Q:

Help Finding the Equation of a Line with a Given a Point and a Slope

I have been receiving a lot of help today and I really appreciate it, I am nearing the end of my class and need help studying for my final. In my review I came across a problem that i am having difficulty figuring out. I need to write an equation of a line that has a slope of $3/4$ and passes through a point of $(-2,4)$. 
So far here is what I have accomplished:
Using the equation $y-y_1= m(x-x_1)$ 
IDK if this equation is the one I should be using.
$y - 4 = \dfrac{3}{4}[x-(-2)]$
to
$y - 4 = \frac{3}{4}(x+2)$
to
$y - 4 = \dfrac{3x}{4} + \dfrac{6}{4}$ 
Do I reduce the $\frac{6}{4}$ to $\frac{3}{2}$? Also, is this the correct step I should be doing here?
$y - 4 = 3x + 6$
Since both values on the right have the same denominator, can I just get rid of it?
to
$y = 3x + 10$
Is this correct?
If this is correct, is there a simpler way of doing this? If not, what should I be doing?

A:

You almost have it with $y-4=(3x/4)+(6/4)$.  Just get the 4 to the right hand side $y=\frac34x+\frac32+4$.  So 
$$y=\frac34x+\frac{11}2$$

