## raebaby420 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.

1. DebbieG

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

2. raebaby420

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

3. 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.

4. raebaby420

It doesn't tell me what the meteors trajectory is

5. DebbieG

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

6. DebbieG

Is there a diagram or something?

7. raebaby420

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

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

9. raebaby420

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

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

11. raebaby420

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

12. 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?

13. DebbieG

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

14. raebaby420

Sorry..

15. raebaby420

I don't remember how to do that.

16. raebaby420

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

17. raebaby420

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

18. 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.

19. DebbieG

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

20. raebaby420

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. raebaby420

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

22. 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".

23. raebaby420

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

24. 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?

25. raebaby420

Yes

26. 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.

27. raebaby420

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

28. 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)

29. 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.

30. 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.

31. DebbieG

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