VCE Methods Question Thread!

[Planck's constant]

For what values do the equations  (m-1)x + 5y = 7 and 3x + (m-3)y = 0.7m have infinitely many solutions?

I rearranged it and solved for m (setting the gradients equal) and got m = 6 and m = -2. Why is 6 an answer but -2 isn't (I plotted it and it gave me parallel lines, but why?)
 

This is a problem for logical thinking.

Every equation of the form ax+by=c represents a straight line.
Therefore a system of 2 simultaneous equations of this type represents 2 straight lines.
This allows the following 3 possibilities ONLY.

1) The lines interscet = the sysem has 1 solution = the lines have different gradients.
2) The lines are parallel = the system has 0 solutions = the lines have the same gradient AND DIFFERENT y-intercepts
3) The lines are the exact same = the system has infinite solutions =  the lines have the same gradient AND THE SAME y-intercepts

By applying those 3 rules, you get all the answers you need

 

[Planck's constant]

help!
solve the simultaneous equations for x and y. I'd plunk it into the calc but i have to show working...

-(b+c)y - ax = ab
(a+b)x + cy = bc

thanks

(+)








You can probably simplify some things in there but I'm just showing you how to go about it.

Do the same for x i.e. multiply each equation such that you will be able to eliminate y from the equation.

edit: replaced ab(a + b) with ac as the coefficient for y in the second last line and hence answer

Things are never that complicated with problems you likely to get asked.
They normally involve some sort of trick, which is not always obvious to see.
I this instance, I first expanded the 2 equations :

-by -cy -ax= ab
ax + bx + cy = bc

Adding those 2 expanded equations gives you:

b(x - y) = b(a + c)
x - y = a + c      (1)

Now you are getting somewhere. I cant think of anything overly clever now, so I just solve equation (1) for x,

x = y + a + c

and substitute x in the original equation (-by-cy-ax=ab),

-by -cy - a(y + a + c) = ab

Solving for y,

y = -a
x = c           .... from (1)

[Phy124]

Ah, I'm an idiot, good spot argonaut! You should certainly go with his method, Kaille 

Anyway, just in case you already put down the working out in your book you can simplify and factorise your answer by doing the following











And then just sub that back into a much simpler equation hahaha

I certainly have a knack for over complicating things, sorry about that  *facepalm*

[mfung95]

Could someone please send me the practise sacs and exams for methods 3/4?  >.<" THANK YOUUUUUUU

Practise tests/SACs

Truetears, are the practise sacs/tests all in the TT document (mediafire)?

[pi]

Yes.

All the files in this entire thread are uploaded here: http://www.mediafire.com/?f41nd5tbo6jjx28

Courtesy to laseredd.

[mfung95]

Thank you, VegemitePi and Truetears!!! ^^

[ashoni]

An investor received $31,000 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. She found that if she interchanged the amounts she invested she could increase her return by $1000 per annum. Calculate the total amount she had invested.

I've never been good at the business side of maths haha ><

[Planck's constant]

I certainly have a knack for over complicating things, sorry about that  *facepalm*

No. You tried a legitimate 'trick'.
But the point I was trying to make was that unless you get an early breakthrough with your trick, you could end up with complex expressions, which is a sure sign that you should go back to the start and try another trick

[TrueTears]

An investor received $31,000 interest per annum from a sum of money, with part of it invested at 10% and the remainder at 7% simple interest. She found that if she interchanged the amounts she invested she could increase her return by $1000 per annum. Calculate the total amount she had invested.

I've never been good at the business side of maths haha ><

there is something wrong with this question due to lack of info

how long is the money invested for?

the part of the sum of money invested at 10% compounded payable when?

earning $31,000 interest per year is definitely wrong, if the 10% is invested at a compounded rate, the interest earned per year is NOT constant.

[ICECOLD]

hi. new to this, is this where I ask a question?

Given X~N(32,16). Fish which are less than 27cm are considered to be undersized.

Find a) Pr fish is undersized?

I got this as 0.1057

b) the expected number of fish that a fisherman could take home if he catches 20fish, ruling out the undersized fish.

I did 20x0.1057 to get 2.2197 fish undersized.

Do I do 20-2 or 20-3. Don't know how to round.

thanks.

[jazza97]

Hi guys, got a couple, would be awesome if anyone could help out

1. y=2x+7/x+4

express this in the form y=a/x-h +k


2. find the equation of a cubic function of the from f(x)=k(x+b)^2(x+a) which has a maximum truning point at (2,0), another x intercept at (-1, 0) and a y intercept at (0, -12). 

Thanks Guys

[Phy124]

1.



Alternatively;

I don't know how to do it with latex but

       ______
x+4|2x + 7 |   2(x+4)
   (-) 2x + 8 |
               -1|

Therefore it is divisible twice with a remainder of , so:

2. If you have a look at the information given, try sketching a graph;

You know it has an x-intercept at -1 and then a y-intercept at -12.The fact that the turning point is at x = 2, means x = -1 can't be a turning point and therefore a = 1.

See - http://i41.tinypic.com/33agifo.png

We also know that we will only have 2 x-intercepts so the turning point must be on the x-axis at the stated value of x = 2, from this we know b = -2.

Finally we know that the y-intercept is at -12, so:









Which gives us the graph http://i40.tinypic.com/33v1zr5.png, which fits the characteristics stated

[Kanon]

Hey guys, just having a bit of trouble with logs.

Solve , correct to three decimal places.

Also, when rearranging logs does it matters where I put the inverse of it?  My wording is horrible, but if i have

does it matter if i put it as

ORRRRR

[jasoN-]

2*2^(-2x) = what?

also for the logs thing, doesn't matter which way, but i prefer keeping everything positive if possible:
ln(x) - 2ln(5) = 2
then ln(x/25) = 2
x = 25e^2

[Kanon]

2*2^(-2x) = what?

also for the logs thing, doesn't matter which way, but i prefer keeping everything positive if possible:
ln(x) - 2ln(5) = 2
then ln(x/25) = 2
x = 25e^2

I'm feeling really ill, so my latex might be off.
* = multiplied