My Math Forum Question about a rational equation

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 6th, 2016, 11:44 PM #1 Newbie   Joined: Sep 2016 From: alaska Posts: 16 Thanks: 0 Question about a rational equation I don't even know where to start. I multiplied everything by the LCM and subtracted hoping that it would get me somewhere, but that's about as far as I got. My program is asking for an exact number (no fractions). How would I go about solving for x? ( x+8 )/( 9 ) − ( x−1 )/( 2 ) = 8 Last edited by MBI; September 6th, 2016 at 11:55 PM.
 September 7th, 2016, 12:21 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics \displaystyle \begin{align*} \frac{x+8}{9} - \frac{x-1}{2} &= 8\\ 2(x+8) - 9(x-1) &= 18 \times 8\\ 2x + 16 - 9x + 9 &= 144\\ 119 &= -7x\\ x &= -17 \end{align*} Thanks from MBI
September 7th, 2016, 12:42 AM   #3
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 Originally Posted by 123qwerty \displaystyle \begin{align*} \frac{x+8}{9} - \frac{x-1}{2} &= 8\\ 2(x+8) - 9(x-1) &= 18 \times 8\\ 2x + 16 - 9x + 9 &= 144\\ 119 &= -7x\\ x &= -17 \end{align*}
I wasn't aware that you should find a LCM for the second part of the equation as well. That's pretty cool thank you.

Last edited by skipjack; October 22nd, 2016 at 03:24 AM.

September 7th, 2016, 01:56 AM   #4
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Quote:
 Originally Posted by MBI I wasn't aware that you should find a LCM for the second part of the equation as well. That's pretty cool thank you.
You don't really have to (you can still get the same answer for x without that manipulation).
But if you want to get rid of the common denominator in the left-hand side, you indeed then have to multiply the right-hand side by the fraction $\displaystyle \frac{\text{denominator}}{\text{denominator}}$ (because by definition, anything divided by itself = 1), so this is just like multiplying the right hand side by 1, which doesn't change its result.

After doing so, every term in your equation will have the same denominator, and this allows you to simply get rid of it.

Hoping this isn't too confusing.

Last edited by skipjack; October 22nd, 2016 at 03:36 AM.

September 7th, 2016, 03:52 AM   #5
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 Originally Posted by sKebess You don't really have to (you can still get the same answer for x without that manipulation). But if you want to get rid of the common denominator in the left hand side, you indeed then have to multiply the right hand side by the fraction $\displaystyle \frac{\text{denominator}}{\text{denominator}}$ (because by definition, anything divided by itself = 1), so this is just like multiplying the right hand side by 1, which doesn't change its result. After doing so, every term in your equation will have the same denominator, and this allows you to simply get rid of it. Hoping this isn't too confusing.
Ah, makes perfect sense! Thank you very much.

Last edited by skipjack; October 22nd, 2016 at 03:36 AM.

September 7th, 2016, 06:31 AM   #6
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Quote:
 Originally Posted by 123qwerty $\displaystyle \frac{x+8}{9} - \frac{x-1}{2} = 8\\$ $\displaystyle 2(x+8) - 9(x-1) = 18 \times 8$

Note: To be good form, the right side of the equation of the second equation as shown above should not be used for two reasons:

1) The "X" looks similar to the x variable, and this is an algebra problem, whereas it would be expected to see that sometimes in an arithmetic problem.

2) It is inconsistent with the rest of the equation. Parentheses should have been continued across for the equation to show multiplication.

$\displaystyle \frac{x+8}{9} - \frac{x-1}{2} = 8\\$

$\displaystyle 2(x+8) - 9(x-1) = 18(8)$

 October 21st, 2016, 11:34 PM #7 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $\dfrac{x+8}{9}-\dfrac{x-1}{2}=8$ $\dfrac{2(x+8 )-9(x-1)}{18}=8$ $2(x+8 )-9(x-1)=8\times 18$ $2x+16−9x+9=144$ $2x-9x=144-16-9$ $-7x=119$ $x=-17$ Last edited by deesuwalka; October 21st, 2016 at 11:44 PM.
October 22nd, 2016, 08:21 AM   #8
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Quote:
 Originally Posted by deesuwalka $\dfrac{2(x+8 )-9(x-1)}{18}=8$ $2(x+8 )-9(x-1)=8\times 18$

Note: To be good form, the right side of the equation of the second equation as shown above should not be used for two reasons:

1) The "X" looks similar to the x variable, and this is an algebra problem, whereas it would be expected to see that sometimes in an arithmetic problem.

2) It is inconsistent with the rest of the equation. Parentheses should have been continued across for the equation to show multiplication.

$\dfrac{2(x+8) \ - \ 9(x-1)}{18} \ = \ 8$

$2(x+8) \ - \ 9(x-1) \ = \ 8(18)$

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