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subtract (3/9x^2-5/6x-14) from (7/9x^2-5/6x+1)

 Aug 1, 2016
 #1
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Simplify the following:
-((3 x^2)/9-5/6 x-14)+7/9 x^2-5/6 x+1

 

3/9 = 3/(3×3) = 1/3:
-(x^2/3-5/6 x-14)+7/9 x^2-5/6 x+1

 

Put each term in x^2/3-5/6 x-14 over the common denominator 6: x^2/3-5/6 x-14 = (2 x^2)/6-(5 x)/6-84/6:
-(2 x^2)/6-(5 x)/6-84/6+7/9 x^2-5/6 x+1

 

(2 x^2)/6-(5 x)/6-84/6 = (2 x^2-5 x-84)/6:
(7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6

 

Put each term in (7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6 over the common denominator 18: (7 x^2)/(9)-(5 x)/(6)+1-(2 x^2-5 x-84)/6 = (14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18:
(14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18

 

(14 x^2)/18-(15 x)/18+18/18+(3 (-2 x^2+5 x+84))/18 = (14 x^2-15 x+18+3 (-2 x^2+5 x+84))/18:
(14 x^2-15 x+18+3 (-2 x^2+5 x+84))/18

 

3 (-2 x^2+5 x+84) = -6 x^2+15 x+252:
(-6 x^2+15 x+252+14 x^2-15 x+18)/18

 

Grouping like terms, 14 x^2-6 x^2+15 x-15 x+18+252 = (14 x^2-6 x^2)+(-15 x+15 x)+(18+252):
((14 x^2-6 x^2)+(-15 x+15 x)+(18+252))/18

 

14 x^2-6 x^2 = 8 x^2:
(8 x^2+(-15 x+15 x)+(18+252))/18

 

18+252 = 270:
(8 x^2+(-15 x+15 x)+270)/18

 

15 x-15 x = 0:
(8 x^2+270)/18

 

Factor 2 out of 8 x^2+270:
2 (4 x^2+135)/18

 

2/18 = 2/(2×9) = 1/9:
Answer: |(4x^2+135) / 9

 Aug 1, 2016
 #2
avatar+14986 
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Hello Guest!

 

\(subtract (3/9x^2-5/6x-14) from (7/9x^2-5/6x+1)\)

 

\({(\frac{7}{9x^{2} }- \frac{5}{6x}+ 1 )} -{(\frac{3}{9x^{2} }- \frac{5}{6x} - 14) }\)

 

 

\(= \frac{7- 3}{9x^{2} }+ 1+ 14 = \frac{4}{9x^{2} }+15 = (\frac{2}{3x}) ^{2} + 15 \)

 

 

Greeting asinus :- ) laugh !

 Aug 1, 2016
 #3
avatar+9665 
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\(\color{aqua}\left(\dfrac{3}{9x^2}-\dfrac{5}{6x}-14\right)-\left(\dfrac{7}{9x^2}-\dfrac{5}{6x}+1\right)\)

\(\color{aqua}=\dfrac{3}{9x^2}-\dfrac{5}{6x}-14 - \dfrac{7}{9x^2}+\dfrac{5}{6x}-1\\ \color{aqua}=\dfrac{3}{9x^2}-\dfrac{7}{9x^2}-14-1\\\color{aqua} =-\dfrac{4}{9x^2}-15\)

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 Aug 3, 2016
edited by MaxWong  Aug 3, 2016

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