linear equations! |
linear equations! |
![]()
Post
#1
|
|
Senior Member ![]() ![]() ![]() Group: Member Posts: 45 Joined: Feb 2008 Member No: 622,075 ![]() |
i need to do this and i don't get it ! i've tried like 100 times! help please!
AB y= -2x + 21 BC y= 1x - 12 CA y=2 a) Choose two of the above equations to form a linear system b) Solve the linear system (substitution method of linear combination method.) c) Show that the point above is on the third line by substituting the coordinates into the equation. thanks :) i'm almost pulling my hair out over this. |
|
|
![]() |
![]()
Post
#2
|
|
![]() AKA RockIt Studios ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Member Posts: 2,286 Joined: Jun 2006 Member No: 421,809 ![]() |
|
|
|
![]()
Post
#3
|
|
![]() in a matter of time ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 ![]() |
Are the A's, B's and C's part of the equations?
|
|
|
![]()
Post
#4
|
|
Senior Member ![]() ![]() ![]() Group: Member Posts: 45 Joined: Feb 2008 Member No: 622,075 ![]() |
oh. no, they aren't,
|
|
|
![]()
Post
#5
|
|
![]() [Insert something Witty Here] ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 363 Joined: Dec 2007 Member No: 598,828 ![]() |
A C and B are variables they are substitutes for numbers in algebra
|
|
|
![]()
Post
#6
|
|
![]() Senior Member ![]() ![]() ![]() ![]() ![]() ![]() Group: Administrator Posts: 2,648 Joined: Apr 2008 Member No: 639,265 ![]() |
Basically, set two of the equations equal to each other and solve for x. I'd pick y=2 as one because it's easy. So something like:
2 = -2x + 21 Then solve for x, and see if that value of x fits in the equation you did not use. If it does, you're golden. |
|
|
![]()
Post
#7
|
|
![]() in a matter of time ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 ![]() |
A C and B are variables they are substitutes for numbers in algebra Thanks, I know what variables are. ![]() Michael, I did that, but failed. I don't really think I'm bad at math if I can get through AP Calculus, but this is what happened: ![]() |
|
|
![]()
Post
#8
|
|
![]() Senior Member ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Member Posts: 1,529 Joined: May 2007 Member No: 523,843 ![]() |
Are you guys assuming that all three lines intersect? Dude it makes a triangle.
![]() |
|
|
![]()
Post
#9
|
|
![]() in a matter of time ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 ![]() |
Then the question is screwed up because it clearly implies that they all share one point.
You have to solve for one point between two lines then prove that same point is on the other line. That just makes no sense. So I'm not completely useless...My previous solution only solves for (9.5,2) |
|
|
![]()
Post
#10
|
|
![]() durian ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 13,124 Joined: Feb 2004 Member No: 3,860 ![]() |
Yea it's definitely not working. But I don't even understand why each equation had to be labeled AB, BC, and CA. That makes no sense, unless it's to form that triangle.
![]() |
|
|
![]()
Post
#11
|
|
![]() AKA RockIt Studios ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Member Posts: 2,286 Joined: Jun 2006 Member No: 421,809 ![]() |
i think AB, BC, and CA are the names of the segments (lines) that you're trying to prove.
![]() remember this could be and is most likely COMPLETELY wrong, i'm just doing what i think would make sense with the given information, so don't harass me! so, the point on y on the graph for line CA would be 2. and since it already tells you that y=2, substitute 2 in for y in the equations...right? wait. don't you graph using a slope? or am I thinking of something completely different? o_O. |
|
|
![]()
Post
#12
|
|
![]() durian ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 13,124 Joined: Feb 2004 Member No: 3,860 ![]() |
^ But if it makes a triangle, the point (on the linear system) would be the intersection at point B (AB to BC), but that point cannot fall in the y=2 line. If it does, then no triangle exists, which doesn't make sense if it's supposed to be line segments. Yea, the problem is most likely wrong.
lol it could be worse though. It could be... VECTOR CALCULUSSSSS |
|
|
![]()
Post
#13
|
|
![]() <joke> inside </joke> ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Member Posts: 2,283 Joined: Oct 2006 Member No: 470,590 ![]() |
quickmath.com
![]() |
|
|
![]()
Post
#14
|
|
![]() Photoartist ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 12,363 Joined: Apr 2006 Member No: 399,390 ![]() |
|
|
|
![]()
Post
#15
|
|
![]() Senior Member ![]() ![]() ![]() ![]() ![]() ![]() Group: Administrator Posts: 2,648 Joined: Apr 2008 Member No: 639,265 ![]() |
Thanks, I know what variables are. ![]() Michael, I did that, but failed. I don't really think I'm bad at math if I can get through AP Calculus, but this is what happened: Well, I mean, you can do it using matrices, too, but it works out the same. The problem does say that you have to determine if the equations actually do work, so they might not. But, I dunno; it's not like I have a degree in computer science or anything like that. ![]() |
|
|
![]()
Post
#16
|
|
![]() Death is a promise given to us at birth ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Designer Posts: 4,757 Joined: Mar 2004 Member No: 7,459 ![]() |
aw damn, I wanna try this.
|
|
|
![]()
Post
#17
|
|
Senior Member ![]() ![]() ![]() Group: Member Posts: 45 Joined: Feb 2008 Member No: 622,075 ![]() |
omg i am sooo sorry, the third line was x=2. & it did work. thanks for all the efforts guys!
|
|
|
![]()
Post
#18
|
|
![]() <joke> inside </joke> ![]() ![]() ![]() ![]() ![]() ![]() Group: Official Member Posts: 2,283 Joined: Oct 2006 Member No: 470,590 ![]() |
|
|
|
![]()
Post
#19
|
|
![]() durian ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 13,124 Joined: Feb 2004 Member No: 3,860 ![]() |
|
|
|
![]()
Post
#20
|
|
![]() in a matter of time ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 ![]() |
^ Yeah, I know right?
Wtf. Let's just give up haha. |
|
|
![]()
Post
#21
|
|
![]() durian ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 13,124 Joined: Feb 2004 Member No: 3,860 ![]() |
|
|
|
![]() ![]() |