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 May 15th, 2015, 03:46 AM #1 Member   Joined: Jan 2015 From: Norway Posts: 95 Thanks: 1 Pull the letters and numbers together Task 1.28 D) (r +2/ r+3)- (r+2/2r+1)= r^2 - 4/2r^2 + 7r + 3 I understand where all the letters and number come from, expect for 7r. Why is it 7r instead 1r+1= 1r + r = 2r ? Except for that i have done the rest of the task correct, please help me Edit : Here is some tasks i don't understand and solved wrong b) (a/x)+ (4a + 5/ 3x) - (1-2a/2x) - (7/6x) = 10a/3x C)(6x + 7y +8/2x + y) - (2x + 26y + 16/4x + 2y) + 6 Task 1.30 b) (a/a^2-b^2) +(a+b)/a-b)^2 = 2a^2 + ab + b^2/(a+b)(a-b)^2 The only thing i don't understand with this task is where 2a comes from, the rest i understand. Last edited by Zman15; May 15th, 2015 at 04:28 AM. May 15th, 2015, 04:57 AM   #2
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 Originally Posted by Zman15 Task 1.28 D) (r +2/ r+3)- (r+2/2r+1)= r^2 - 4/2r^2 + 7r + 3 I understand where all the letters and number come from, expect for 7r. Why is it 7r instead 1r+1= 1r + r = 2r ? Except for that, I have done the rest of the task correct, please help me Edit : Here is some tasks I don't understand and solved wrong b) (a/x)+ (4a + 5/ 3x) - (1-2a/2x) - (7/6x) = 10a/3x C) (6x + 7y +8/2x + y) - (2x + 26y + 16/4x + 2y) + 6 Task 1.30 b) (a/a^2-b^2) +(a+b)/a-b)^2 = 2a^2 + ab + b^2/(a+b)(a-b)^2 The only thing I don't understand with this task is where 2a comes from, the rest I understand.
D) (x^2 - 121/x^2-144) * (3x+36/2x+222) = 3(x-11)(3(x-12)
I don't understand anything of this task. Should I divide/subtract, what? Help me, please.

Last edited by skipjack; May 17th, 2015 at 02:32 AM. May 15th, 2015, 06:37 PM   #3
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Quote:
 Originally Posted by Zman15 Task 1.28 D) (r +2/ r+3)- (r+2/2r+1)= r^2 - 4/2r^2 + 7r + 3 I understand where all the letters and number come from, expect for 7r. Why is it 7r instead 1r+1= 1r + r = 2r ?
(r + 3)(2r + 1)
= r(2r) + r(1) + 3(2r) + 3(1)
= 2r^2 + r + 6r + 3
= 2r^2 + 7r + 3 May 15th, 2015, 06:58 PM   #4
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Quote:
 Originally Posted by Zman15 D) (x^2 - 121/x^2-144) * (3x+36/2x+222) = 3(x-11)(3(x-12) I don't understand anything of this task. Should I divide/subtract, what? Help me, please.
Frankly, what you wrote is simply wrong.

What you wrote:
$\displaystyle \left ( x^2 - \frac{121}{x^2} - 144 \right ) * \left ( 3x + \frac{36}{2x} + 222 \right ) = 3(x - 11) (3(x - 12))$

What you meant:
$\displaystyle \left ( \frac{x^2 - 121}{x^2 - 144} \right ) * \left ( \frac{3x + 36}{2x + 222} \right ) = 3(x - 11) (3(x - 12))$

And the LHS and RHS aren't equal even after it is written correctly. Is there a typo?

The RHS of your original question with the r's was even worse. Please use parentheses!

-Dan

Last edited by skipjack; May 17th, 2015 at 02:30 AM. May 16th, 2015, 12:03 AM   #5
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 Originally Posted by Zman15 b) (a/x)+ (4a + 5/ 3x) - (1-2a/2x) - (7/6x) = 10a/3x With minimal attention to writing, these things can be:

$\displaystyle \color{red}{a/x+(4a+5)/(3x)-(1-2a)/(2x)-7/(6x)}$

It can be even more clearly:

$\displaystyle \color {red}{\dfrac{a}{x}+\dfrac{4a+5}{3x}-\dfrac{1-2a}{2x}-\dfrac{7}{6x}}$

This calculation is very simple:

1) The common denominator is 6x.

2) Attention to the minus (in the penultimate fraction).

Last edited by skipjack; May 17th, 2015 at 02:54 AM. May 16th, 2015, 02:54 AM   #6
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 Originally Posted by topsquark Frankly what you wrote is simply wrong. What you wrote: $\displaystyle \left ( x^2 - \frac{121}{x^2} - 144 \right ) * \left ( 3x + \frac{36}{2x} + 222 \right ) = 3(x - 11) (3(x - 12))$ What you meant: $\displaystyle \left ( \frac{x^2 - 121}{x^2 - 144} \right ) * \left ( \frac{3x + 36}{2x + 222} \right ) = 3(x - 11) (3(x - 12))$ And the LHS and RHS aren't equal even after it is written correctly. Is there a typo? The RHS of your original question with the r's was even worse. Please use parenthesis! -Dan
$\displaystyle 3(x - 11) (2(x - 12))$is what I meant.

Last edited by skipjack; May 17th, 2015 at 02:40 AM. May 16th, 2015, 03:46 AM   #7
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 Originally Posted by aurel5  With minimal attention to writing, these things can be: $\displaystyle \color{red}{a/x+(4a+5)/(3x)-(1-2a)/(2x)-7/(6x)}$ It can be even more clearly: $\displaystyle \color {red}{\dfrac{a}{x}+\dfrac{4a+5}{3x}-\dfrac{1-2a}{2x}-\dfrac{7}{6x}}$ This calculation is very simple: 1) The common denominator is 6x. 2) Attention to the minus (in the penultimate fraction).
You are right, it was simple. I got 20/6 as a answer and then I divided it with 2 and got 10/3x. Thanks for the help. Quote:
 Originally Posted by Monox D. I-Fly (r + 3)(2r + 1) = r(2r) + r(1) + 3(2r) + 3(1) = 2r^2 + r + 6r + 3 = 2r^2 + 7r + 3
Thanks, that wasn't hard Last edited by skipjack; May 17th, 2015 at 02:53 AM. May 16th, 2015, 05:44 AM   #8
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Quote:
 Originally Posted by Zman15 You are right, it was simple. I got 20/6 as a answer and then I divided it with 2 and got 10/3x. Thanks for the help. Actually, no. What you "got as an answer" was 20/(6x) and canceling 2 in numerator and denominator you get 10/(3x)

Had you actually "got 20/6 as a answer and then divided it with 2" you would have got 20/12= 5/3.

You are still being very careless in the way you write numbers and expressions.
Quote:
 Thanks, that wasn't hard. Last edited by skipjack; May 17th, 2015 at 02:42 AM. May 16th, 2015, 07:36 AM   #9
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 Originally Posted by Country Boy Actually, no. What you "got as an answer" was 20/(6x) and canceling 2 in numerator and denominator you get 10/(3x) Had you actually "got 20/6 as a answer and then divided it with 2" you would have got 20/12= 5/3. You are still being very careless in the way you write numbers and expressions.
I did write 6x on paper. And what you said is what I meant. I am just bad with English grammar, so I can't explain myself properly. Sorry. :P

Last edited by skipjack; May 17th, 2015 at 02:43 AM. May 17th, 2015, 02:51 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 $\displaystyle \frac{x^2 - 121}{x^2 - 144} \cdot \frac{3x + 36}{2x + 22} = \frac{(x - 11)(x + 11)}{(x - 12)(x + 12)} \cdot \frac{3(x + 12)}{2(x + 11)} = \frac{3(x - 11)}{2(x - 12)}$ Tags letters, numbers, pull Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bethewind Probability and Statistics 1 March 6th, 2015 03:19 PM xichyu Differential Equations 0 February 2nd, 2015 04:40 PM AzraaBux Calculus 4 July 19th, 2013 06:59 AM

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