# Thread: some y=mx+b help?

1. ## some y=mx+b help?

okay well i have some algebra homework to do and there are a couple things that have me confused. you see, in this section of our book we are finding slope and y-intercepts, as well as graphing them. i could do just fine if they're already in slope-intercept form, but im having trouble changing a few of them. for example:

y=3x+1

this is easy, as the slope is right there. slope is 3. and y-intercept is 1. and some problems like this:

y-9x=0

i could put that into slope intercept form without major complications.

y-9x=0
+9x +9x

the -9x and the +9x cancel eachother out, and you have to do the same thing to the other side. so it's now changed to:

y=9x

because the 0 is understood in that. so slope is 9 and intercept is 0 or the origin. but im kind of lost for a couple of these:

3x+4y=16

y= 6-x
3
for the first one, im not sure if you would want to:

3x+4y=16
-3x -3x

4y=16-3x
(i also dont know if you'd write it as 4y=3x-16 or 4y=16-3x)

or if you'd want to:

3x+4y=16
4 4

in this, the four would cancel out on the y side and just divide as usual on the other side, making it:

3x+y=4

and then you'd subtract the 3x on both sides.

or does that order even matter?

and on the other one:

y=6-x
3

im just lost on that one. i tried looking back in the book and i can't really find any help for those.

so is there anybody out there who could tell me what exactly to do in these problems? thanks!

2. Originally Posted by bespinboy
3x+4y=16
$3x+4y=16$

Subtract $3x$ from both sides, $4y=-3x + 16$.

Divide through by 4, $y=\frac{-3}{4}x + 4$.

There you have your equation in the form you have above.

Gradient of $\frac{-3}{4}$ and intercept of $4$.

3. thanks craig!

4. How did you manage with the second question did that go ok?