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
… … $y = 2x + 1 \qquad \Big[ \textbf{1} \Big]$
… … $y = 2x - 4 \qquad \Big[ \textbf{2} \Big]$
Solution:
… … $\Big[\textbf{1} \Big] = \Big[\textbf{2}\Big]$
… … $2x + 1 = 2x - 4 \qquad \{ -2x \}$
… … $1 = -4$
This is impossible so no solutions.
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
… … $y = 2x - 1 \qquad \Big[ \textbf{1} \Big]$
… … $2x - y = 1 \qquad \Big[ \textbf{2} \Big]$
Solution:
… … Sub [1] into [2]
… … $2x - (2x - 1) = 1 \qquad \{Simplify\}$
… … $1 = 1$
This is always true so infinite number of solutions.
Matrices and No Unique Solutions
Recall that for the simultaneous equations
… … $ax + by = e$
… … $cx + dy = f$
This can be written in matrix form as
And the solutions can be obtained by multiplying both sides by the inverse matrix.
Notice that if the determinant equals zero:
… … $\Big| \text{A} \Big| = ad-bc=0$
there can be no unique (or particular) 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
… … $-2x + y = -1$
… … $\; \; \; 2x - y = 1$
Solution:
These can be written in matrix form as:
The determinant is
… … $\Big| \text{A} \Big| = -2 \times -1 - 1 \times 2$
… … $\quad \; \, = \, 2 - 2$
… … $\quad \; \, = \, 0$
Hence there are no particular solutions.
{Graphing or manipulating algebraically shows they are coincident so there is an infinite number of solutions.}
Simultaneous Equations with Pronumeral Coefficients
Example 4a
Consider a set of simultaneous equations
… … $ax + 5y = 4 \qquad \qquad \Big[ \textbf{1} \Big]$
… … $2x + \Big( a - 3 \Big) y = 1 \qquad \Big[ \textbf{2} \Big]$
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.
- We need to find the value of a where they have the same gradient.
- There will be one unique solution for all of R excluding those values of a.
Rearranging equation [1]
… … $ax + 5y = 4$
… … $5y = -ax + 4$
… … $y = -\dfrac{a}{5}x + \dfrac{4}{5}$
… … so $\qquad m_1 = -\dfrac{a}{5}$
Rearranging equation [2]
… … $2x + \Big( a - 3 \Big) y = 1$
… … $\Big( a - 3 \Big) y = -2x + 1$
… … $y = - \dfrac{2}{a-3} x + \dfrac{1}{a - 3}$
… … so $\qquad m_2 = - \dfrac{2}{a-3}$
Gradients are the same where $m_1 = m_2$
… … $- \dfrac{a}{5} = - \dfrac{2}{a-3} \qquad \Big\{ \times -1 \Big\}$
… … $\dfrac{a}{5}=\dfrac{2}{a-3} \qquad \qquad \Big\{ \text{Cross Multiply} \Big\}$
… … $a \Big( a-3 \Big) = 2 \times 5$
… … $a^2 - 3a = 10$
… … $a^2 - 3a - 10 = 0 \qquad \Big[ \textbf{3} \Big]$
… … $\Big( a-5 \Big) \Big( a+2 \Big) = 0$
… … $a - 5 = 0 \qquad \textit{or} \qquad a + 2 = 0$
… … $\quad a = 5 \qquad \textit{or} \qquad a = -2$
Hence there are no unique solutions when $a = –2,\; 5$
Hence there is a unique solution for $\; a \in R \backslash \big\{ -2, \; 5 \big\}$
Note: The type of situation at $a = –2,\; 5$ can be discovered by substituting these values into the original equations.
When $a = –2$
… … $-2x+5y=4 \qquad \Big[ \textbf{1} \Big]$
… … $2x - 5y = 1 \qquad \Big[ \textbf{2} \Big]$
These are parallel lines.
When $a = 5$
… … $5x+5y=4 \qquad \Big[ \textbf{1} \Big]$ \
… … $2x+2y=1 \qquad \Big[ \textbf{2} \Big]$
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
… … $ax + 5y = 4 \qquad \qquad \Big[ \textbf{1} \Big]$
… … $2x + \Big( a - 3 \Big) y = 1 \qquad \Big[ \textbf{2} \Big]$
Find the values of a for which there is one unique solution.
Solution 2:
Express these equations in matrix form.
… … $\left[ \begin{matrix} a&5 \\ 2 & a-3 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 4 \\ 1 \\ \end{matrix} \right]$
There will be no unique solutions when the determinant is zero.
… … $a \Big( a-3 \Big) - 2 \times 5 = 0$
… … $a^2 - 3a - 10 = 0$
This brings us to the same point as at equation [3] in the first solution (above).
So proceed exactly the same as above.
… … $a^2 - 3a - 10 = 0 \qquad \Big[ \textbf{3} \Big]$
… … $\Big( a-5 \Big) \Big( a+2 \Big) = 0$
… … $a - 5 = 0 \qquad \textit{or} \qquad a + 2 = 0$
… … $\quad a = 5 \qquad \textit{or} \qquad a = -2$
Hence there are no unique solutions when $a = –2, \; 5$
Hence there is a unique solution for $\; a \in R \backslash \big\{ -2, \; 5 \big\}$
Example 5 (Eg 32 p36)
For the linear simultaneous equations given below, determine the values of t for which there are:
- Infinitely many solutions
- No solutions
- A unique solution.
… … $\quad tx - 3y = 6 \qquad \Big[ \textbf{1} \Big]$
… … $2x + \big( t-5 \big) y = 3t \qquad \Big[ \textbf{2} \Big]$
Solution:
Write in matrix form:
… … $\left[ \begin{matrix} t & -3 \\ 2 & t-5 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] = \left[ \begin{matrix} 6 \\ 3t \\ \end{matrix} \right]$
There will be no unique solutions when the determinant is zero.
… … $t \big( t-5 \big) - \big( -3 \big) \big( 2 \big) = 0$
… … $t^2 - 5t + 6 = 0$
… … $\big( t-2 \big) \big( t - 3 \big) = 0$
… … $t = 2 \qquad or \qquad t = 3$
When $t = 2$
… … $2x - 3y = 6 \qquad \Big[ \textbf{1} \Big]$
… … $2x - 3y = 6 \qquad \Big[ \textbf{2} \Big]$
These are coincident so,
when $t = 2$ there are infinitely many solutions (1)
When $t = 3$
… … $3x - 3y = 6 \qquad \Big[ \textbf{1} \Big]$
… … $2x - 2y = 9 \qquad \Big[ \textbf{2} \Big]$
These are parallel so,
when $t = 3$ there are no solutions (2)
(3) There will be a unique solution when $t \in R \backslash \Big\{ 2, \; 3 \Big\}$
.