My Math Forum Pull the letters and numbers together

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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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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 ? 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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Quote:
 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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Quote:
 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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Quote:
 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 $\displaystyle \frac{x^2 - 121}{x^2 - 144} \cdot \frac{3x + 36}{2x + 22} = \frac{(x - 11)(x + 11)}{(x - 12)(x + 12)} \cdot \frac{3(x + 12)}{2(x + 11)} = \frac{3(x - 11)}{2(x - 12)}$

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