Chapter 6.1, Problem 43PPS

To determine

Find a system of linear equations for the given graph and if it has one solution then name it.

Answer to Problem 43PPS

The system of linear equation:

$y=3x+4$ and $y=3x-3$

Explanation of Solution

Given information:

It is given in the question that,

The graph is:

Formula used:

Here,

Use the formula of two point form of a line that is,

  $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

(where $\left(x_1,y_1\right)\&(x_2,y_2)$ are the co-ordinate of the two point on line)

Calculation: Co-ordinate on the $1^{st}$ line: $(0,4)\&(-2,-2)$ .

Co-ordinate on the $2^{nd}$ line: $(1,0)\&(0,-3)$ .

So, for finding the equation of $1^{st}$ line,

By putting $(0,4)\&(-2,-2)$ in place of $\left(x_1,y_1\right)\&(x_2,y_2)$ in the formula that is

  $\Rightarrow \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}\Rightarrow \frac{y-4}{x-0}=\frac{-2-4}{-2-0}\\ \Rightarrow \frac{y-4}{x}=\frac{-6}{-2}\\ \Rightarrow \frac{y-4}{x}=\frac{3}{1}\\ \Rightarrow (y-4)=3x\\ \Rightarrow y=3x+4\end{array}$

So, the equation is $y=3x+4$

And for finding the equation of $2^{nd}$ line,

By putting $(1,0)\&(0,-3)$ in place of $\left(x_1,y_1\right)\&(x_2,y_2)$ in the formula that is,

  $\Rightarrow \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}\Rightarrow \frac{y-0}{x-1}=\frac{-3-0}{0-1}\\ \Rightarrow \frac{y}{x-1}=\frac{-3}{-1}\\ \Rightarrow \frac{y}{x-1}=\frac{3}{1}\\ \Rightarrow y=3(x-1)\\ \Rightarrow y=3x-3\end{array}$

So, the equation is $y=3x-3$ .