VCE Methods Question Thread!

[knightrider]

They are equivalent.



Edit: beaten by Eulerfan (you are too fast )

Because that's easier. Watch closely:

Thanks eulerfan101 and Zealous 

[knightrider]

Hi guys and girls what are your tips on doing well in maths methods and how did you go about preparing for sacs and the exam and how many practice exams did you end up doing and when do you recommend doing the vcaa ones.

Also what are your thoughts on finishing  the methods course in the holidays and at what pace did you work through the course

[AirLandBus]

Hi guys and girls what are your tips on doing well in maths methods and how did you go about preparing for sacs and the exam and how many practice exams did you end up doing and when do you recommend doing the vcaa ones.

Also what are your thoughts on finishing  the methods course in the holidays and at what pace did you work through the course

In terms of holiday work, i think the best is to go over stuff from 1/2 and practice the stuff you messed up on, clear up any doubts and probs do chapter 1 and 2 for the holidays. Can do more but i think thats over kill. Doing chapter 2 is probably over kill tbh. However, ive been talking to sir.jonse and hes smashing the textbook out. So i guess each to their own. Just do as much as you can and feel like doing. He loves maths so doing the whole textbook is probably enjoyable for him. Where as for others, they enjoy maths but not "that" much, so doing a couple of chapters ahead is sufficient otherwise they may burn out. The way i see it, say your a chapter or two ahead, when the class goes through the work in class it will be additional revision and clearing things up. And because youve already done the textbook questions, you can work on the harder questions and extending your knowledge. Thats the way I see it, differs from person to person. Just start and see what works for you.

Also, i your interested in joining a skype study group for methods (and possibly other subjects) add sir.jonse on skype and he will add you to the group convo. That applies to anyone BTW.

[Talia2144]

Hi,can anyone please help with this question : it is a simultaneous eq)uation. ax+by=p and bx-ay=q I know how to solve the equation but I want to know when you solve for x you get y=bp-aq divided by by a^2+b^ 2. Can you sub this y value into one of.the simultaneously and get x=ap+bq  divides. by a^2+B^2. Thank you

[AirLandBus]

Hi,can anyone please help with this question : it is a simultaneous eq)uation. ax+by=p and bx-ay=q I know how to solve the equation but I want to know when you solve for x you get y=bp-aq divided by by a^2+b^ 2. Can you sub this y value into one of.the simultaneously and get x=ap+bq  divides. by a^2+B^2. Thank you

Might be helpful if you post the background info - ie the entire question.

[Talia2144]

It is questions 2 I want solve it without using simultaneous twice but use substitution method if.its possible

[kevviinn95]

It is questions 2 I want solve it without using simultaneous twice but use substitution method if.its possible

ax+by=p.........  equation 1
bx-ay=q.......... equation 2

bx-ay=q (using equation 2, rearrange for x)
bx=q+ay
x=(q+ay)/b

a(q+ay/b)+by=p (sub x value into equation 1)
(aq+a^2y)/b + by=p
aq+a^2y+b^2y=bp (multiplied previous equation by b)
y(a^2+b^2)+aq=bp(factorised by common factor y)
y(a^2+b^2)=bp-aq
y=(bp-aq)/(a^2+b^2) (solved for y)


bx-ay=q (using equation 2, rearrange for y)
-ay=-bx+q
ay=bx-q
y=(bx-q)/a

ax+b(bx-q/a)=p (sub y value into equation 1)
ax+(b^2x-bq)/a=p
a^2x+b^2x-bq=ap (multiplied previous equation by a)
x(a^2+b^2)-bq=ap (factorised by common factor x)
x=(ap+bq)/(a^2+b^2) (solved for x)

[SE_JM]

Hello!

I have a question about graphing logarithms.

find the x-intercept of:
y=2 + log_e(3x-2)

I got the answer:
x= (e^-2 + 2)/3

When I put that into the CAS, it comes up as 0.711778... but the answer is x=0.79.

WHY??

Please help

Thanks!!

[ikiwi]

Hmm I got 0.711778 as well. Maybe the answers are wrong?

[SE_JM]

Hmm I got 0.711778 as well. Maybe the answers are wrong?

It gives me comfort to know someone got the same answer as me 
thanks perhaps the answer is wrong

[catherine_mimi]

I keep getting the answer wrong...

3^2x-3^x+2+8=0,

Somebody please help me!

Thanks,
Catherine

[brightsky]

3^(2x) - 3^(x+2) + 8 = 0
3^(2x) - 9*3^x + 8 = 0
Let A = 3^x.
A^2 - 9A + 8 = 0
(A - 1)(A - 8 ) = 0
A = 1 or A = 8
3^x = 1 or 3^x = 8
x = 0 or x = log_3(8 )

[catherine_mimi]

you're the best! 

3^(2x) - 3^(x+2) + 8 = 0
3^(2x) - 9*3^x + 8 = 0
Let A = 3^x.
A^2 - 9A + 8 = 0
(A - 1)(A - 8 ) = 0
A = 1 or A = 8
3^x = 1 or 3^x = 8
x = 0 or x = log_3(8 )

[catherine_mimi]

Hey all,
I have a question that I couldn't solve
Solve for x:
e^-x + 3= x
Thanks!
 

[catherine_mimi]

It's in the methods 3 & 4 Essential Cambridge book
There is a specific question where they ask you for a intersection concerning two graphs: e^-x +3 and -log_e(x+3)
I figured that if I made e^-x +3 =x, I'd find the answer the easier way.