Systems of linear equations: An equation of a line
Solution linear equation with two unknowns
We have just seen that a solution to a linear equation with two unknowns of the form #\blue p \cdot x + \green q\cdot y +\purple r=0# is a point #\rv{x,y}#. In general, there are multiple solutions of a linear equation, we will now see what these solutions look like. For this we will use the same rules of reduction as for a linear equation with one unknown.
We solve the equation #\blue 3 \cdot x + \green 5 \cdot y +\purple 5=0# in the following way:
\[\begin{array}{rcl}3 \cdot x + 5 \cdot y +5&=&0\\&& \blue{\small\text{the equation}}\\
5 \cdot y+5&=&-3 \cdot x\\ && \blue{\small\text{both sides minus }3 \cdot x}\\5\cdot y &=& -3 \cdot x -5 \\ && \blue{\small\text{both sides minus }5}\\ y &=& -\frac{3}{5}x-1\\ && \blue{\small\text{both times divided by }5}\end{array}\]
All points on the line #{y}=-\tfrac{3}{5} {x}-1# are solutions to the equation.
We solve the equation #\green 5 \cdot {y}+\purple 5=0# in the following way:
\[\begin{array}{rcl}
5y + 5 &=& 0 \\
&&\blue{\small\text{the equation}} \\
5y &=& -5 \\
&&\blue{\small \text{both sides minus \(5\)}} \\
y &=& -1 \\
&&\blue{\small \text{both sides divided by \(5\)}} \\
\end{array}\]
All points on the horizontal line #{y}=-1# are solutions to the equation.
We solve the equation #\green 3 \cdot {x}+\purple 5=0# in the following way:
\[\begin{array}{rcl}
3x + 5 &=& 0 \\
&&\blue{\small\text{the equation}} \\
3x &=& -5 \\
&&\blue{\small \text{both sides minus \(5\)}} \\
x &=& -\frac{5}{3} \\
&&\blue{\small \text{both sides minus \(3\)}} \\
\end{array}\]
All points on the vertical line #{x}=-\tfrac53# are solutions to the equation.
The equation contains both variables #x# and #y#. Therefore, the solution is an oblique line of the form #y=a \cdot x +b#. We will find the solution to the equation by means of reduction:
\[\begin{array}{rcl}
4\cdot x-y+6&=&0\\&& \phantom{xxx}\blue{\text{the given equation}}\\
4\cdot x-y&=&-6 \\ && \phantom{xxx}\blue{\text{left and right hand side plus }-6}\\
-y&=&-4\cdot x-6 \\&& \phantom{xxx}\blue{\text{left and right hand side }4\cdot x\text{ subtracted }}\\
y&=&\displaystyle 4\cdot x+6\\&& \phantom{xxx}\blue{\text{left and right hand side divided by the coefficient of }y}
\end{array}\]
Hence, the solutions to #4\cdot x-y=-6# are equal to the oblique line #y=4\cdot x+6#.
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