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.
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`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`
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`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`
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`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)
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`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)`

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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)`