Solution linear equation with two unknowns

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.

geogebra plaatje

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.

geogebra plaatje

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.

geogebra plaatje

Give the solutions to $6\cdot x-5\cdot y-7=0$ .
Express your answer in the form $y=a \cdot x +b$ or $x=b$ for the correct values of $a$ and $b$ .

$y={{6\cdot x}\over{5}}-{{7}\over{5}}$

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:
$\begin{array}{rcl} 6\cdot x-5\cdot y-7&=&0\\&& \phantom{xxx}\blue{\text{the given equation}}\\ 6\cdot x-5\cdot y&=&7 \\ && \phantom{xxx}\blue{\text{left and right-hand side plus }7}\\ -5\cdot y&=&-6\cdot x+7 \\&& \phantom{xxx}\blue{\text{left and right-hand sides }6\cdot x\text{ subtracted }}\\ y&=&\displaystyle {{6\cdot x}\over{5}}-{{7}\over{5}}\\&& \phantom{xxx}\blue{\text{left and right-hand sides divided by the coefficient of }y} \end{array}$
Hence, the solutions to $6\cdot x-5\cdot y=7$ are equal to the oblique line $y={{6\cdot x}\over{5}}-{{7}\over{5}}$ .

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