Tough equations no.1

$\frac{x+1}{x+2}+\frac{x+8}{x+9}=\frac{x+2}{x+3}+\frac{x+7}{x+8}$

or, $1-\frac{x+1}{x+2}+1-\frac{x+8}{x+9}=1-\frac{x+2}{x+3}+1-\frac{x+7}{x+8}$

or, $\frac{1}{x+2}+\frac{1}{x+9}=\frac{1}{x+3}+\frac{1}{x+8}$

or, $\frac{1}{x+2}-\frac{1}{x+3}=\frac{1}{x+8}-\frac{1}{x+9}$

or, $\frac{1}{(x+2)(x+3)}=\frac{1}{(x+8)(x+9)}$

or, $x^2+5x+6=x^2+17x+72$

or, $12x=-66$ or, $x=\boxed{-5.5}$

$\displaystyle \frac {x+1}{x+2}+\frac {x+8}{x+9} = \frac {x+2}{x+3} +\frac {x+7}{x+8}$

$\displaystyle \Rightarrow \frac {x+2-1}{x+2}+\frac {x+9-1}{x+9} = \frac {x+3-1}{x+3} +\frac {x+8-1}{x+8}$

$\displaystyle \Rightarrow \left( 1-\frac {1}{x+2} \right) + \left( 1-\frac {1}{x+9} \right) = \left( 1-\frac {1}{x+3} \right) + \left( 1-\frac {1}{x+8} \right)$

$\displaystyle \Rightarrow \frac {1}{x+2} + \frac {1}{x+9} = \frac {1}{x+3} + \frac {1}{x+8}$

$\displaystyle \Rightarrow \frac {1}{x+2} - \frac {1}{x+3} = \frac {1}{x+8} - \frac {1}{x+9}$

$\displaystyle \Rightarrow \frac {x+3-x-2} {x^2+5x+6} = \frac {x+9-x-8} {x^2+17x+72}$

$\displaystyle \Rightarrow x^2+5x+6 = x^2+17x+72 \quad \Rightarrow 12x = -66 \quad \Rightarrow x = \boxed{5.5}$

Sujoy Roy's method is the best. However, the method below may help in other problems. Hence I give them.
$\dfrac{X + 8}{X+ 9} -\dfrac{X + 7}{X+ 8} = \dfrac{X +2}{X+ 3} -\dfrac{X + 1}{X+ 2} \\ \therefore~\dfrac{(X + 8)^2 - (X + 9)(X + 7)}{(X+ 9)(X + 8)} =\dfrac{(X + 2)^2 - (X + 3)(X + 1)}{(X+ 9)(X + 8)} \\\implies \dfrac{1}{X^2+ 17X + 72)} = \dfrac{1}{X^2+ 5X + 6} \\ Equating ~the ~reciprocals~ and ~simplifying, we~ get~X=\boxed{- 5.5} \\ OR\\$
$\dfrac{X + 1}{X+ 2} +\dfrac{X + 8}{X+ 9} = \dfrac{X +2}{X+ 3} +\dfrac{X + 7}{X+ 8} \\ \therefore \dfrac{X^2+ 10X + 9~+~X^2+ 10X + 16}{X^2+11X +18} = \dfrac{X^2+10X +16~+~X^2+10X+ 21}{X^2+ 11X + 24} \\$
$\implies \dfrac{2(X^2+ 10X + \dfrac{25}{2})}{X^2+11X +18} = \dfrac{2(X^2+10X + \dfrac{37}{2})}{X^2+ 11X + 24}\\ \implies \dfrac{2(X^2+ 10X + \dfrac{25}{2})}{2(X^2+10X + \dfrac{37}{2})} = \dfrac{X^2+11X +18}{X^2+ 11X + 24} \\ \implies 1~~- \dfrac{ \dfrac{12}{2}}{X^2+10X + \dfrac{37}{2}} = 1~~- \dfrac{6}{X^2+ 11X + 24} \\ Equating ~the ~reciprocals~ and ~simplifying, we~ get~X=\boxed{- 5.5}$

The equation can be put in the form

1 - 1/( x + 2 ) + 1 - 1/( x + 9) = 1 - 1/( x + 3 ) + 1 - 1/( x + 8 )

1/( x + 2 ) + 1/( x + 9) = 1/( x + 3 ) + 1/( x + 8 )

( 2 x + 11 )/( x + 2 )( x + 9 ) = (2 x + 11 )/( x+ 3 )( x + 8 )

Since ( x + 2 )( x + 9 ) not equal ( x+ 3 )( x + 8 )

Then

2 x + 11 = 0

x = - 5.5

1-1/(x+2)+1-1/(x+9) = 1-1/(x+3)+1-1/(x+8),

1/(x+2)+1/(x+9) = 1/(x+3)+1/(x+8),

(2x+11)/(x^2+11x+18) = (2x+11)/(x^2+11x+24),

(2x+11)[(x^2+11x+24)-(x^2+11x+18)] = 0,

x = -11/2.

Let x+5 = a a-4/a-3 + a+3/a+4 = a-3/a-2 + a+2/a+3 (a^2 - 16 + a^2 -9) / (a-3)(a+4) = (a^2 - 9 + a^2 -4) / (a-2)(a+3) ( 2 * a^2 - 25 ) / (a-3)(a+4) = ( 2 * a^2 - 13) / (a-2)(a+3) after cross multiplication and expansion of brackets , ( 2 * a^2 - 13 -12) / ( 2 * a^2 - 13) = a^2 + a - 12 / a^2 + a - 6 1 - [12 / ( 2 * a^2 - 13)] = 1 - [6 / a^2 + a - 6] -12 / ( 2 * a^2 - 13) = - 6 / a^2 + a - 6 -12 / ( 2 * a^2 - 13) = -12 / 2( a^2 + a - 6) 2 * a^2 - 13 = 2( a^2 + a - 6) 2 * a^2 - 13 = 2*a^2 + 2a - 12 -13 = 2a -12 a = -1/2 = 0.5 x = -5+a x = -5 - 0.5 = -5.5