How do you find the equation of a line that Contains point (-1, 2) and is parallel to x-2y=-3?

Method:
Convert the given equation into standard form for ease of recognition.

Use the given information to determine the equation of the new line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Converting the equation into standard form")#

#color(green)("Step 1")#

Given: #color(brown)(x-2y=-3)#

Subtract #color(blue)(x)# from both sides (isolates the term with #y# in it)

#" "color(brown)(x color(blue)(-x)-2y" "=" "-3color(blue)(-x))#

#" "0-2y" "=" "-x-3#
'........................................................................................................

#color(green)("Step 2")#

Change the equation so that the #y# term is positive

Multiply both sides by #color(blue)(-1)# giving:

#" "+2y=+x+3" "->" "2y=x+3#
'...............................................................................................
#color(green)("Step 3")#

Isolate #y# so that it is on the left of = and everything else is on the right.

Divide both sides by #color(blue)(2)#

#" "color(brown)(2/(color(blue)(2)) xxy=x/(color(blue)(2))+3/(color(blue)(2))#

But #2/2=1" and "1xxy=y#

#" "=>y=x/2+3/2#..............................Equation (1)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Building the new equation")#

Standard form is #y=mx+c" whare m is the gradient (slope)"#

From equation (1) the gradient is #1/2#

We are given that the new line passes through the point:
#" "P->(x,y)=(-1,2)#

'.....................................................................................................
Substituting into equation (1)

#color(brown)(y=mx+c)color(blue)(" "->" "2=1/2(-1)+c)#

#" "2=-1/2+c#

Add #1/2# to both sides

#" "2 1/2=c#

Thus the new equation is:

#" "color(magenta)(y=1/2x+2 1/2)#