If you need to ask this many questions here, you really need to talk to your maths teacher to get some help!
You should know that y= mx + c is the equation of a straight line with a gradient of m and a y-intercept of c.
Let's rearrange 2x = 3y = 5 into that form:
2x + 3y = 5
<=>
3y = -2x + 5
<=>
y = -2/3 x + 5/3
So we know that the line in the question must have a gradient of -2/3
Any line which is parallel to it must also have the same gradient.
So we're seeking a line with an equation in the form y = mx + c
where m = -2/3
and where x=2 and y = -1 (from the coordinates in the question) satisfy that equation.
That gives us
-1 = (-2/3 times 2) + c
Adding 4/3 to both sides of the equation gives
c = 1/3
So the equation we require is y = -2/3 x + 1/3
To tidy that up, multiply throughout by 3:
3y = -2x +1
and then add 2x to both sides:
2x + 3y =1
Chris