Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 91399 by jagoll last updated on 30/Apr/20

Commented by Prithwish Sen 1 last updated on 30/Apr/20

$∵ (x + 8) + (x + 7) = (x + 9) + (x + 6) = 2 x + 15$ $∴ 2 x + 15 = 0$ $\Rightarrow x = - \frac{15}{2}$
Commented by jagoll last updated on 30/Apr/20

$h o w y o u p r o v e t h a t x + 8 + x + 9 = 0 ?$
Commented by $@ty@m123 last updated on 30/Apr/20

$C u t e S o l u t i o n .$ $C o u l d u p l . e x p l a i n t h e t r i c k ?$
Commented by Prithwish Sen 1 last updated on 30/Apr/20

$it is vedic maths .$
Commented by jagoll last updated on 30/Apr/20

$v e d i c m a t h ?$
Commented by Prithwish Sen 1 last updated on 30/Apr/20

$\frac{1}{x + 8} + \frac{1}{x + 7} = \frac{1}{x + 9} + \frac{1}{x + 6}$
Commented by jagoll last updated on 30/Apr/20

$w h a t i s v e d i c m a t h ?$
Commented by Prithwish Sen 1 last updated on 30/Apr/20

$(2 x + 15) [\frac{1}{(x + 8) (x + 7)} - \frac{1}{(x + 9) (x + 6)}] = 0$ $∵ 8 + 7 = 9 + 6 and the term x^{2} vanishes$ $then 2 x + 15 = 0$
Commented by jagoll last updated on 30/Apr/20

$i f \frac{x + 3}{(x + 6)} + \frac{x + 4}{x + 9} = \frac{x + 6}{x + 7} + \frac{x + 1}{x + 8}$ $\frac{1}{x + 6} + \frac{1}{x + 9} = \frac{1}{x + 7} + \frac{1}{x + 8} ?$
Commented by Prithwish Sen 1 last updated on 30/Apr/20

Commented by Prithwish Sen 1 last updated on 30/Apr/20

Commented by jagoll last updated on 30/Apr/20

$w a w . . g r e a t . . . t h a n k y o u$
Commented by $@ty@m123 last updated on 30/Apr/20

$T h a n k s$
	Answered by jagoll last updated on 30/Apr/20

	Answered by Rasheed.Sindhi last updated on 30/Apr/20
	$((x+7)/(x+8))−1+((x+6)/(x+7))−1=((x+8)/(x+9))−1+((x+5)/(x+6))−1 ((−1)/(x+8))+((−1)/(x+7))=((−1)/(x+9))+((−1)/(x+6)) (1/(x+8))+(1/(x+7))=(1/(x+9))+(1/(x+6)) ((2x+15)/((x+8)(x+7)))−((2x+15)/((x+9)(x+6)))=0 (2x+15){(1/((x+8)(x+7)))−(1/((x+9)(x+6)))}=0 •2x+15=0⇒x=−15/2 •(x^2 +15x+54)−(x^2 +15x+56)=0 ⇒ No root$
	$\frac{x + 7}{x + 8} - 1 + \frac{x + 6}{x + 7} - 1 = \frac{x + 8}{x + 9} - 1 + \frac{x + 5}{x + 6} - 1$ $\frac{- 1}{x + 8} + \frac{- 1}{x + 7} = \frac{- 1}{x + 9} + \frac{- 1}{x + 6}$ $\frac{1}{x + 8} + \frac{1}{x + 7} = \frac{1}{x + 9} + \frac{1}{x + 6}$ $\frac{2 x + 15}{(x + 8) (x + 7)} - \frac{2 x + 15}{(x + 9) (x + 6)} = 0$ $(2 x + 15) {\frac{1}{(x + 8) (x + 7)} - \frac{1}{(x + 9) (x + 6)}} = 0$ $∙ 2 x + 15 = 0 \Rightarrow x = - 15 / 2$ $∙ (x^{2} + 15 x + 54) - (x^{2} + 15 x + 56) = 0$ $\Rightarrow N o r o o t$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum