01Hsolving2equations

= Solving Systems of Two Equations =

A ** System of Equations ** is a different term for ** Simultaneous Equations **.

Simultaneous Equations with Two Variables.
To solve two simultaneous equations means to find any points (x, y) where both equations are true at the same time.

Recall that the graph of a function shows all the points (x, y) where that function is true. {A line is made up of an infinite number of points}.

If we graph two different linear functions, the lines will intersect at one point {unless they are parallel}. At the point where they intersect, both functions are true at the same time.

This means the coordinates of the point of intersection are the solution to the simultaneous equations.

Solving Simultaneous Equations
You have previously learned to solve simultaneous equations using four methods.
 * Substitution Method
 * Elimination Method
 * Using Matrices
 * Using CAS calculator

It is important to be comfortable using all four of those methods.

Parallel Lines
Two linear functions produce parallel lines if they have the same gradient.

If we attempt to solve two simultaneous equations that are parallel, there will be ** no solution **.

If solving algebraically, you will get an impossible equation (eg 1 = –4)

Example
Solve simultaneously
 * Solution:**

Coincident Lines
If two linear functions create exactly the same line, they are called **coincident**.

If we attempt to solve two simultaneous equations that are coincident, then there is an ** infinite number of solutions **.

{ie the line has an infinite number of points and every point on the line satisfies both functions.}

If solving algebraically, you will get an equation that is always true (eg 1 = 1)

Example
Solve simultaneously


 * Solution:**

Matrices and No Unique Solutions
Recall that for the simultaneous equations



This can be written in matrix form as

And the solutions can be obtained by multiplying the inverse matrix.

Notice that if the ** determinant ** equals zero:

math . \qquad \Big| \text{A} \Big| = ad-bc=0 math

there can be no __**unique**__ solutions {because we are dividing by zero}.

If the determinant is zero, it could be that
 * there are __no solutions__ OR
 * there could be an __infinite number of solutions__.

** Example 3 **
Solve simultaneously


 * Soution:**

These can be written in matrix form as:

The determinant is Hence there are no particular solutions.

{Graphing or manipulating algebraically shows they are coincident so there is an infinite number of solutions.}

** Example 4a **
Consider a set of simultaneous equations

Find the values of a for which there is one unique solution.


 * Solution 1:**

There will be one unique solution __unless__ the two lines are __parallel__ or __coincident__. This will occur when the lines have the same gradient.
 * We need to find the value of a where they have the same gradient.
 * The answer will be all of R excluding those values of a.

Rearranging equation ** [1] **

math \\ . \qquad ax + 5y = 4 \\. \\ . \qquad 5y = -ax + 4 \\. \\ . \qquad y = -\dfrac{a}{5}x + \dfrac{4}{5} \\. \\ \textbf{so} \qquad m_1 = -\dfrac{a}{5} math

Rearranging equation ** [2] **

math \\ . \qquad 2x + \Big( a - 3 \Big) y = 1 \\. \\ . \qquad \Big( a - 3 \Big) y = -2x + 1 \\. \\ . \qquad y = - \dfrac{2}{a-3} x + \dfrac{1}{a - 3} \\. \\ \textbf{so} \qquad m_2 = - \dfrac{2}{a-3} math

Gradients are the same where m 1 = m 2

math \\ . \qquad - \dfrac{a}{5} = - \dfrac{2}{a-3} \qquad \Big\{ \times -1 \Big\} \\. \\ . \qquad \dfrac{a}{5}=\dfrac{2}{a-3} \qquad \qquad \Big\{ \text{Cross Multiply} \Big\} \\. \\ . \qquad a \Big( a-3 \Big) = 2 \times 5 \\. \\ . \qquad a^2 - 3a = 10 math

math \\ . \qquad a^2 - 3a - 10 = 0 \qquad \Big[ \textbf{3} \Big] \\. \\ . \qquad \Big( a-5 \Big) \Big( a+2 \Big) = 0 \\. \\ . \qquad a - 5 = 0 \qquad \textit{or} \qquad a + 2 = 0 \\. \\ . \qquad \quad a = 5 \qquad \textit{or} \qquad a = -2 math

Hence there are __**no**__ unique solutions when a = –2, 5

math \text{Hence there is a unique solution for } \; a \in R \backslash \big\{ -2, \, 5 \big\} math


 * Note: ** The type of situation at a = –2, 5 can be discovered by substituting these values into the original equations.

When a = –2 math \\ . \qquad -2x+5y=4 \qquad \Big[ \textbf{1} \Big] \\ . \qquad 2x - 5y = 1 \qquad \Big[ \textbf{2} \Big] math

These are parallel lines.

When a = 5 math \\ . \qquad 5x+5y=4 \qquad \Big[ \textbf{1} \Big] \\ . \qquad 2x+2y=1 \qquad \Big[ \textbf{2} \Big] math

These are parallel lines.

Alternate Solution Using Matrices.
The same problem could have been solved using matrices.

** Example 4b **
Consider a set of simultaneous equations Find the values of a for which there is one unique solution.


 * Solution 2:**

Express these equations in matrix form.

math . \qquad \left[ \begin{matrix} a&5 \\ 2 & a-3 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 4 \\ 1 \\ \end{matrix} \right] math

There will be no unique solutions when the __**determinant**__ is zero.

math \\ .\qquad a \Big( a-3 \Big) - 2 \times 5 = 0 \\. \\ . \qquad a^2 - 3a - 10 = 0 math

This brings us to the same point as at equation ** [3] ** in the first solution (above). So proceed exactly the same as above.

math \\ . \qquad a^2 - 3a - 10 = 0 \qquad \Big[ \textbf{3} \Big] \\. \\ . \qquad \Big( a-5 \Big) \Big( a+2 \Big) = 0 \\. \\ . \qquad a - 5 = 0 \qquad \textit{or} \qquad a + 2 = 0 \\. \\ . \qquad \quad a = 5 \qquad \textit{or} \qquad a = -2 math

Hence there are no unique solutions when a = –2, 5

math \text{Hence there is a unique solution for } \; a \in R \backslash \big\{ -2, \, 5 \big\} math


 * Example 5 ** (Eg 32 p36 new text)

For the linear simultaneous equations given below, determine the values of t for which there are:
 * 1) Infinitely many solutions
 * 2) No solutions
 * 3) A unique solution.

math \\ . \qquad \quad tx - 3y = 6 \qquad \Big[ \textbf{1} \Big] \\ . \qquad 2x + \big( t-5 \big) y = 3t \qquad \Big[ \textbf{2} \Big] math


 * Solution:**

Write in matrix form: math . \qquad \left[ \begin{matrix} t & -3 \\ 2 & t-5 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 6 \\ 3t \\ \end{matrix} \right] math

There will be no unique solutions when the determinant is zero.

math \\ . \qquad t \big( t-5 \big) - \big( -3 \big) \big( 2 \big) = 0 \\. \\ . \qquad t^2 - 5t + 6 = 0 \\. \\ . \qquad \big( t-2 \big) \big( t - 3 \big) = 0 \\. \\ . \qquad t = 2 \qquad or \qquad t = 3 math

When t = 2 math \\ . \qquad 2x - 3y = 6 \qquad \Big[ \textbf{1} \Big] \\ . \qquad 2x - 3y = 6 \qquad \Big[ \textbf{2} \Big] \\ math

These are coincident so,
 * ** when t = 2 there are infinitely many solutions (1) **

When t = 3 math \\ . \qquad 3x - 3y = 6 \qquad \Big[ \textbf{1} \Big] \\ . \qquad 2x - 2y = 9 \qquad \Big[ \textbf{2} \Big] \\ math

These are parallel so,
 * ** when t = 3 there are no solutons (2) **

math . \qquad t \in R \backslash \Big\{ 2, \, 3 \Big\} math
 * (3) There will be a unique solution when **

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