Question

How do you write $2x-3y=15$ in slope – intercept form and graph it?

Answer 1

Hint: We know that slope – intercept of a straight line is as follows: $y=mx+c$. In this form, “m” is the slope of the straight line and “c” is the intercept of the line. So, in order to write the given equation $2x-3y=15$ in slope – intercept form, we have to rearrange in such a manner so that we will get this equation of the following form: $y=mx+c$. Now, to graph, this equation, we are going to first put x as 0 in the above equation and then see what value of y we are getting and make this as point A then put y as 0 in the above equation and see what value of x we are getting and mark it as point B. And then plot these two points on the graph paper and joining of these two points will give you straight line.

Complete step-by-step solution:
The equation given above which we have to write in slope – intercept form is as follows:
$2x-3y=15$
We know that the slope – intercept form for straight line is as follows:
$y=mx+c$
In the above equation, “m” is the slope and “c” is the intercept.
And as the equation given above is a straight line so we are going to rearrange the above equation in this slope – intercept form.
Adding 3y on both the sides of the given equation we get,
$\begin{align}
  & \Rightarrow 2x-3y+3y=15+3y \\
 & \Rightarrow 2x=15+3y \\
\end{align}$
Now, subtracting 15 on both the sides we get,
\[\Rightarrow 2x-15=3y\]
Dividing 3 on both the sides we get,
$\Rightarrow \dfrac{2x-15}{3}=y$
Rewriting the above equation we get,
$\begin{align}
  & \Rightarrow \dfrac{2x}{3}-\dfrac{15}{3}=y \\
 & \Rightarrow \dfrac{2x}{3}-5=y \\
\end{align}$
Now, we can write -5 as follows:
$\Rightarrow \dfrac{2x}{3}+\left( -5 \right)=y$
Hence, we have written the given equation in the slope – intercept form which is:
$\Rightarrow y=\dfrac{2x}{3}+\left( -5 \right)$
Now, drawing the graph of the above equation by putting x as 0 in the above equation and the see the value of y we are getting:
$\begin{align}
  & \Rightarrow y=\dfrac{2}{3}\left( 0 \right)-5 \\
 & \Rightarrow y=-5 \\
\end{align}$
Let us name this point (0, -5) as A.
Now, plotting this point A on the graph we get,

Now, putting y as 0 in the above equation we get,
$\begin{align}
  & \Rightarrow 0=\dfrac{2x}{3}+\left( -5 \right) \\
 & \Rightarrow 5=\dfrac{2x}{3} \\
 & \Rightarrow x=\dfrac{15}{2} \\
\end{align}$
Let us name the above point as $B\left( \dfrac{15}{2},0 \right)$ and then plotting this point on the graph we get,

Now, joining point A to point B we get,

Hence, we have drawn the graph of the given equation.

Note: The negative intercept which we are getting from the slope – intercept form shows that the straight line is cutting the negative side of the y axis.
In the above, the intercept of the above equation $y=\dfrac{2x}{3}+\left( -5 \right)$ is -5 which is negative so the straight line is cutting the negative y axis which is shown above as follows: