How do you graph the following equation and identity y-intercept \[2y+3x=-2\]?

Hint: We are asked to draw the graph of the equation \[2y+3x=-2\]. The degree of an equation is the highest power of the variable present in it. So, for this equation, the highest power present \[x\] is 1, the degree is also 1. From this, it can be said that this is a linear equation. The graph of a linear equation represents a straight line.

Complete step-by-step solution:
The general equation of a straight line is \[ax+by+c=0\], where \[a,b,c\] are any real numbers. The given equation is \[2y+3x=-2\], the equation can also be written as \[3x+2y+2=0\], comparing with the general equation of straight line, we get \[a=3,b=2\And c=2\].
To plot the graph of an equation of the straight line, we should know at least two points, through which the line passes.
To make things simple, let’s take the X-intercept and Y-intercept as the two points. X-intercept is the point where the line crosses X-axis, this means that the Y-coordinate will be \[0\], similarly Y-intercept is the point where the line crosses Y-axis, so X-coordinate will be \[0\]. We will use this property now.
We substitute \[y=0\] in the equation \[3x+2y+2=0\], we get
\[\begin{align}
  & \Rightarrow 3x+2(0)+2=0 \\
 & \Rightarrow 3x+2=0 \\
\end{align}\]
Subtracting 2 from both sides of the equation we get,
\[\begin{align}
  & \Rightarrow 3x+2-2=0-2 \\
 & \Rightarrow 3x=-2 \\
\end{align}\]
We divide 3 to both sides we get,
\[\begin{align}
  & \Rightarrow \dfrac{3x}{3}=\dfrac{-2}{3} \\
 & \therefore x=\dfrac{-2}{3} \\
\end{align}\]
So, the coordinates of the X-intercept are \[\left( \dfrac{-2}{3},0 \right)\].
Similarly, now we substitute \[x=0\] in the equation, we get
\[\begin{align}
  & \Rightarrow 3(0)+2y+2=0 \\
 & \Rightarrow 2y+2=0 \\
\end{align}\]
Subtracting 2 from both sides of the equation, we get
\[\begin{align}
  & \Rightarrow 2y+2-2=0-2 \\
 & \Rightarrow 2y=-2 \\
\end{align}\]
Dividing both sides of above equation by 2, we get
\[\begin{align}
  & \Rightarrow \dfrac{2y}{2}=\dfrac{-2}{2} \\
 & \therefore y=-1 \\
\end{align}\]
So, the coordinates of the Y-intercept are \[\left( 0,-1 \right)\].
Thus, we get the Y-intercept of the line as \[-1\].
 Using these two points we can plot the graph of the equation as follows:

We can identify the Y-intercept from the graph as the point where the line crosses the Y-axis.

Note: Here, we found the two points which are X-intercept and Y-intercept by substituting either x or \[y\] to be zero, one at a time. We can also find these values by converting the straight-line equation to the equation in intercept form which is, \[\dfrac{x}{a}+\dfrac{y}{b}=1\]. Here, \[a\And b\] are X-intercept and Y-intercept respectively.