BEC'S methods questions

[bec]

How do I do this simultaneous equation question?

(a-2)x - 3y = 2
     8x + by = 4
What are the values of a and b where there are a) no solutions, b) infinite number of solutions, c) a unique solution.

I've done a and b - is this right?
a) a=6, b=-3                b)  a=6, b=-6

and i'm stuck with c), finding the values of a and b to form a unique solution.

thanks!

[Collin Li]

These are probably ridiculously hard addition of ordinates problems

Not really. You can factorise them quite easily to find the x-intercepts. You can use calculus to find the turning points too (addition of ordinates wont find you the turning points).

For example, the last one, which would be horrendous with addition of ordinates is easily done as:

[Mao]

rearranging:






a) no solutions, so both lines are parallel but do NOT overlap i.e. m₁=m₂ but c₁<>c₂





and



when

so there are infinitely many sets of a and b for no solutions. (a curve describes this relationship, as a curve has infinite number of points on it, there are infinite sets of a and b to "no solutions")


b) infinite number of solutions is lust like no solutions, except this time m₁=m₂ but c₁=c₂ as they do overlap

hence , substituting this gives us




c) a unique solution is EVERYTHING ELSE, practically, where m₁<>m₂



for

a big honking set of numbers (think of it as, shading the whole graph except for one curve, so...)



so it's not that simple as a couple numbers, but a whole LOAD of them

[Collin Li]

The case for no solutions is when on the LHS (with the variables) of both of the equations are exactly the same, but the RHS is different.

Example:





Equation one can be changed into , but there are no solutions to these two equations because they are inconsistent (try subtracting equation one from equation two, you get something nonsensical like ).


The case for infinite solutions is when the two equations (usually can solve two unknowns) are actually the same thing.

Example





Equation one can be transformed into equation two by multiplying by two. This means we only have one piece of information for two variables - an infinite number of solutions arise.


For unique solutions, this is when neither of those cases above occur. This is the reason why the question gets you to find the case for infinite solutions, and the case for no solutions first.

In linear equations, you can only get three cases: the three listed above.

[Matt The Rat]

A multiple choice question of this kind has come up on the last 2 years (possibly more, I'm not sure but def. last 2) Methods exam 2 and the '06 results stated that it was done poorly so make sure that you know how to do this type of question!!!

[bec]

ok so i've done this question a few times and come up with different answers every time...

for

can anyone do it? if you don't want to go through the painful process of writing it all in latex, i'd still really appreciate it if you could tell me what answer i should be getting..
thanks

[Four]

Step 1: Rearrange in terms of sin(2x)


Step 2:Ask yourself: when does sin equal the right hand term? Keeping in mind that the basic angle is pi/3 and sin is negative in quadrants three and four, realise that:


Therefore...



Step 3: Now add and subtract the period (2pi/k, where k is the coefficient of x: here, the period is just plain pi) to 'fill' the domain.

[bec]

that is actually the clearest explanation i've ever had for trig equations, from any teacher/ book! thank you!

[Four]

No problem. Just remember to approach every trig problem systematically, and you'll be fine. 

[Collin Li]

if you don't want to go through the painful process of writing it all in latex

Hey, it's easy

[Chocolate]

How do I do this simultaneous equation question?

(a-2)x - 3y = 2
     8x + by = 4
What are the values of a and b where there are a) no solutions, b) infinite number of solutions, c) a unique solution.

I've done a and b - is this right?
a) a=6, b=-3                b)  a=6, b=-6

and i'm stuck with c), finding the values of a and b to form a unique solution.

thanks!

Can I ask which chapter/area of study this is this from? I haven't really seen anything like it before

[bec]

Can I ask which chapter/area of study this is this from? I haven't really seen anything like it before

it's from chapter 1 (around p21 in mathsworld if that's the book you've got)...possibly only for cas though, but i'm not sure.

[Collin Li]

Oh, by the way, are you meant to use matrices for these? (not familiar with the CAS course)

[bec]

well see i'm not too sure - usually, i use rref (i'm pretty sure that's on the other calculator too isn't it?)
but i didn't know how to use rref for this one, so doing it by hand seemed like a good option. although i guess you could do it with matrices by hand?

[bec]

Solve for
i'm thinking it has something to do with substituting 2x-pi/6 for X...and possibly some factorising and using the null factor law...and that's as far as i can get.
help?