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.

- anonymous

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- DebbieG

Do you have an equation for the meteor's path? You're missing some information here.

- anonymous

No, it didn't give me it.. let me double check

- DebbieG

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.

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## More answers

- anonymous

It doesn't tell me what the meteors trajectory is

- DebbieG

Post the whole problem. This is not a stand-alone problem. There is missing info.

- DebbieG

Is there a diagram or something?

- anonymous

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

- skullpatrol

It could be simply asking where y=4x + 1 intercepts the x-axis.

- anonymous

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.

- DebbieG

Ah, ok. Just DESCRIBE how to FIND the solution.

- anonymous

It just wants to know how would I solve to find out where the would cross..

- DebbieG

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?

- DebbieG

Oh, and it helps that you are told that the meteor's path is modeled by a LINEAR equation. So both are lines.

- anonymous

Sorry..

- anonymous

I don't remember how to do that.

- anonymous

Id do it all in my calculator if i could..

- anonymous

butt im assuming that i would make the x and y chart and graph my lines.. am i correct?

- DebbieG

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.

- DebbieG

Graphing is one possible method, although it's not very precise.

- anonymous

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.

- anonymous

Well ok, i want the most precise way. Would that be solving the liner equation? Im not sure...

- DebbieG

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".

- anonymous

Ok.. sounds sorta familiar how do i find the pair?

- DebbieG

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?

- anonymous

Yes

- DebbieG

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.

- anonymous

I know of substitution.. but i don't remember how to do it. ive done it before though, quite a few times.

- DebbieG

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)

- DebbieG

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.

- DebbieG

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.

- DebbieG

Graphically, it looks like this:
|dw:1378209299923:dw|

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