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anonymous

  • 2 years ago

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.

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  1. DebbieG
    • 2 years ago
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    Do you have an equation for the meteor's path? You're missing some information here.

  2. anonymous
    • 2 years ago
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    No, it didn't give me it.. let me double check

  3. DebbieG
    • 2 years ago
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    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.

  4. anonymous
    • 2 years ago
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    It doesn't tell me what the meteors trajectory is

  5. DebbieG
    • 2 years ago
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    Post the whole problem. This is not a stand-alone problem. There is missing info.

  6. DebbieG
    • 2 years ago
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    Is there a diagram or something?

  7. anonymous
    • 2 years ago
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    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

  8. skullpatrol
    • 2 years ago
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    It could be simply asking where y=4x + 1 intercepts the x-axis.

  9. anonymous
    • 2 years ago
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    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.

  10. DebbieG
    • 2 years ago
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    Ah, ok. Just DESCRIBE how to FIND the solution.

  11. anonymous
    • 2 years ago
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    It just wants to know how would I solve to find out where the would cross..

  12. DebbieG
    • 2 years ago
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    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?

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

  14. anonymous
    • 2 years ago
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    Sorry..

  15. anonymous
    • 2 years ago
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    I don't remember how to do that.

  16. anonymous
    • 2 years ago
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    Id do it all in my calculator if i could..

  17. anonymous
    • 2 years ago
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    butt im assuming that i would make the x and y chart and graph my lines.. am i correct?

  18. DebbieG
    • 2 years ago
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    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.

  19. DebbieG
    • 2 years ago
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    Graphing is one possible method, although it's not very precise.

  20. anonymous
    • 2 years ago
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    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.

  21. anonymous
    • 2 years ago
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    Well ok, i want the most precise way. Would that be solving the liner equation? Im not sure...

  22. DebbieG
    • 2 years ago
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    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".

  23. anonymous
    • 2 years ago
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    Ok.. sounds sorta familiar how do i find the pair?

  24. DebbieG
    • 2 years ago
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    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?

  25. anonymous
    • 2 years ago
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    Yes

  26. DebbieG
    • 2 years ago
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    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.

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

  28. DebbieG
    • 2 years ago
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    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)

  29. DebbieG
    • 2 years ago
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    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.

  30. DebbieG
    • 2 years ago
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    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.

  31. DebbieG
    • 2 years ago
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    Graphically, it looks like this: |dw:1378209299923:dw|

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