no solution.

Overlayed Lines

A system of equations has an infinite number of solutions if the equations of the lines have the same gradient and the same \( y \) intercept. In this case the two equations in the system eseentially just express multiples of the same equation.

For the system of equations

\[ ax + by = m \]

\[ cx + dy = n \]

The system has infinite solutions

If \( ad - cb = 0 \) and \( dm - bn = 0 \)

We can build this rule from the above parallel line formula that was already described. For this rule we also take into account the \( y \) intercept which is expressed as \( \frac{m}{b} \) and \( \frac{n}{d} \). Therefore, the lines share the same intercept if

\[ \frac{m}{b} = \frac{n}{d} \]

Which can be re-written using cross multiplication in the subtractive form as

Lines share a y intercept if \( dm - bn = 0 \)

For example, the two equations below are just multiple forms of the equation \( y = 2x + 1 \)

\[ 2y - 4x = 2 \]

\[ 4y - 8x = 4 \]

If they are converted to \( y \) intercept form

\[ y = \frac{4}{2}x + \frac{2}{2} \]

\[ y = \frac{8}{4}x + \frac{4}{4} \]

It becomes evident that these are the same equation because they both share a gradient of \( 2 \) and a \(y \) intercept of \( 1 \). If they are graphed they overlay at each other and have a solution at every point.