VCE Methods Question Thread!

[Jimmmy]

Suppose you're dividing 1359 by 11. We need to pick a first digit, and it is going to be 1 because 1x11 = 11, closest to 13. Same idea above. When dividing 2x+1 by 5x+3, 2/5 gets you the 'closest' match as you then match the x's.

Yeah I get what you mean, so it's sort of an inference? I'm looking for those little 1%ers to increase my efficiency, so having this knowledge when doing long division will no doubt be very helpful.  and by matching the x's, you mean being able to cancel them out?

Thanks so much for your help 

[Evolio]

I would highly recommend watching some videos to help you understand topics. Learning from a textbook is sometimes quite hard.

Two channels that come to my mind are Khan Academy and Worm’s maths academy.

Worm’s directed at VCE methods and can be found here

https://www.youtube.com/user/andrewworm

I absolutely loved his videos and they were super useful.

Don't worry, I'm also currently pre-learning methods 1/2 and also spending at the least 30 minutes to complete the harder questions.
What I did is just do the ones I know how to complete and if I do not know some certain things, for e.g Logarithms, I would search it up and have a look.

I know this is late but thank you!
I really appreciate your help!
 

[Evolio]

Hi.
I was wondering if someone could please help me with questions d,e,f?
I tried to factorise for x and y but I keep getting weird answers.
Thanks.

[aspiringantelope]

Hi.
I was wondering if someone could please help me with questions d,e,f?
I tried to factorise for x and y but I keep getting weird answers.
Thanks.

Hey I'm probably going to get it wrong but I'm going to have a go cause I'm probably going to be learning this as well -_-
d) \(ax+by=a^2+2ab-b^2\)
\(bx+ay=a^2+b^2\)
Top line
\(ax=a^2+2ab-b^2-by\)
=\(x=\frac{a^2}{a}+\frac{2ab}{a}-\frac{b^2}{a}-\frac{by}{a}\)
Can simplify to
= \(x=a+2b-\frac{b^2}{a}-\frac{by}{a}\)
I'm not sure if that's right?
Anyways to get y you plug it in another equation.
Let's use the bottom line cause it's not being used at all xD
\(ay=a^2+b^2-bx\) Moved bx
= \(y=\frac{a^2}{a}+\frac{b^2}{a}-\frac{bx}{a}\)
= \(y=a+\frac{b^2}{a}-\frac{bx}{a} \)
\(y=a+\frac{b^2}{a}-\frac{b\left(a+2b-\frac{b^2}{a}-\frac{by}{a}\right)}{a}\)
and i got  \( y=a+\frac{b^2}{a}-\frac{ba+2b^2-\frac{b^3}{a}-\frac{b^2y}{a}}{a}\)
LOL im not sure -_- but i know the you can connect the last two fractions cause same denominator.
which is let me edit
\(y=a-\frac{ba+3b^2-\frac{b^3}{a}-\frac{b^2y}{a}}{a}\)
I think? LOL if this is all wrong some1 pls tell me i want to delete before people all see

[maxk]

Hey I'm probably going to get it wrong but I'm going to have a go cause I'm probably going to be learning this as well -_-
d) \(ax+by=a^2+2ab-b^2\)
\(bx+ay=a^2+b^2\)
Top line
\(ax=a^2+2ab-b^2-by\)
=\(x=\frac{a^2}{a}+\frac{2ab}{a}-\frac{b^2}{a}-\frac{by}{a}\)
Can simplify to
= \(x=a+2b-\frac{b^2}{a}-\frac{by}{a}\)
I'm not sure if that's right?
Anyways to get y you plug it in another equation.
Let's use the bottom line cause it's not being used at all xD
\(ay=a^2+b^2-bx\) Moved bx
= \(y=\frac{a^2}{a}+\frac{b^2}{a}-\frac{bx}{a}\)
= \(y=a+\frac{b^2}{a}-\frac{bx}{a} \)
\(y=a+\frac{b^2}{a}-\frac{b\left(a+2b-\frac{b^2}{a}-\frac{by}{a}\right)}{a}\)
and i got  \( y=a+\frac{b^2}{a}-\frac{ba+2b^2-\frac{b^3}{a}-\frac{b^2y}{a}}{a}\)
LOL im not sure -_- but i know the you can connect the last two fractions cause same denominator.
which is let me edit
\(y=a-\frac{ba+3b^2-\frac{b^3}{a}-\frac{b^2y}{a}}{a}\)
I think? LOL if this is all wrong some1 pls tell me i want to delete before people all see

Yeah slightly off

 
 

I'm trying to type out the solution thingy here

[Jim_Bob]

Anyone know anything about the KIS maths team?

[maxk]

Hi.
I was wondering if someone could please help me with questions d,e,f?
I tried to factorise for x and y but I keep getting weird answers.
Thanks.

d)

1. \(ax+by=a^2+2ab-b^2\)
2. \(bx+ay=a^2+b^2\)

Rearrange 1. for x:
\(ax+by=a^2+2ab-b^2\)
\(ax=a^2+2ab-b^2-by\)
\(x=\frac{a^2+2ab-b^2-by}{a}\)

Sub \(x=\frac{a^2+2ab-b^2-by}{a}\) into 2. and solve for y:

\(bx+ay=a^2+b^2\)
\(b\frac{a^2+2ab-b^2-by}{a}+ay=a^2+b^2\)
\(\frac{b(a^2+2ab-b^2-by)}{a}+ay=a^2+b^2\)
\(\frac{a^2b+2ab^2-b^3-b^2y)}{a}+ay=a^2+b^2\)
\(ab+2b^2-\frac{b^3}{a}-\frac{b^2y}{a}+ay=a^2+b^2\)

Multiply both sides by a:

\(a^2b+2ab^2-b^3-b^2y+a^2y=a^3+ab^2\)
\(-b^2y+a^2y=a^3+ab^2-a^2b-2ab^2+b^3\)
\(-b^2y+a^2y=a^3-ab^2-a^2b+b^3\)
\(y(a^2-b^2)=a^3-ab^2-a^2b+b^3\)
\(y=\frac{a^3-ab^2-a^2b+b^3}{a^2-b^2}\)
\(y=\frac{a^3-a^2b-ab^2+b^3}{a^2-b^2}\)
\(y=\frac{a^2(a-b)-ab^2+b^3}{a^2-b^2}\)
\(y=\frac{a^2(a-b)+b^2(b-a)}{a^2-b^2}\)
\(y=\frac{a^2(a-b)-b^2(b+a)}{a^2-b^2}\)
\(y=\frac{(a^2-b^2)(a-b)}{a^2-b^2}\)
\(y=a-b\)

Sub \(y=a-b\) into \(x=\frac{a^2+2ab-b^2-by}{a}\):

\(x=\frac{a^2+2ab-b^2-b(a-b)}{a}\)
\(x=\frac{a^2+2ab-b^2-ab+b^2}{a}\)
\(x=\frac{a^2+2ab-ab}{a}\)
\(x=\frac{a^2+ab}{a}\)
\(x=\frac{a(a+b)}{a}\)
\(x=a+b\)

So,
\(x=a+b\)
\(y=a-b\)

[darkz]

Hi.
I was wondering if someone could please help me with questions d,e,f?
I tried to factorise for x and y but I keep getting weird answers.
Thanks.

\[
\text{d.}\\
\text{Multiply the first equation by a and the second equation by b}\\
a^2x+aby=a^3+2a^2b=ab^2...\boxed{1}\\
b^2x+1by=a^2b+b^3...\boxed{2}\\
\boxed{1}-\boxed{2}:\\
x(a^2-b^2)=a^3+a^2b-ab^2-b^3\\
\begin{aligned}
x&=\frac{a^3+a^2b-ab^2-b^3}{a^2-b^2}\\
&=\frac{a^2(a+b)-b^2(a+b)}{a^2-b^2}\\
&=\frac{(a^2-b^2)(a+b)}{a^2-b^2}\\
&=a+b\\
\end{aligned}\\
\text{Then sub it back into }\boxed{2}\text{ and solve for y}\\
y=a-b\\
\text{ }\\
\text{ }\\

\text{e. Rewrite the second equation, then multiply the first equation by }b+c\text{ and the second equation by }c\\
(a+b)(b+c)x+c(c+c)y=bc(b+c)...\boxed{1}\\
acx+x(b+c)y=-abc...\boxed{2}\\
\boxed{1}-\boxed{2}:\\
\begin{aligned}
x((a+b)(b+c)-ac)&=bc(b+c)+abc\\
x(ab+ac+b^2+bc-ac)&=bc(b+c+a)\\
x(ab+b^2+bc)&=bc(a+b+c)\\
xb(a+b+c)&=bc(a+b+c)\\
x&=c\\
\end{aligned}\\
\text{Sub that into }\boxed{1}\text{ and solve for y}\\
y=-a\\

\text{ }\\
\text{ }\\
\text{f. First simplify the equations}\\
\begin{aligned}
3x-3a-2y-2a&=5-4a\\
3x-2y&=5-4a+3a+2a\\
3x-2y&=a+5...\boxed{1}\\
2x+2a+3y-3a&=4a-1\\
2x+3y&=4a-1-2a+3a\\
2x+3y&=5a-1...\boxed{2}\\
\end{aligned}\\
\text{Multiply }\boxed{1}\text{ by 3 and }\boxed{2}\text{ by 2}\\
\begin{aligned}
9x-6y&=3a+15...\boxed{3}\\
4x+6y&=10a-2...\boxed{4}\\
\end{aligned}\\
\boxed{3}+\boxed{4}:\\
\begin{aligned}
13x&=13a+13\\
x&=a+1\\
\end{aligned}\\
\text{Sub into }\boxed{2}\\
y=a-1\\
\]

[Evolio]

Thank you guys for all your help!

[Evolio]

I was wondering if you guys knew how to do question h?

[darkz]

I was wondering if you guys knew how to do question h?

\[\text{Make h the subject of the second equation}\\
\begin{aligned}
as+2b&=3a\\
2h&=4a-as\\
h&=\frac{3a-as}{2}
\end{aligned}\\
\text{Sub into the first equation}\\
\begin{aligned}
3s-ah&=a^{2}\\
3s-\frac{a(3a-as)}{2}&=a^{2}\\
6s-a(3a-as)&=2a^2\\
6s-3a^{2}+a^{2}s&=2a^{2}\\
a^{2}s+6s&=2a^{2}+3a^{2}\\
s(a^{2}+6)&=5a^{2}\\
s&=\frac{5a^2}{a^2+6}
\end{aligned}\\
\]

Edit: Fixing Latex

[Evolio]

Thank you!
I was wondering if anyone knew how to solve this question?
I showed that the value of y=bp-aq/a^2+b^2.
But when I subbed it in, I'm not getting the right answer.

[fun_jirachi]

Hey there!

You solve this question by multiplying one equation by a, and the other by b, and it doesn't matter which one, as long as you pick one for a and one for b. You then add or subtract accordingly to solve simultaneously. You should get both results given by the question. I think your problem with substituting back in stems from the fact that you haven't substituted back in both x and y? Or maybe you had a computation error? It seems okay for both equations when substituting back in. Just try again, remembering to sub in both, ie a(x) + b(y) ((x) and (y) indicating the solutions for x and y), then making sure it equates to p in this case.

Hope this helps

[Jimmmy]

Thank you!
I was wondering if anyone knew how to solve this question?
I showed that the value of y=bp-aq/a^2+b^2.
But when I subbed it in, I'm not getting the right answer.

Yeah, I made the same mistake! I ended up finding the solution with jirachi's answer, but if you sub both x & y into equation 1 and/or equation 2, you should get p for eq. 1 and q for eq. 2.

[Evolio]

Thank you guys!
I got the correct answers.
Basically what I did is that I found x by multiplying the first equation by a and the second by b.
Then I did the opposite again for y. I found y by multiplying the first equation by b and the second by a.
Is this correct or did I do an unnecessary step?
Also, just to clarify with these questions you should sub both x and y to get the answer right?