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.

this and this might help.

I haven't done this since middle school, though, so I can't be much more help. but also remember google is your friend! happy solving. =D

Are the A's, B's and C's part of the equations?

oh. no, they aren't,

A C and B are variables they are substitutes for numbers in algebra

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.

Thanks, I know what variables are. And she just said that they aren't part of the equation.

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:

Are you guys assuming that all three lines intersect? Dude it makes a triangle.

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)

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.

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.

^ 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

quickmath.com

lol Asians + Michael arguing

I hate linear equations and the rest of mathematics.
Holy crap, I hope that actually works. If it does, you would have just made a new best friend.

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.

aw damn, I wanna try this.

omg i am sooo sorry, the third line was x=2. & it did work. thanks for all the efforts guys!

it does! i use it all the time. shhh

But the point's not on the third line, if the lie is x=2 instead of y=2. It would still form a triangle...

^ Yeah, I know right?

Wtf.

Let's just give up haha.

Yea screw this. I stopped taking math a year and a half ago.

*gets back to political science* SCREW MATH!!!