Simplifying

[joshdelayslahsc]

Hi all,

I'm so confused on when and where to distribute stuff on algebra.

Example:

2(4x) = 8x; however, if I'm trying to simplify stuff and it looks like this: 3(x^2y^2)^3; it would be 3(x^6y^6) then you don't need to distribute it. The final would be 3x^6y^6.

My question is when should you distribute and when to just remove the brackets.

Cheers

[fun_jirachi]

Welcome to the forums!!!!
I reckon in the examples provided you've just distributed as you say because you haven't got a coefficient in the term in the brackets? I think its best to distribute all the time, as even when you just need to remove the brackets, you'd still get the same answer. Removing the brackets isn't that much faster, it's just a matter of seeing if there's a coefficient there or not.

That's my take, hope I helped!

[RuiAce]

Hi all,

I'm so confused on when and where to distribute stuff on algebra.

Example:

2(4x) = 8x; however, if I'm trying to simplify stuff and it looks like this: 3(x^2y^2)^3; it would be 3(x^6y^6) then you don't need to distribute it. The final would be 3x^6y^6.

My question is when should you distribute and when to just remove the brackets.

Cheers

The problem with the word distribute in this context is that it's basically too vague. Distribute can mean many things.

Your first example, \(2 (4x) = 8x\), is an example of the distributive law \(a(b+c) = ab + ac\). You let \(a = 2\), \(b = 4x\) and \(c = 0\), so that \( 2(4x + 0) = 8x + 0\). Of course, then +0's then disappear.

The next one features an index law. The index law relevant to this question is \( (ab)^m = a^m b^m\). It looks like "distributing", but it's not the same type of distribution, rather this is a theorem.
Of course, in your case, you have \( (x^2y^2)^3 \) = \( (x^2)^3 (y^2)^3 \).

(In particular, this becomes \(x^6y^6\) like you have stated. This is because you've now used yet another index law: \( (a^m)^n = a^{mn} \).)

That aside, just see the above post for a response to your main question

[joshdelayslahsc]

The problem with the word distribute in this context is that it's basically too vague. Distribute can mean many things.

Your first example, \(2 (4x) = 8x\), is an example of the distributive law \(a(b+c) = ab + ac\). You let \(a = 2\), \(b = 4x\) and \(c = 0\), so that \( 2(4x + 0) = 8x + 0\). Of course, then +0's then disappear.

The next one features an index law. The index law relevant to this question is \( (ab)^m = a^m b^m\). It looks like "distributing", but it's not the same type of distribution, rather this is a theorem.
Of course, in your case, you have \( (x^2y^2)^3 \) = \( (x^2)^3 (y^2)^3 \).

(In particular, this becomes \(x^6y^6\) like you have stated. This is because you've now used yet another index law: \( (a^m)^n = a^{mn} \).)

That aside, just see the above post for a response to your main question

Welcome to the forums!!!!
I reckon in the examples provided you've just distributed as you say because you haven't got a coefficient in the term in the brackets? I think its best to distribute all the time, as even when you just need to remove the brackets, you'd still get the same answer. Removing the brackets isn't that much faster, it's just a matter of seeing if there's a coefficient there or not.

That's my take, hope I helped!

Thank you so much. It makes sense now.

Also, I've got this question and I can't really figure out.

The question is: simplify, write the answers with positive indices.

3^2n*25^2n-1 / 15^n-1

I don't know how to do this.

[RuiAce]

Thank you so much. It makes sense now.

Also, I've got this question and I can't really figure out.

The question is: simplify, write the answers with positive indices.

3^2n*25^2n-1 / 15^n-1

I don't know how to do this.

It's very hard to interpret what's going on. When I read this I see \( 3^2 n \times 25^2 n-\frac{1}{15^n} - 1 \).

Please use brackets to indicate what goes with what

[joshdelayslahsc]

It's very hard to interpret what's going on. When I read this I see \( 3^2 n \times 25^2 n-\frac{1}{15^n} - 1 \).

Please use brackets to indicate what goes with what

it's 3^2n*25^2n-1 over 15^n-1

[S200]

it's 3^2n*25^2n-1 over 15^n-1

So...


That equation right?

[joshdelayslahsc]

So...


That equation right?

yes

[S200]

Ok, so you split the equation.



You can split the bottom term to make it \(3^{n-1} \times 5^{n-1}\)

SO that gives you two different bases to equate...
You should be OK form there, just ask again if you're not.

[Duarashid]

Hi, could you complete it? I just want to make sure I understand it properly. Thanks!

[RuiAce]

If that was what you meant, in the future you should type something like: (3^(2n) * 25^(2n-1)) / (15^(n-1))
Note the use of the brackets.
\begin{align*} \frac{3^{2n} \times 25^{2n-1}}{15^{n-1}} &= \frac{3^{2n} \times \left(5^2\right)^{2n-1}} {(3\times 5)^{n-1}} \\ &= \frac{3^{2n} \times 5^{2(2n-1)} } {3^{n-1} \times 5^{n-1}}\\ &= \frac{3^{2n}}{3^{n-1}} \times \frac{5^{4n-2}}{5^{n-1}}\\ &= 3^{2n - (n-1)} \times 5^{(4n-2) - (n-1)} \\ &= 3^{n+1}\times 5^{3n - 1}\end{align*}