Questions about algebraic transformation

Hi, I have a few problems understanding the following;

1-n = x/y becomes n = 1-(x/y)

I don't understand this, I would have expected this: -n=(x/y)-1

It's the n that's negative not the 1, so why it would suddenly not be the case is confusing me.

--------------------

again, I have problems with this;

1-(p/q) = (q-p)/q

:mad:

any help would be greatly appreciated, I've enrolled onto an HNC and this is a big jump up to the maths I remember doing.

Reply 1

Franzen

16

Original post by Brainfrozen

Hi, I have a few problems understanding the following;

1-n = x/y becomes n = 1-(x/y)

I don't understand this, I would have expected this: -n=(x/y)-1

It's the n that's negative not the 1, so why it would suddenly not be the case is confusing me.

If you take -n=(x/y)-1 and multiply both sides by -1 what do you end up with?

again, I have problems with this;

1-(p/q) = (q-p)/q

In order to manipulate the LHS we need both parts to have the same denominator i.e. we want them both as fractions with q on the bottom. How can you rewrite 1 so it has a q on the bottom? (Or maybe look at it as, what fraction with q as the denominator/bottom cancels to 1)

(edited 7 years ago)

Reply 2

Brainfrozen

OP

1

Original post by Elzar

If you take -n=(x/y)-1 and multiply both sides by -1 what do you end up with?

In order to manipulate the LHS we need both parts to have the same denominator i.e. we want them both as fractions with q on the bottom. How can you rewrite 1 so it has a q on the bottom? (Or maybe look at it as, what fraction with q as the denominator/bottom cancels to 1)

n = -(x/y) +1

I understand now :smile:

(q/q) = 1

If only my lecturers were as rapid at answering I wouldn't have so many problems!

Expect more silly questions! :tongue:

Reply 3

Brainfrozen

OP

1

Another problem with transformation, I've moved onto series and understand this ok, however its the transformation that's doing my head in;

408=(n/2)(2×12+(n−1)×4)

I just don't understand how I'm doing this so wrong each time, you think you get somewhere and hit a brick wall :frown:

I'll show you how I have been trying it;

408 x 2 = n(2×12+(n−1)×4)

816/n = 24+(n−1)×4

n = 816/(24+(n−1)×4)

4n = 816/(24+n−1)

(4n)n = 816/234

n^2 = etc

very wrong :frown:

(edited 7 years ago)

Original post by Brainfrozen

Another problem with transformation, I've moved onto series and understand this ok, however its the transformation that's doing my head in;

408=(n/2)(2×12+(n−1)×4)

I just don't understand how I'm doing this so wrong each time, you think you get somewhere and hit a brick wall :frown:

I'll show you how I have been trying it;

408 x 2 = n(2×12+(n−1)×4)

816/n = 24+(n−1)×4

n = 816/(24+(n−1)×4)

4n = 816/(24+n−1)

(4n)n = 816/234

n^2 = etc

very wrong :frown:

You're correct up to the third line. The problem with the fourth is that you multiply both sides with 4 but the 24 is not affected.

\displaystyle 24+4(n-1)= 4(\frac{24}{4})+4(n-1)=4(\frac{24}{4}+n-1)=4(6+n-1)=4(n+5)

and this would be the denominator. And NOW you can multiply freely by 4 on both sides.

Though of course, your method is way too long winded.

(edited 7 years ago)

Reply 5

Brainfrozen

OP

1

Original post by RDKGames

You're correct up to the third line. The problem with the fourth is that you multiply both sides with 4 but the 24 is not affected.

\displaystyle 24+4(n-1)= 4(\frac{24}{4})+4(n-1)=4(\frac{24}{4}+n-1)=4(6+n-1)=4(n+5)

and this would be the denominator. And NOW you can multiply freely by 4 on both sides.

Though of course, your method is way too long winded.

Thanks for that RDK, but I'm not moving forward with this, I understand what you mean, but I'm struggling to understand how to isolate n when there's two n terms and one is squared. Factorising doesn't seem to help the issue.

Could you advise me on the methodology to solve for n? I'd rather understand the thinking behind the method rather than see how to solve it if you know what I mean.

(edited 7 years ago)

Original post by Brainfrozen

Thanks for that RDK, but I'm not moving forward with this, I understand what you mean, but I'm struggling to understand how to isolate n when there's two n terms and one is squared. Factorising doesn't seem to help the issue.

You can factorise it since you're solving it. Otherwise, complete the square and use the

\pm

sign to display the two solutions.

Reply 7

Brainfrozen

OP

1

I don't understand this at all I'm afraid.

Original post by Brainfrozen

I don't understand this at all I'm afraid.

You need to get a quadratic in the form

an^2+bn+c=0

from what you are given, and solve it for n. You don't really make it clear on what you need to do with it (considering you mentioned transformations, unless you meant algebraic manipulations instead).

408=\frac{n}{2}[2\cdot 12+4(n-1)]

\Rightarrow 816=n[24+4n-4]

\Rightarrow 816=n(4n+20)

\Rightarrow 816=4n^2+20n

\Rightarrow 0=4n^2+20n-816

\Rightarrow n^2+5n-204=0

Then complete the square on it, or factorise it.

(edited 7 years ago)

Reply 9

Brainfrozen

OP

1

Don't know mate, I just need to solve for n, it should equal 12 as u know.

I have gaps in my understanding, can understand medium difficulty but tend to get tripped up on silly things.

Eg, polynomial division, once I went back and remembered how to do long division it was easy.

Original post by Brainfrozen

Don't know mate, I just need to solve for n, it should equal 12 as u know.

I have gaps in my understanding, can understand medium difficulty but tend to get tripped up on silly things.

Eg, polynomial division, once I went back and remembered how to do long division it was easy.