# Thread: slope-intercept to standard form

1. ## slope-intercept to standard form

I am a little confused with this one and I have a test tomorrow

Why does y = -3x -3 (slope-intercept form) become 3x + y + 3 = 0 in standard form instead of -3x-y-3 = 0?

and - 3 = y-(-10)/x-1 should be - 3x-y-7 = 0 so why is it 3x+y+7 = 0?

Mind explaining? Thanks so much!

2. They've just multiplied everything by -1 to make it a bit cleaner.

3. do you always have to time it by (-1)?
Why does is my asnwer always the opposite charge.

4. Originally Posted by dizzycarly
I am a little confused with this one and I have a test tomorrow

Why does y = -3x -3 (slope-intercept form) become 3x + y + 3 = 0 in standard form instead of -3x-y-3 = 0?

and - 3 = y-(-10)/x-1 should be - 3x-y-7 = 0 so why is it 3x+y+7 = 0?

Mind explaining? Thanks so much!
$\displaystyle y=-3x-3$
Subtracting y from both sides
$\displaystyle y-y=-3x-y-3$
$\displaystyle 0=-3x-y-3$
$\displaystyle -3x-y-3=0$

It is good to note here that if we would have added (3x+3) to both sides, the problem would have been more aesthetically pleasing while maintaining equivalency:

$\displaystyle y=-3x-3$
$\displaystyle y+(3x+3)=-3x-3+(3x+3)$
$\displaystyle y+3x+3=0$

This form is not only equivalent to the other, but more conventional, and therefore what your teacher is going to be looking for first. Both answers are correct.

By the way, I like your avatar.

5. Originally Posted by dizzycarly
do you always have to time it by (-1)?
Why does is my asnwer always the opposite charge.
Nope, this is only done to make things look less complex. Multiply through by -1 if there is an abundance of negative terms.