hello i need help in math can anyone help me.

- anonymous

hello i need help in math can anyone help me.

- schrodinger

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- anonymous

sure

- anonymous

thanks

- anonymous

ok here it is

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## More answers

- anonymous

also i have a few is that ok

- anonymous

I need to see it to see if i am able to help

- anonymous

ok

- 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

also i only have 5 questions ok

- anonymous

i will give medal after

- anonymous

even if i get some wrong just because you helped me i will give medal at the end

- anonymous

hello are you still there

- 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

So this is no bueno, we want infinitely many solutions,
this only happens when both variables disappear in the operation.

- anonymous

so this is a single soulution

- 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

but thats not what we want we want infinate soulutions correct

- zepdrix

yes, correct :)

- anonymous

ok got it

- zepdrix

How bout we try the first one:|dw:1440539063251:dw|

- anonymous

ok

- zepdrix

We need the coefficients to match up, so what can we do to the first equation?

- 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

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

ok i think i understand were your going with this

- zepdrix

|dw:1440539255066:dw|Gives us something like this, ya?

- zepdrix

So um

- anonymous

ok so we added the 2 right

- anonymous

and then mutyplye

- 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

ok

- zepdrix

|dw:1440539392434:dw|Careful though, you have to subtract everything.

- zepdrix

So what dare you left with?
Try to deal with each column.
2x - 2x?

- anonymous

would that make everything 0

- zepdrix

2x - 2x = 0
2y - 2y = 0
The left side is clearly 0, good.
-2 - 2 = ?

- anonymous

so then the answer would be 0

- zepdrix

Noooo :O
Check that last line again carefully!

- anonymous

ok sorry

- zepdrix

\(\large\rm -2-2\ne0\)

- anonymous

ok so 0

- anonymous

lol im sorry im super confused

- zepdrix

If you're at negative 2,
and you take 2 away from that,
you become even more negative.

- anonymous

wait would it be 4

- anonymous

-4

- 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

ok lol so now on to the other aswer right

- zepdrix

Let's check out the 3rd option:
x + y = -1
2x + 2y = −2

- anonymous

also im sorry math is the only thing i realy suck at

- anonymous

ok

- anonymous

so would we do the same thing we did last time

- zepdrix

Yah, that seems like a good idea.

- anonymous

ok

- 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

And then do some subtraction or addition to combine them.
What are we gonna get this time? :o

- anonymous

umm the end parts on both sides are nagative so those be 0s

- 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

so thats the answer

- zepdrix

yay team \c:/ we got it!

- anonymous

yayaya

- anonymous

also wich question was this

- zepdrix

The only one that you asked :D lol
I dunno, you didn't number it :d

- anonymous

no i mean the question to the question we just worked on is it x + y = -1
2x + 2y = −2

- 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

ok cool next question! :)

- 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

i think its B or A

- 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

ok thaks

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