3x - 2y = 6
6x - 4y = 12
Show that the simultaneous equations have an infinite number of solutions.
Hello, Joker37!
$\displaystyle \begin{array}{cccc}3x - 2y &=& 6 & {\color{blue}[1]} \\ 6x - 4y &=& 12 & {\color{blue}[2]} \end{array}$
Show that the system has an infinite number of solutions.
$\displaystyle \begin{array}{cccc}\text{Multiply {\color{blue}[1]} by -2:} & \text{-}6x + 2y &=& \text{-}12 \\ \text{Add {\color{blue}[2]}:} & 6x - 4y &=& 12 \end{array}$
And we get: . . . . . . . . $\displaystyle 0 \:=\:0$
This is true, regardless of the values of $\displaystyle x$ and $\displaystyle y$.
Therefore, there are an infinite number of solutions.
Using an augmented matrix?
Perform some elementary row operations on the augmented matrix to get to reduced row echelon form.
$\displaystyle \left[\begin {array}{cc|c}3 & -2 & 6 \\ 6 & -4 & 12 \end{array}\right]$
Divide R1 by 3
$\displaystyle \left[\begin {array}{cc|c}1 & -\frac{2}{3} & 2 \\ 6 & -4 & 12 \end{array}\right]$
Multiply R1 by -6 and add to R2
$\displaystyle \left[\begin {array}{cc|c}1 & -\frac{2}{3} & 2 \\ 0 & 0 & 0 \end{array}\right]$
If there is a row in the augmented matrix containing all zeros (to the left and right of the line), then it is as if that row could be completely ignored, as it represents 0 = 0, which by itself is inconclusive. Whether or not there are solutions depends on the remaining row. The remaining row consists of an equation in 2 non-zero variables. We have more variables than we have equations, thus indicating an infinite number of solutions.