Not quite.
Adding to both sides:
So how would you finish?
Hello everyone, I have a small issue while solving an equation in the form of slope-intercept, they give me an equation and ask me to solve it and graph it.
Formula Given for Graphing: y=mx+b
Equation: -3x - 6y = 12
So I know I have to solve for y, so here's what I do:
add -3x to 12, and that's where I get lost. I saw a video on youtube that said when you do that, you don't really add the -3x to 12, it just becomes the slope in the next step so for example it would be: -6y=-3x + b? Not sure if that's right.
Help is appreciated, thanks.
Can it be done like this?
-3x -6y = 12
add 3x to both sides, and it becomes:
-6y= 3x + 12
divide by -6:
y=-3x/-6 + 12/-6
y= y = -1/2 x - 2?
Assuming that's right, what it is, is that whatever your first step is becomes the slope, then to leave y by itself, you divide everything, but ONLY simplify the slope?
Be careful with your signs. Since you added to both sides, the should be positive.
You are correct with
But I don't agree with the last part. Questions won't always just be dealing with the slope. There will be times when you need to know things about the intercepts or equation in general, so simplify where possible, unless asked otherwise.
Yeah sorry, that might of been confusing, thanks! Sigh, you explained this way better than my algebra teacher in under 20 minutes.
edit: in the case that the initial -3x, wasn't negative and it was positive, the first step would be:
3x - 6y = 12 ---> -6y = 3x - 12? or is it -3x - 12?
Ah, yes, how bad of me hehe, well thanks for all the help got it now.
edit: Just to solve it,
3x - 6y = 12
subtract -3x
-6y = 12 - 3x
divide by -6
y = 12/-6 - -3x/-6
y = -2 - 1/2
A graphing would look like: start at -2 for the y-intercept, then go down -1, and run over 2 spaces?
edit: Also since I'm more used to reading y=mx + b than reading y=b + mx, how can I write this: -6y = 12 - 3x in the y=mx+b form, I know it's the same but just to make it easier on me.
Addition is commutative: the order in which things are added doesn't matter. To graph , set , and deduce that , then set and deduce that
You now have points, and and you can just draw the line connecting these points. You could think of it as going down space, and horizontally right spaces.
Edit: Corrected a typo.
Alright, got it.
And just to make sure since I always write the equation on paper as y=mx + b, where the y intercept is always on the right, the addition between the mx and b never changes right? what changes is whether the mx is positive or negative right?
What I meant was, the equation made after the first step to solving it in y=mx+b form will always be mx will either be negative or positive depending on the initial equation, and the + b will always remain right?
edit: let me clarify, b will ALWAYS be positive, then mx can be either positive or negative? So an equation for me would always be written y= mx (can be negative or positive) then + b?
That equation you posted in y=mx+b form would be: -4y= 8x + 3.