Question

How do you graph the equation $y=5x-3$?

Answer 1

Change of form of the given equation will give the x-intercept and y-intercept of the line $y=5x-3$. We change it to the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x-intercept, and y-intercept of the line as $p$ and $q$ respectively. then we place the points on the axes and from there we draw the line on the graph.

Complete step by step answer:
We are taking the general equation of a line to understand the slope and the intercept form of the line $y=5x-3$. The given equation is in the form of $y=mx+k$. m is the slope of the line. The slope of the line is $5$.
We have to find the x-intercept and y-intercept of the line $y=5x-3$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be$p$ and $q$ respectively. The points will be $\left( p,0 \right),\left( 0,q \right)$.
The given equation is $y=5x-3$. Converting into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$, we get
$\begin{align}
  & y=5x-3 \\
 & \Rightarrow 5x-y=3 \\
 & \Rightarrow \dfrac{x}{{}^{3}/{}_{5}}+\dfrac{y}{-3}=1 \\
\end{align}$
Therefore, the x intercept, and y intercept of the line $y=5x-3$ is $\dfrac{3}{5}$ and 3 respectively. The axes intersecting points are $\left( \dfrac{3}{5},0 \right),\left( 0,-3 \right)$.

Note:
 A line parallel to the X-axis does not intersect the X-axis at any finite distance. Hence, we cannot get any finite x-intercept of such a line. The same goes for lines parallel to the Y-axis. In the case of the slope of a line, the range of the slope is 0 to $\infty $.