anonymous
  • anonymous
What quadrant are the equations in? A. 3x + 2y = -15 x - 4y = -20 B. 10x - y = -30 x - y = -2 C. 2x - 5y = 10 x + 2y = 10 D. 3x + y = 9 -7x - 7y = -14
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
http://lhh.tutor.com/presession.aspx?ProgramGUID=c2f98cc5-be2f-42f3-abbd-891c960a6bd5&TopicId=1&InterfaceId=k12&InterfaceType=k12
anonymous
  • anonymous
go there for some accuarate help
anonymous
  • anonymous
I don't have a library card #. That's not going to help me.

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anonymous
  • anonymous
240100447204
anonymous
  • anonymous
try that
anonymous
  • anonymous
lemme know what happens
anonymous
  • anonymous
It worked but I don't see it helping me too much.
anonymous
  • anonymous
ok you shouldve pressed the math section
anonymous
  • anonymous
I did.
anonymous
  • anonymous
ok then you fill out the information on the bottom it shouldve let you type in your qestion
anonymous
  • anonymous
Would this be algebra I or II?
anonymous
  • anonymous
algebra 2
anonymous
  • anonymous
What subtopic?
anonymous
  • anonymous
what are your choices
anonymous
  • anonymous
would it be graphing functions?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I just need to know how to graph them.
anonymous
  • anonymous
girl go the bottom of tha page and get a tutor a live tutor will help you out
anonymous
  • anonymous
I don't like that site.
anonymous
  • anonymous
lol ok then well their helpful i jsut uesd them rite now and they gave me the rite answer to everyting
anonymous
  • anonymous
Brizy, here's something else you can also try. Go to http://fooplot.com. It will graph the equations for you and you can see what quadrants they are in. You do have to solve for y first. So your first equation in A would be: 3x + 2y = -15 2y = -3x - 15 y = (-3/2)x - (15/2) Then you'd enter everything right of the "=" sign on that website. Hope that helps.
anonymous
  • anonymous
now thats coool im gonna try that
anonymous
  • anonymous
Yeah that seems complicated :-(
anonymous
  • anonymous
my god girl you are hard to please
anonymous
  • anonymous
Lol I've herd that before :-)
anonymous
  • anonymous
Haha. It does take some work, but that process should at least make it easier.
anonymous
  • anonymous
ya try it
anonymous
  • anonymous
oh shadowfiend was here
shadowfiend
  • shadowfiend
Still is.
anonymous
  • anonymous
aaaaahhhhhh
anonymous
  • anonymous
have i metntioned you look wonderful today:)
shadowfiend
  • shadowfiend
BrizyBaby—the easiest way to answer these question is usually to graph them, honestly.
anonymous
  • anonymous
I don't know how to graph the equations....that's what I need help with.
shadowfiend
  • shadowfiend
Got it! Now we're at the heart of the problem :)
shadowfiend
  • shadowfiend
Let's start with A then. We have two equations: 3x + 2y = -15 x - 4y = -20
anonymous
  • anonymous
well i give up nice meeting you thoug Brizzy
anonymous
  • anonymous
I need a easy way to know how to graph them.
anonymous
  • anonymous
You too aminah.love18
shadowfiend
  • shadowfiend
I find it easiest to graph equations when they're in y = mx + b (slope-intercept) form. So first things first, we try to solve the equations for y: \[\begin{align} 3x + 2y &= -15\\ 2y &= -3x - 15\\ y &= \frac{-3x - 15}{2} = -\frac{3}{2}x - \frac{15}{2} \end{align}\] And the other one: \[\begin{align} x - 4y &= -20\\ -4y &= -20 - x\\ y &= \frac{-20 - x}{4} = \frac{-20}{4} - \frac{1}{4}x\\ y &= -5 - \frac{1}{4}x = -\frac{1}{4}x - 5 \end{align}\]
shadowfiend
  • shadowfiend
So, we now have: y = -3/2 x - 15/2 y = -1/4x - 5 The first thing to notice is that both of these have a negative slope. Do you know what a negative slope means on a graph?
anonymous
  • anonymous
I don't get how you did the equations but yeah a negative slope goes down left to right.
shadowfiend
  • shadowfiend
Whoops. Sorry then, let's go back to that :) Let's start with 3x + 2y = -15. We're trying to solve for y. That means we need y alone. Remember that an equation doesn't change if we subtract the same thing from both sides. So first things first, we move the 3x to the right: 3x + 2y = -15 3x - 3x + 2y = -15 - 3x Subtract 3x from both sides, 3x - 3x is 0: 0 + 2y = -15 - 3x 2y = -15 - 3x (We leave off the 0.) Then we need to get rid of the 2. Dividing the same thing from both sides also leaves the equation unchanged, so we'll divide by 2: y = (-15 - 3x) / 2 y = -15/2 - 3/2 x Does that make more sense?
anonymous
  • anonymous
yeah. So the points are -7.5 and -1.5?
shadowfiend
  • shadowfiend
Close! :) This means that the line is, like we said, going top left to bottom right. It means for every 3 units down that it goes, it goes two units to the right. And it means that it meets the y axis at -7.5. A quick, semi-accurate drawing: |dw:1329863330467:dw|
anonymous
  • anonymous
What did I get wrong?
shadowfiend
  • shadowfiend
So you can see here that that line is in quadrants II, III, and IV: |dw:1329863470553:dw|
shadowfiend
  • shadowfiend
-1.5 isn't a point, it's the slope, that's all.
anonymous
  • anonymous
So, it's in the third Quadrant?
shadowfiend
  • shadowfiend
Well, that's what I'm not 100% clear on in the question. Obviously this line goes through three quadrants. Is the question asking which quadrant both lines are in or..?
anonymous
  • anonymous
Well does it mean what quadrant the slope is in?
shadowfiend
  • shadowfiend
Sorry -> slope isn't really in a quadrant. It just tells you how the line moves from one position to another (i.e., what angle it's at).
shadowfiend
  • shadowfiend
It probably means what quadrant the two lines meet in, so let's look at the other line.
shadowfiend
  • shadowfiend
y = -1/4x - 5 So we have a slope of 1/4, meaning for every one unit we go down, we move to the right 4 units. And -5 is the y-intercept, which means the line meets the y axis at -5. Another rough drawing: |dw:1329864223848:dw|
shadowfiend
  • shadowfiend
If we put the two together: |dw:1329864317169:dw|
shadowfiend
  • shadowfiend
You can see there that they do in fact intersect in quadrant III.
anonymous
  • anonymous
So for A it's Quadrant III?
shadowfiend
  • shadowfiend
I would assume so. Again, the question isn't *super* clear, but I think that's what it's asking for.
anonymous
  • anonymous
Okay thanks :-)
shadowfiend
  • shadowfiend
Glad I could help :)

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