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I haven't done Algebra in over 5 years, so I need help with this problem: What’s the equation of a line that passes through points (0, -1) and (2, 3)?

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First to find the slope of the line you use the equation: \[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\] pick one of the pairs to plug into \[y _{2}\] and \[x _{2}\] just make sure to keep the order after you start plugging numbers in. Does this make sense so far?
you want to obtain y=mx+b and that formula is the formula for slope (aka "m") @masukasu
How do I know what numbers to assign to y and x?

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Other answers:

so this is what it will look like once you plug the points in: \[\frac{ (3) - (-1) }{ (2)-(0) }\]
You have the points (0, -1) and (2, 3); you know that 0 and 2 are x-values and -1 and 3 are y-values, correct?
I used the second set of points to use first only so the equation won't look messy. Putting the -1 in the \[y _{1}\] spot allows it to become positive because two negatives make a positive.
So, (0, -1) and (2, 3) would look like this with their appropriate values: (x, -y) and (x, y) right?
But you will get the same answer no matter which set of points you plug first, but always be sure to keep the order and correct to your above question
so go ahead and solve for the slope (m) and tell me what you get
But, the example of the order of my values(in my last comment) are correct, right?
yes you are correct = )
Ah! Thank you. I'll try to solve the problem. One minute (:
now that is only part of it, there is another equation to arrive to the final equation
once you find the slope I can tell you the next part = )
And I would multiply \[x{2}\] and \[y{2}\] like any other number with power, right? i.e. 2 x 2 and 3 x 3
Oh I'm sorry, the subscript 1 and 2 are only there to decipher between the different values those aren't exponents
Ohhh ok, thank you. I couldn't remember the right terminology either, sorry. But, I think it's: \[\frac{ 4 }{ -1 }\]
here let me do this \[\frac{ y-y }{ x-x }\] do you see how this can look like the answer would be 0 right? that's why they put the 1 and 2 on the x and y to determine between the two sets of points
your numerator is correct double check your denominator (2-0) = ?
Pff... Sorry. I mixed my numbers up >.<
no apologies = ) I was just on here earlier getting help and I did the same thing lol ; D
Essentially it comes out to \[\frac{ 2 }{ 1 }\]
lol I guess we all make mistakes haha! It's how we learn.
correct = ) so that is your "m" or slope
yay! Alright, so that should be written out like so: \[y =2x +b\]
And b is the 'base', right?
the next equation you will need is: \[y-y _{1} = m (x-x _{1})\] Now don't get confused that the \[y _{1}\] and \[x _{1}\] are the same as the ones above you can choose either set of points to plug in. The regular y and x are the ones that will remain in the equation after plugging in the set of points which will give you the y=mx+b. Does this make sense so far?
I can't remember if it is called the base but that will be your y-intercept once you graph it
Ok so i chose the points (2,3) to use bc I like positive numbers lol Plug in the set of points and what you got for m into the equation \[y-(3)=(2)(x-(2))\] Tell me what you get after solving that
Sorry. Someone had to use my laptop.
that's ok = )
Ok, I'm going to solve that. One minute (:
Wait... Do I NEED the parenthesis around the numbers? Or does it not really change anything?
It's just easier to see if you have any sign changes. Now say you used the other set of points instead of the positive numbers you would end up with this: \[y-(-1) = (2)(x-0) \] It's safe to have that parentheses around the -1 bc the sign of it changes to y+1 which will screw things up when you have to solve for y
but it will only screw things up when you have to solve for y if you forget to change the sign of the -1.
This equation is a little confusing. I'm used to it looking like \[2\left( x -2 \right)=y -\left( 3 \right)\] but even then, it's a little difficult.
ok so the only difference is the sides of the equation are on opposite sides = ) do you know how to factor? I'll help you step by step = )
I would bring the 3 over and it would become a +3, then it would look like this:\[2\left( x-2 \right)+3=y\] I think
Factoring... Uh lol Is that factoring? I can't remember. Am I able to reverse the equation like I did though?
that is a correct move but that's not factoring. Take a look at the 2(x-2): when you see something like this you use factoring where you take the front number and multiply it by both numbers inside the parentheses. So you would have 2*x - 4 then tack on what's left in the equation: 2x-4+3=y Then simplify
OH! I remember now!
Well... that part at least
And then that would turn into \[y =2x-12\] right?
No wait.... Ugh! I think that's wrong actually.
lol Yeah, I loved the Illustration haha!
no look at it this way: \[y= 2x+(-4) +3\] \[y=2x +(-1) \] \[y=2x-1\] Does that make sense? = D
Trying to rework the problem so I can better understand it...
(2 x X)=2x (2 x 2)=4 y=2x ???????????? I can't figure out how you got the -4
I'm sorry I keep doing it backwards from what you are used to = ( I have my blonde moments lol I'm going to draw it for you and instead try not moving the 3 as the first move
I uh.... Well, I have blonde moments too lol And I've black hair and I'm Asian haha! So, no worries :P
I'm a brunette lmfao
It's the 2x+(-4) part that gets me
lawl xD
this is the way you're used to right?
ok here we go lol
Hopefully I can get this done before I have to go to work xD I have a history in high school of a single math problem taking me almost 2-3 hours to figure out.
it's always good to make an equation have positives so y-(+3) can also be written as y+(-3) its the same thing
Trying to understand this... I always had a hard time understanding these problems because of all the positives switching to negatives and vice versa.
so back to the full equation and i'll make it fast = ) a lot of people have problems with this that's why I'm here to help although my explanations seem long at times lol I apologize = )
Ohhh so that's what you did +(-4), right?
Ok so we have: |dw:1359323569725:dw||dw:1359323806202:dw|
yes correct! = D
\[y=2x + \left( -4+3 \right)\] right? Which becomes \[y =2x\] With the +(-4), that + sign pretty much comes out of nowhere, right?
Correction: \[y =2x +1\] Sorry
lol yeah pretty much. that's what I've been learning as I've gotten through to later math classes [calculus 2 now = ( lol] is that you can just put random stuff in an equation that isn't even in it just as long as it doesn't change the equation haha and yes you are correct I think you got it = D So do you understand it now?
Do you feel confident that you can do another one by yourself?
I would like to try (: So, shoot! Let's see if I can do this.
ok let me just quickly work one out to make sure its not messy lol like no fractions i hate fractions lol
And I had to repeat Pre-Algebra once, went on to Algebra 1, and didn't do too well there.
Ok! And you're 50,000,000 times smarter than me haha!
While you're doing that, I'm going to take a super fast shower. Like... 5-10 minutes.
ok when you get back I will have the equation already up = )
Ok (: Be back soon!
What is the equation of a line that passes through points (-1,6) and (3,-2)?
Back! I'll figure that out real quick.
kk = )
So, I want to figure out y+mx+b right?
Correction: y=mx+b
correct and you will use two formulas to figure that out: \[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\] \[m(x-x _{1})=y-y _{1}\]
correction: \[m=\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\]
Ok, so does that second formula come from somewhere or do you have to remember it? lol
you have to remember those
Sorry. Does it come from somewhere within the first formula, or does it come from nowhere, except memory?
it sucks bc there is so much memorization in math = (
Yeahhhhh lol >.<
just memory bc they give you the formula in class
Uhhhhhhhhhhhhhhhhhhhhhhhhhh..... Ok here goes >.<\[y =2x -1\]
no but good try let's go through it step by step. I admit it's a little challenging with all the sign changes
what did you get for the slope?
Where's my Fail Whale when I need it? xD
= )
ok that's probably where it went wrong
so how did you set it up? which set of points did you put first?
We have to be fast though. I have to leave for work in 15 minutes o.o
I'll be sure of it = )
This is hard. One sec.
I see what you did you put the values in the wrong locations
Ahhhhhhhhhhhhhhhh! I feel dumb haha! Help me with that please?
ok you have the points: (-1,6) and (3,-2) |dw:1359327131329:dw|
that's supposed to be a 2 at the end of that second y (subscript)
Oops.... Blonde moment? Or just blind? xD
y1 and x1 are the same set is what that says lol like I said we all have 'em lol
what i usually do is when I'm given the two sets of points I label the numbers so I don't get confused later in the problem
just as I did here |dw:1359327396584:dw|
I did that too lol
Did you re-work it out?
If you have to leave for work I'll post how to do the full equation and the answer at the end = )
One sec. I wish they'd call in so I could study more for the ASVAB :( I'll try to figure it out real quick
lol kk = )
Ya? lol
yes! = D
YES! Ok, let me solve the rest ;D
lol kk
Or try haha! I hope this is right.
I'm sure you got it = )
I uh.... I'm doubtful, but here goes o.o y=-2x-5 -_-
very extremely close which points did you plug in?
OH! When I brought over the -6, I left it as -6 and didn't switch it to 6
y=2x+7 ??????????????????
take a deep breath you're using (-1,6) right?
= )
Yeah I am lol
Have a good day at work = )
SORRY! My laptop actually overheated. I was also kind of late to work ^^;

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