Simplifying a fraction (block reduction, control)

[Bringitondown]

Homework Statement

2*3*4*5/4+3*4*4*6

Homework Equations

2*3*4*5/4+3*4*4*6

The Attempt at a Solution

2*3*5/1+3*4*6

Knowledge gap 1 - I do not understand why if top and bottom are divided through by 4 there should be a 4 left in the denominator and why the rest of the numerators and denomenators are not divided through by 4.

I think I am just needing a prompt in the right direction here

Any help would be greatley apprecited.

I have attched photos to show where this comes from. Th blue pen working (337) is where I have changed the original problem (335) from letters to numbers to allow me to attempt to make more snse of it.

 

[Samy_A]

Knowledge gap 1 - I do not understand why if top and bottom are divided through by 4 there should be a 4 left in the denominator and why the rest of the numerators and denomenators are not divided through by 4

How many 4's are there in 3*4*4*6?

 

[Bringitondown]

How many 4's are there in 3*4*4*6?

2 obviously

 

[Bringitondown]

How many 4's are there in 3*4*4*6?

I was thinking why two of the 4's come out on the bottom line and only one on the top. Is this because the two 4's on the bottom line are from different expresions?

 

[BvU]

The 30/72 on the lower right is not correct. if you strike 4 it means you divide by four. The quotient is 1. So you get 30/(1+72) = 120/292 and all is right.

 

[Samy_A]

2 obviously

So when you simplify your fraction by removing one 4 in each term, why should the second 4 in 3*4*4*6 just disappear?

 

[Bringitondown]

The 30/72 on the lower right is not correct. if you strike 4 it means you divide by four. The quotient is 1. So you get 30/(1+72) = 120/292 and all is right.

yes I have since noticed this, thanks. Appreciated

 

[BvU]

Let $$A = {G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3}$$ and we want to look at $$ {A\over 1 + {H_2\over G_4}A}\ ,$$ right ? So we have $$
{{ G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3} \over 1 + { G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3}\ { H_2\over G_4}} \quad
= \quad {{ G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3} \over 1 + { G_2\,G_3\,G_4\, H_2 \over G_4+ G_3\,G_4\,G_4\,H_3}} \quad \equiv B
$$ the block where your pen is pointing at (I call it $B$ for now). Then the dividing out of $G_4$:

$$ { G_2\,G_3\,G_4\, H_2 \over G_4+ G_3\,G_4\,G_4\,H_3}\quad = \quad { G_2\,G_3\,G_4\, H_2 \over G_4 ( 1 + G_3\,G_4\,H_3) } \quad = \quad
{G_4\over G_4}\, { G_2\,G_3\, H_2 \over 1 + G_3\,G_4\,H_3 }
\quad = \quad
{ G_2\,G_3\, H_2 \over 1 + G_3\,G_4\,H_3 }
$$
Then he multiplies B with 1 : $$ B\ { 1 + G_3\,G_4\,H_3 \over 1 + G_3\,G_4\,H_3 } \quad = \quad
{G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3 + G_2\,G_3\, H_2 } $$

and this times $G_1$ yields the next rectangular block.

etc.
--

 

[Bringitondown]

Crystal clear, thanks very much BvU.

 

[Bringitondown]

--[/QUOTE]
Please if you have the time BvU can you help me fill in the step in the attached.
Sorry that should also read C(S)/R(S) =
Thanks

 

[Mark44]

2*3*4*5/4+3*4*4*6

The above does not agree with what you wrote in the image you attached. Because of the higher precedence of division over addition, the above is the same as $2\cdot 3 \cdot 4 \cdot \frac 5 4 + 3 \cdot 4 \cdot 4 \cdot 6$

To match what you wrote in the image, you need parentheses around the entire denominator, like so:
2 * 3 * 4 * 5 / (4 + 3 * 4 * 4* 6)

 

[Bringitondown]

The above does not agree with what you wrote in the image you attached. Because of the higher precedence of division over addition, the above is the same as $2\cdot 3 \cdot 4 \cdot \frac 5 4 + 3 \cdot 4 \cdot 4 \cdot 6$

To match what you wrote in the image, you need parentheses around the entire denominator, like so:
2 * 3 * 4 * 5 / (4 + 3 * 4 * 4* 6)

Ok fair enough.

 

[BvU]

the step in the attached picture C(S)/R(S) =

Sure: multiplying with 1 is allowed, so here goes:$$
{C\over R} =
{ I\, S^2 \over I\, S^2\, } \ { {K_1\,K_2 \over I\, S^2} \over
1 + {K_1\,K_2\,K_3 \over I \, S } + {K_1\,K_2\over I\, S^2} } =
{ { I\, S^2 }\ {K_1\,K_2 \over I\, S^2}
\over {I\, S^2\, } \left ( 1 + {K_1\,K_2\, K_3 \over I\, S} + {K_1\,K_2\over I\, S^2}\right ) } =
{K_1\,K_2\over I\, S^2 + {K_1\,K_2\,K_3\, S} + {K_1\,K_2} } $$

 

[Bringitondown]

Sure: multiplying with 1 is allowed, so here goes:$$
{C\over R} =
{ I\, S^2 \over I\, S^2\, } \ { {K_1\,K_2 \over I\, S^2} \over
1 + {K_1\,K_2\,K_3 \over I \, S } + {K_1\,K_2\over I\, S^2} } =
{ { I\, S^2 }\ {K_1\,K_2 \over I\, S^2}
\over {I\, S^2\, } \left ( 1 + {K_1\,K_2\, K_3 \over I\, S} + {K_1\,K_2\over I\, S^2}\right ) } =
{K_1\,K_2\over I\, S^2 + {K_1\,K_2\,K_3\, S} + {K_1\,K_2} } $$

Thats fantastic BvU, once again crystal clear,

 

[Bringitondown]

So when you simplify your fraction by removing one 4 in each term, why should the second 4 in 3*4*4*6 just disappear?

Just rusty on the syplifying. Thanks