I have a little assignment here that I’m trying to do and I need a little bit of guidance on how to do two questions…I know how to do them, but as soon as I get home I seem to forget how.
<b>Given the following system of equations, find the point of intersection.</b>
4m-4n=23
2m+5n=5
If someone could steer me in the right direction on how to start solving, that’d be great. You can give me the answer if you want, but I’ll try and understand if anyone chooses to help me by just helping with setting up the question.
Do I have to set the M’s equal to each other to start solving?I know how to do that, if so. Does it have to be in y=mx+b form?
After that, I know what I’m doing…
Second question.
<b>Given the following system of equations, solve for x, y and z.</b>
-6x + 4y +3z = 0
x + y -5z = -23
7x -10y + z =34
Which variable should I cancel out first? I tried all three at some poitn and time and I keep getting large decimal numbers. Since this is an assignment my teacher has tried to keep it as simple as possible and my answers should be coming out to full numbers, or one decimal spot and no more.
I use elimination to find these variables, so don’t try to explain any other method to me as you’ll confuse me 
Pleeeeease help, anyone.
4m-4n=23
2m+5n=5
m = slope, so yeah, make them equal. Unless this is a “cancelling” problem. But since it’s asking you for a coordinate, I’m guessing no on cancelling. I have no idea what “n” is. What kind of math is this?
6x + 4y +3z = 0
x + y -5z = -23
7x -10y + z =34
It doesn’t matter which you cancel out first, just whichever is the easiest. You should probably go for cancelling out Z.
Originally posted by Dark Paladin
m = slope, so yeah, make them equal. Unless this is a “cancelling” problem. But since it’s asking you for a coordinate, I’m guessing no on cancelling. I have no idea what “n” is. What kind of math is this?.
That question…the M is basically the Y and the N is the X. They’re just using different variables. A+B, M+N, X+Y, all the same thing.
Alright so then I have to multiply my second question to it becomes equal with the first? That’s the way we do it…so it’ll look like this:
4m-4n=23
2m+5n=5 (x2)
New equations:
4m-4n=23
4m+10n=10
elimination
-14n=13
I could go on from there, but let me ask, are my integers right? We’re encouraged to use calculators, so my mad positive/negative skillz aren’t quite…developed.
The second problem is not so hard you just eliminate the variables
using simultaneous (i hope you know what i mean) and yes like Paladin said try eliminating z first
Yeah, that’s right.
n = -[u]13[/u]
14
Just plug that in to one of the equations to find m
I know the second one isn’t so hard…I did an even more difficult question in the same set up by myself and got the whole thing right, even after doing a check. I don’t know why this one is stumping me, sometimes my brain just refuses to work.
So the second question, I’m picking two random equations out of that combo. Actually, i’m choosing the 3rd one for sure, so it will be easier to multiply. I’m going to set Z equal to 5, like the one above. My equations are now:
x + y -5z = -23
30x -50y + 5z =170 (because I multiplied this by 5)
Am I good so far?I am adding these two together obviously to get rid of Z.
EDIT:DP, the answer comes out to an insanely long decimal number…not the right answer obviously 
I’m doing something seriously wrong…
Correct me if I’m wrong, but don’t you have to use all three equations at once? (Unless the integer for the variable you’re solving for is zero in the third equation)
EDIT: I got
m = [u]135[/u]
28
Does it say to put in decimal form?
I always eliminate the one that seems easier but if it gives me to much trouble i eliminate another one fist so if z gives you to much problems then try eliminating z or y first
Originally posted by Dark Paladin
[b]Correct me if I’m wrong, but don’t you have to use all three equations at once? (Unless the integer for the variable you’re solving for is zero in the third equation)
EDIT: I got
m = [u]135[/u]
28
Does it say to put in decimal form? [/b]
Umm the way we do, or the way we were taught is that we take two equations from the three, and set a certain variable equal to each other. Then we take a different pair of equations from the three and set the same variables equal to each other so they cancel out. Then we take, those two new equations that we got, and set a different variable to each other (say we cancelled out Z the first two times, we now set X equal to each other and cancel them) and then we do the division and get our first variable.
Then we take one of those two new equations and plug the known variable in to get a second.
THEN we take one of the original equations, from the three, and plug both known variables in to get the last unknwon variable (essentially, it’s a check).
phew
Trouble is, I keep getting bad answers. They just don’t work out and I don’t know what I’m doing wrong.
I’ll work out the equation, and you tell me what you got.
2nd prob: x= 0 , y=-3, z = 4
to find y
-6x + 4y +3z = 0
6x + 6y -30z = -138
(1)10y - 27z= -138
-4x -4y+20z=92
4x-10y+z=34
(2)-14y+21z=126
7(1) + 9(2)
70y-189z = -966
-126y=189z = 1134
-56y =168
y=-3
Alright DP, give me awhile I tend to get frustrated. Show me your methods and how you worked it out…
Thanks Sin, hopefully keeping the answers in mind I can spot what I’m doing wrong.
Are you sure that’s how you work it, Sin? Don’t you have to combine two of the equations every way you can? Meaning you get three “y+z=”
And wouldn’t the answers vary depending on which two equations you chose to work with?
I put everything in terms of z.
That causes me to get
x = 2.3z - 9.2
y = 2.7z - 13.8
z = z
Just for completion sake:
7x -10y + z = 34
x + y -5Z = -23
so
7x +30 + z = 34
7x + z =4
x -3 -5z = -23
x-5z = -20
35x +5z =20
x -5z =-20
36x=0
x =0
so 0+ z = 4
z=4
that’s how they teach it in alg 2. I charge 15 bucks an hour to show people how to do this after school :ah-ha!: I love my job
If you plug in 4 for z in my equations, then you get Sinistral’s answers ::shrug::
you can plug numbers in any equation and potentially get answers, but the thing is , that is harder and longer to do or it implies you know some answers which you usually don’t, esp on tests. Also, not showing work will get you points docked on a test.
Also, from the description of how she was taught to do those problems, the way I did prob 2 was right.
Anyway, problem 1:
eliminate m like you did, n = -13/14. Plug that into either equation, and m =135/28. I graphed it to verify and it is the point of intersection. When you’re asked the point of intersection, they’re asking for a point, like an (x,y) co-ordinate, where both lines will cross. Its a point where both equations have (x,y) co-ordinates in common.
Thanks muchly guys you’ve been a big help, I’m going over the answers and plugging em in until I understand what I did wrong, and how I got confused.