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# -RM- watch

1. -RM-
2. (Original post by UCASLord)
The simultaneous equations in x,y,

(cosθ)x - (sinθ)y = 2
(sinθ)x + (cosθ)y = 1

are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.

What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.
Multiply both equations by sinθ or multiply one by sinθ and the other by cosθ and try then adding both equations together - try something like that - in far too lazy to bother thinking about the details.
3. (Original post by UCASLord)
The simultaneous equations in x,y,

(cosθ)x - (sinθ)y = 2
(sinθ)x + (cosθ)y = 1

are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.

What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.
If you know about determinants (of 2x2 matrices), this is the quickest approach here.

Otherwise, just solve for x and y as you would any other simultaneous equations (effectively treating sin θ and cos θ as constants).
4. If you've done FM, it's a lot quicker to solve with matrices, and also "nicer" in the solution, I believe.

Otherwise it seems pretty ugly.
5. (Original post by UCASLord)
The simultaneous equations in x,y,

(cosθ)x - (sinθ)y = 2
(sinθ)x + (cosθ)y = 1

are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.

What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.
To clarify what the others mean by "use matrices", write the system as . As DFranklin said, you can then use the determinant to check where the system is consistent.
6. (Original post by artful_lounger)
If you've done FM, it's a lot quicker to solve with matrices, and also "nicer" in the solution, I believe.
It only takes a few lines to solve the simultaneous equations, and the final solution is equivalent. (On the other hand, someone used to matrices can do this question without any lines at all, but it's the difference between the question taking 2 minutes and it taking 30 seconds - it's not a huge deal).
7. -RM-
8. (Original post by UCASLord)
Thanks for all your replies guys. I have a couple of questions here, one about this specific question, and one about the FP1/C3/C4 units.

In regards to this question, I tried to solve it by defining x, and then substituting the definition of x for x, and then trying to find y. I can post my full working if needed, but what I pretty much get to is this
The solution you get should not be anything like this messy. I'm not saying what you've done is wrong, but I can't recognize the correct answer from it, which doesn't look good.

When you wish to solve simultaneous equations, particularly ones with non-numeric coefficients (i.e. things like x sin t + y cos t = 1, rather than 3 sin t + 4 cos 4 = 1), you should generally avoid dividing until you absolutely have to.

Folliowing is a big hint about the correct approach, but not a full solution
Multiply the first equation by cos and the 2nd by sin and add the two resultant equations to find x. Do something similar to find y.

In regards to my FP1, C3, C4: is it worth learning these units before doing the MAT? I've noticed quite a few of them offer faster solutions for a number of questions.
I think it can't hurt, but only if you have the time. I think one of the reasons people struggle with the MAT is that although it's "only" C1, C2 material, you don't generally get a good understanding of the material until you've used it in the context of the later modules.
9. First of all, the question is not asking you to solve the equations. The question is asking you about whether or not it is solvable for a range of values of theta.

Think about it. You have two lines. If there is a solution, then the lines must intersect (as the point of intersection is the solution). Any pair of lines will intersect unless they are parallel. Therefore, you just need to work out the gradients of the two lines and equate them (i.e. make the lines parallel) and the values of theta that solve that equation (if there are any solutions) will be the values of theta for which the lines don't intersect and thus don't have a solution.
10. (Original post by OnceUponAThyme)
First of all, the question is not asking you to solve the equations. The question is asking you about whether or not it is solvable for a range of values of theta.

Think about it. You have two lines. If there is a solution, then the lines must intersect (as the point of intersection is the solution). Any pair of lines will intersect unless they are parallel. Therefore, you just need to work out the gradients of the two lines and equate them (i.e. make the lines parallel) and the values of theta that solve that equation (if there are any solutions) will be the values of theta for which the lines don't intersect and thus don't have a solution.
The one issue with this is that although it doesn't happen here, for many problems of this type, you may need to distinguish between the case where the lines are parallel and there is no solution and the case where they are parallel and there are infinite solutions. At which point you're back to solving the equations.

The other point I'd make is that the OP should definitely make sure he does know how to solve equations of this form before taking the MAT. So at this point I recommend he press on with that approach.
11. (Original post by DFranklin)
The one issue with this is that although it doesn't happen here, for many problems of this type, you may need to distinguish between the case where the lines are parallel and there is no solution and the case where they are parallel and there are infinite solutions. At which point you're back to solving the equations.

The other point I'd make is that the OP should definitely make sure he does know how to solve equations of this form before taking the MAT. So at this point I recommend he press on with that approach.
Surely if the lines are parallel and there are infinite solutions then the lines are the same? In which case, you would just have to show if there's a common point or not (e.g. by letting x = 0 and finding y for both lines).

And for the MAT, I think it is very useful to think about the problem from different angles and carefully think about what the problem is actually asking. Of course, both approaches are valid.
12. -RM-
13. -RM-
14. (Original post by OnceUponAThyme)
Surely if the lines are parallel and there are infinite solutions then the lines are the same? In which case, you would just have to show if there's a common point or not (e.g. by letting x = 0 and finding y for both lines).
Yes, for 2 lines it's fairly straightforward. I'm perhaps biased from seeing too many people make a complete hash of the case with >=3 variables by trying to be too clever with determinants etc. and failing to analyse the special cases.

And for the MAT, I think it is very useful to think about the problem from different angles and carefully think about what the problem is actually asking. Of course, both approaches are valid.
In this case, my attitude is "yes, there are shortcuts you can take, but the obvious thing to try to do is to solve the simultaneous equations, and you should be able to solve them. So if you can't solve them, that's something that needs to be fixed, irregardless of whether there's a slick method that doesn't require solving them".
15. (Original post by UCASLord)
Hey, only just saw your reply. Thanks for your advice. I've managed to confirm the solution to the equation using the "big hint." (plus it's easier when you already know what the answer should be :3)
OK, now you've got a solution, let me waffle about solving simultaneous equations for a bit.

If you have a pair of simultaneous equations to solve for X and Y:

aX + bY = P
cX + dY = Q

then rather than doing something like writing X = (P-bY) / a (which involves dividing by a, which might be 0), the thing to do is multiply the 1st equation by d and the 2nd by b so the "Y" terms are the same:

ad X + bd Y = dP
bc X + bd Y = bQ

then subtracting the 2nd from the first gives

(ad-bc) X = dP - bQ

Now I've said" don't divide", but now you've eliminated Y, you need to "sit or get off the pot". More specifically, at this point, either (ad-bc) = 0 (and then if dP - BQ = 0, any value of X will work, otherwise no value of X will work), or (ad-bc) =/= 0 and you can divide by it to find X.

If you do similar calculations for Y, you'll find you have the same (ad-bc) factor.

ad-bc is the "determinant" we were talking about earlier, if it's non-zero, then the equations will always have a unique solution. Note in particular that it's possible for any one of a, b, c, d to be 0 and there to still be a unique solution, so dividing by any of them when you don't know if they might be 0 is not a good idea.

I doubt you'd get this in the MAT, but If you had 3 variables X, Y, Z, you would use a similar approach to first eliminate Z and then Y.
16. (Original post by UCASLord)
Made it all too easy there, but thanks in any case. :P By the end point, I can spot the answer: it's solvable for any value of theta.

However, I do have a question in regards to how you arrived at the gradient of the first line being cos(theta)/sin(theta). From what I know, there's two main ways to find the gradient. One of them is m=difference in y/difference in x, and the other method is differentiation. I can't quite tell how you've found cos(t)/sin(t) being the gradient. Could you explain a bit more on this point?
try getting the equation in the form y = mx + c
then m is the gradient

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