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raebaby420
Please help. c: If the linear equation is Y=4x+1, describe how to solve where your equation and the meteor’s path will cross. Explain any possible methods used in discovering a solution.
Do you have an equation for the meteor's path? You're missing some information here.
No, it didn't give me it.. let me double check
I know that y=4x+1 is your equation from the last problem, but you're going to need to post all the problem info. This is the first I've heard of a meteor, lol.
It doesn't tell me what the meteors trajectory is
Post the whole problem. This is not a stand-alone problem. There is missing info.
Is there a diagram or something?
I think what its asking me is if I had that info what would I do to solve where they would cross since I have my linear equation
It could be simply asking where y=4x + 1 intercepts the x-axis.
Heres the question: If the linear equation of the meteor’s path is known, describe to the captain how to solve where your equation from question 1 and the meteor’s path will cross. Explain any possible methods used in discovering a solution.
Ah, ok. Just DESCRIBE how to FIND the solution.
It just wants to know how would I solve to find out where the would cross..
What do you think? You would have 2 equations... you want to find a point that is on the graph of BOTH equations. how do you do that?
Oh, and it helps that you are told that the meteor's path is modeled by a LINEAR equation. So both are lines.
I don't remember how to do that.
Id do it all in my calculator if i could..
butt im assuming that i would make the x and y chart and graph my lines.. am i correct?
Don't remember? isn't this topic being covered in your class? No, don't resort to your calculator all the time. It's a useful tool, but it doesn't teach you the CONCEPT.
Graphing is one possible method, although it's not very precise.
Mm im doing credit recovery.. so im trying to understand it. i failed algebra 2 because i wasn't paying attention in school and now im a senior and struggling.
Well ok, i want the most precise way. Would that be solving the liner equation? Im not sure...
If you have two equations, say: y=4x+1 y=-3x+5 and you want to find a point that is on BOTH of the lines, then you are looking for an (x,y) pair that satisfies BOTH equations. This is called a "system of 2 linear equations in 2 unknowns".
Ok.. sounds sorta familiar how do i find the pair?
Now, I conveniently wrote both of these in the form y={some stuff}... but that's OK, because you can ALWAYS transform a linear equation into that form (so long as it isn't a VERTICAL line, that's the only time you can't). So since y={some stuff} and y={some other stuff}.... the easiest thing in this case is probably to set the "stuffs" equal to one another. That is, they are both =y, so they are both = each other, see that?
That's called "substitution", because I have one equation (actually both in this case, but one is enough) solved for ONE variable, and then I'm PLUGGING THAT IN ("substituting") in the other equation where I see that variable.
I know of substitution.. but i don't remember how to do it. ive done it before though, quite a few times.
So, in my made-up example, I would have: 4x+1=-3x+5 And now I would solve THAT for x, just by moving the terms around (add 3x to both sides, subtract 1 from both, and then divide by the coefficient on the x term)
OK, that's fine... that's why I'm walking you through it. :) You should know how to do this, so it's good that you are going back to re-learn it now.
Now, once I solve that equation, I'll have a VALUE, a number, for x. That's my x for my solution. Now I just need y. Well, I have two equations, and again - remember what this solution must be: an (x,y) pair that satisfies BOTH equations. So NOW I'm going to take that x I just found, and plug it into either equation, and solve for y. Check the solution (plug both x and y into both equations, and they should be true). After you confirm that it works, THAT is your solution to the system.
Graphically, it looks like this: |dw:1378209299923:dw|