Question

How do you write an equation of a line through (-5,3) parallel to `y=2/3x+3`?

Answer 1

`y=2/3x+19/3`

Explanation:

Parallel lines have the same slopes. So your new line has slope `2/3`

Apply point-slope formula, `y-y_1=m(x-x_1)` where `(x_1,y_1)` is `(-5,3)`:
`y-3=2/3(x-(-5))`
`y-3=2/3(x+5)`
`y-3=2/3x+10/3`
`y=2/3x+19/3`
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Answer 2

`y=2/3x+19/3`

Explanation:

We're going to use two things here, point-slope form and slope-intercept form. Slope intercept is `y=mx+b`, whereas point-slope is `y-y_1=m(x-x_1)`. The point that we know we go through can be written as `(x_1,y_1)` so `x_1=-5` and `y_1=3`. We know the slope from the equation given in the question by applying it to slope-intercept. The `m` is the slope, so the slope of the equation in question is `2/3`.

Just to organize this better:
`x_1=-5`
`y_1=3`
`m=2/3`

Now we plug this information into point-slope `y-y_1=m(x-x_1)`
`y-3=2/3(x-(-5)) -> y-3=2/3(x+5)`

This is a good answer but I'm just going to bring it to y intercept because that's what I'm used to personally.

Add `3` to both sides.

`y=2/3(x+5)+3`

Distribute `2/3`

`y=2/3x+10/3+3`

Add `10/3` and `3`

`y=2/3x+19/3`