VCE Methods Question Thread!

[ShortBlackChick]

I'm an arts student, but i'm gonna give it a shot

3y=x                     ------------1
y+32=x                 ------------2

3y=y+32
3y-y-32=0
2y=32
y=16                 

by substitution
3x16 = x = 48 or some shit.

[Nato]

Why did no one teach me this!!!! 
I need help with this question please.

Sharon is three times as old as her daughter.  If Sharon is 32 years old than her daughter, what are their ages?
My working out… Sharon=x Daughter=y

3x+y
32x

Thats all i can do, now what that looks like crap so i don't even know what to do this is pathetic!  This is a year 9 question.

So, just extract information from the question.
"Sharon is three times as old as her daughter". Let's put this in some type of equation:
What this is saying is that if you multiply her daughter's age by three, you get sharons age. this can be represented as:
  which is

Now for the second part
If Sharon is 32 years old than her daughter I could reword as, if you have her daugther's age, and you add 32, you get sharon's age. Hence, she's 32 years older. this can be represented as which is

Now, as usual you have simultaneous equation and . Just solve for x and y, which will corresponsd to Sharon's and the daughter's age respectively.

There is a suitable methods, posted above by "ShortBlackChick" outlining the steps to solve for x and y using substitution.

[Only Cheating Yourself]

Yea thanks guy, still confused but will work on it

[Nato]

Yea thanks guy, still confused but will work on it

I'll assume, you're confused not by the solving of simultaneous equations, but the nature of the problem. These extended response/worded questions can get quite tricky. But as I mention in my previous post, just try to extract the information from the question. Especially with worded simultaneous equations you sorta know what to expect - two equations with two unknowns. So looking at the question, there are statements basically giving you hints to form what into an equations.
hope this help in some way

[Only Cheating Yourself]

I'll assume, you're confused not by the solving of simultaneous equations, but the nature of the problem. These extended response/worded questions can get quite tricky. But as I mention in my previous post, just try to extract the information from the question. Especially with worded simultaneous equations you sorta know what to expect - two equations with two unknowns. So looking at the question, there are statements basically giving you hints to form what into an equations.
hope this help in some way

Actually the question above I'm having trouble solving, but once i do get the information i don't know what to.  Do i work it out simultaneously or what…

[Nato]

Actually the question above I'm having trouble solving, but once i do get the information i don't know what to.  Do i work it out simultaneously or what…

oh okay. Well, since we had the equations and to solve for x and y, you could either use substitution or elimination. These are just the usual simultaneous equations that you would've covered mayyybe year 9 or 10.

By substitution:
--------[1]
----------- [2]

quite simply we can substitute equation [1] into equation [2]. we have and that should fit nicely into , obviously replacing the x in equation two.
this would result in :
and solve for y like you would do a normal linear equation, and you would get . since represents the daughters age, clearly the daughter is years old.

Now since we have the y-value, it becomes a little easier, as in we can just substitute the y value we have recevied back into any equation - it doesn't matter which one we choose, we will get the same answer for the result x-value.
Due to simplicity, let's substitute our new found y-value, into equation [1]:
we'd get which is . Since 'x' represents Sharon's age, we have found out that she is years old.



You could have initially done the substitution methods the other way around, as in, find the y-value first, sub back into one the equations and get the x-value last. As long as you method is correct, you will end up with the same answer.

[Yacoubb]

Let P(x) = x⁵ - 3x⁴ + 2x³ - 2x² + 3x + 1

Given that P(x) can be written in the form (x² - 1)Q(x) + ax + b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by x² - 1.


Help would be appreciated thanks!
Btw - how do I know when I need to resort to my calculator :/?? Thanks

[Nato]

Let P(x) = x⁵ - 3x⁴ + 2x³ - 2x² + 3x + 1

Given that P(x) can be written in the form (x² - 1)Q(x) + ax + b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by x² - 1.


Help would be appreciated thanks!
Btw - how do I know when I need to resort to my calculator :/?? Thanks

Hey Yacoubb, look at the third response:

methods3/4 q's

[soNasty]

how would i show that k=3 in


i let
then had
i know that if instead of 2ku, 6u would allow the equation to be factorised to

am i on the right track? or is there another way of 'showing' that k=3

[eagles]

Let P(x) = x⁵ - 3x⁴ + 2x³ - 2x² + 3x + 1

Given that P(x) can be written in the form (x² - 1)Q(x) + ax + b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by x² - 1.


Help would be appreciated thanks!
Btw - how do I know when I need to resort to my calculator :/?? Thanks

We can use both the polynomial long division method and the calculator to solve this question.
1. Polynomial long division
(See attachments)

2. Calculator method
Using a ti-nspire, simply input expand((x⁵ - 3x⁴ + 2x³ - 2x² + 3x + 1) over (x² - 1)). Hopefully this results in the answer in the second attachment.

The remainder is the fraction part at the end.

[Yacoubb]

We can use both the polynomial long division method and the calculator to solve this question.
1. Polynomial long division
(See attachments)

2. Calculator method
Using a ti-nspire, simply input expand((x⁵ - 3x⁴ + 2x³ - 2x² + 3x + 1) over (x² - 1)). Hopefully this results in the answer in the second attachment.

The remainder is the fraction part at the end.

Thanks so much! That's so much clearer now!

My next question:

Show that the equation ax² - (a + b)x + b = 0 has a solution for a all values of a and b.

How do I go about questions like this? A thorough explanation would be much appreciated!

[Phy124]

Thanks so much! That's so much clearer now!

My next question:

Show that the equation ax² - (a + b)x + b = 0 has a solution for a all values of a and b.

How do I go about questions like this? A thorough explanation would be much appreciated!

A quadratic will have at least one solution if its discriminant is greater than or equal to zero.

The discriminant for a quadratic of equation is given by .

So we have:





Hence has at least one solution for all

[Yacoubb]

A quadratic will have at least one solution if its discriminant is greater than or equal to zero.

The discriminant for a quadratic of equation is given by .

So we have:





Hence has at least one solution for all

Thank you I knew we had to show the discriminant as being greater than or equal to 1, I just didn't know how to go about setting it out. Thanks for that!

[Blondie21]

how would i show that k=3 in


i let
then had
i know that if instead of 2ku, 6u would allow the equation to be factorised to

am i on the right track? or is there another way of 'showing' that k=3

Yep I did it the same way and we know that it's right too. The factorised equation tells us that 5 and 1 are the x intercepts, so when put into the original equation, y should equal 0.

which equals to zero
and
, which also equals to zero

[soNasty]

 A patient suffering with hypothermia after being lost in the snow for 2 days is covered in
thermal blankets and warmed so that his body temperature can return to normal levels.The change in body temperature (T°C) from the initial temperature that occurs during thermal treatment is modelled by the equation T = A loge (t − b) where t represents the time in minutes. The initial body temperature is 35.7°C and it reaches 36.1°C ten minutes after treatment is initiated.
(a)   What is the value ofT when t = 0?
(b)   Find the value of b.
(c)   Find the value of A.

help please! question is bugging me.