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**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:

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