How to convert from slope-intercept to standard form?

Question

Hey, I have this equation, in slope-intercept form it's -2 = -3(-1) + b

how do I convert that to standard form?!

Answer 1

Slope intercept form

- 2 = - 3(- 1) + b

- 2 = 3 + b

Transpose 3

- 2 - 3 = 3 + b - 3

- 5 = b

- - - - -

The equation

y = - 3x - 5

Transpose 3x

3x + y = - 3x - 5 + 3

3x + y = - 5

Transpose 5

3x + y + 5 = - 5 + 5

3x + y + 5 = 0. . .←. .Standard Form

- - - - - - - s-

Answer 2

you have the form: y = mx + b

where m = slope, b = y-intercept

standard form: y - (some y value) = m (x - (some x value))

you're given the x, y, and m...

So...

y - (-2) = -3(x - (-1))

y + 2 = -3 (x + 1)

Answer 3

I think there's an error in your question. It is vague...lack details.