anonymous
  • anonymous
hello i need help in math can anyone help me.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
sure
anonymous
  • anonymous
thanks
anonymous
  • anonymous
ok here it is

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anonymous
  • anonymous
also i have a few is that ok
anonymous
  • anonymous
I need to see it to see if i am able to help
anonymous
  • anonymous
ok
anonymous
  • anonymous
Line M is represented by the following equation: x + y = −1 What is most likely the equation for line P so the set of equations has infinitely many solutions? 2x + 2y = 2 2x + 2y = 4 2x + 2y = −2 x − y = 1
anonymous
  • anonymous
also i only have 5 questions ok
anonymous
  • anonymous
i will give medal after
anonymous
  • anonymous
even if i get some wrong just because you helped me i will give medal at the end
anonymous
  • anonymous
hello are you still there
zepdrix
  • zepdrix
You want to find a line that is a parallel to the given line. Mechanically what we'll be doing is, Taking our line \(\large\rm x + y = −1 \) Adding each of the options to it, and seeing which one gives us a result of 0=0. Example: I'll add our line and the 4th option:|dw:1440538891099:dw|This gave us 2x=0 which will lead to a `unique solution` of x=0. This is a single solution.
zepdrix
  • zepdrix
So this is no bueno, we want infinitely many solutions, this only happens when both variables disappear in the operation.
anonymous
  • anonymous
so this is a single soulution
zepdrix
  • zepdrix
I'm saying that the "solution" or "intersection" of these two lines is at a single place, so it is clearly the wrong answer for us.
anonymous
  • anonymous
but thats not what we want we want infinate soulutions correct
zepdrix
  • zepdrix
yes, correct :)
anonymous
  • anonymous
ok got it
zepdrix
  • zepdrix
How bout we try the first one:|dw:1440539063251:dw|
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
We need the coefficients to match up, so what can we do to the first equation?
anonymous
  • anonymous
im not sure im realy bad at math do you think you can give me like a hint to understand how to find the coefficients
zepdrix
  • zepdrix
Let's just pay attention to the x's. I want the x's to disappear through some type of addition. I have 2x's in the bottom so I'm going to need 2x's in the top equation as well. So let's multiply the entire top equation by 2.
anonymous
  • anonymous
ok i think i understand were your going with this
zepdrix
  • zepdrix
|dw:1440539255066:dw|Gives us something like this, ya?
zepdrix
  • zepdrix
So um
anonymous
  • anonymous
ok so we added the 2 right
anonymous
  • anonymous
and then mutyplye
zepdrix
  • zepdrix
We `multiplied` by 2! :O not added. I guess in this case, in order to make the x's disappear, we'll need to `subtract` the bottom equation, not add it.
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
|dw:1440539392434:dw|Careful though, you have to subtract everything.
zepdrix
  • zepdrix
So what dare you left with? Try to deal with each column. 2x - 2x?
anonymous
  • anonymous
would that make everything 0
zepdrix
  • zepdrix
2x - 2x = 0 2y - 2y = 0 The left side is clearly 0, good. -2 - 2 = ?
anonymous
  • anonymous
so then the answer would be 0
zepdrix
  • zepdrix
Noooo :O Check that last line again carefully!
anonymous
  • anonymous
ok sorry
zepdrix
  • zepdrix
\(\large\rm -2-2\ne0\)
anonymous
  • anonymous
ok so 0
anonymous
  • anonymous
lol im sorry im super confused
zepdrix
  • zepdrix
If you're at negative 2, and you take 2 away from that, you become even more negative.
anonymous
  • anonymous
wait would it be 4
anonymous
  • anonymous
-4
zepdrix
  • zepdrix
Ok good! So we end up with this result: \(\large\rm 0=-4\) Which is `false`. Therefore the system containing these two lines has NO solution. So again, this is not the one we're looking for. Your princess is in another castle :o
anonymous
  • anonymous
ok lol so now on to the other aswer right
zepdrix
  • zepdrix
Let's check out the 3rd option: x + y = -1 2x + 2y = −2
anonymous
  • anonymous
also im sorry math is the only thing i realy suck at
anonymous
  • anonymous
ok
anonymous
  • anonymous
so would we do the same thing we did last time
zepdrix
  • zepdrix
Yah, that seems like a good idea.
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
Again, we multiply by 2 to match up the x's. 2(x + y = -1) 2x + 2y = −2 Giving us: 2x + 2y = -2 2x + 2y = −2
zepdrix
  • zepdrix
And then do some subtraction or addition to combine them. What are we gonna get this time? :o
anonymous
  • anonymous
umm the end parts on both sides are nagative so those be 0s
zepdrix
  • zepdrix
Well notice that both equations are exactly the same now, ya? What happens when you subtract something from itself? You're gonna get 0. So good, in this case we end up with 0=0. So this system of two lines gives us infinitely many solutions!
anonymous
  • anonymous
so thats the answer
zepdrix
  • zepdrix
yay team \c:/ we got it!
anonymous
  • anonymous
yayaya
anonymous
  • anonymous
also wich question was this
zepdrix
  • zepdrix
The only one that you asked :D lol I dunno, you didn't number it :d
anonymous
  • anonymous
no i mean the question to the question we just worked on is it x + y = -1 2x + 2y = −2
zepdrix
  • zepdrix
Line M was `given` to be x+y=-1. And our `answer` that we obtained for line P is the third one, 2x+2y=-2
anonymous
  • anonymous
ok cool next question! :)
anonymous
  • anonymous
Line Q is represented by the following equation: 2x + y = 11 Which equation completes the system that is satisfied by the solution (3, 5)? x + 2y = 15 x + y = 15 x + y = 8 x − y = 2
anonymous
  • anonymous
i think its B or A
zepdrix
  • zepdrix
I gotta go make some foods. If you need further help, close this and open up another thread. This one is getting too long and messy. You can use the @ symbol to page someone for help. Example @zepdrix The people at the top of the lobby are usually a bunch of smarty pants. Include your question in title, it will help to get it answered faster. Don't just call it "need help with math". Just a little tip :) I'll try to come by and help when I can :) But the belly is rumblin at the moment!
anonymous
  • anonymous
ok thaks

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