Math, asked by sahanacmallikapeg4v6, 1 year ago

a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35 What is the value of a+b+c+abc? _________

Answers

Answered by rohitkumargupta
3

HELLO DEAR,



GIVEN:-



a + b + ab = 8



\Rightarrow b + a(1 + b) = 8



\bold{\Rightarrow a = \frac{8 - b}{1 + b}}-------( 1 )



b + c + bc = 15



\Rightarrow b + c(1 + b) = 15



\bold{\Rightarrow c = \frac{15 - b}{1 + b}} ------( 2 )



a + c + ac = 35



\bold{\Rightarrow \frac{8 - b}{1 + b} + \frac{15 - b}{1 + b} + \frac{8 - b}{1 + b}\times \frac{15 - b}{1 + b} = 35}



From----------( 1 ) &-----------( 2 )



\bold{\Rightarrow \frac{8 - b + 15 - b}{1 + b} + \frac{120 - 8b - 15b + b^2}{(1 + b)^2} = 35}



\Rightarrow \bold{(23 - 2b)(1 + b) + (120 - 23b + b^2) = 35(1 + b)^2}



\Rightarrow \bold{23 + 23b - 2b - 2b^2 + 120 - 23b + b^2 = 35 + 35b^2 + 70b}



\Rightarrow \bold{143 - 35 - 2b - b^2 = 35b^2 +  70b}



\Rightarrow \bold{108 - 72b - 36b^2 = 0}



\Rightarrow \bold{b^2 + 2b - 3 = 0}



\Rightarrow \bold{b^2 + 3b - b - 3 = 0}



\Rightarrow \bold{b(b + 3) - 1(b + 3)}



\Rightarrow \bold{(b + 3)(b - 1)}



so, b = 1 , b = -3[neglect]



because a , b , c ,are positive,



now,



from----------( 1 )



a = (8 - b)(1 + b)



\Rightarrow a = (8 - 1)(1 + 1)



\Rightarrow a = 7/2



from-----------( 2 )



c = (15 - b)(1 + b)



\Rightarrow c = (15 - 1)(1 + 1)



\Rightarrow c = 14/2



\Rightarrow c = 7



thus, the value of "a + b + c + abc" is



\Rightarrow 7/2 + 1 + 7 + (7 * 7/2 * 1)



\Rightarrow {(7 + 2 + 14 )/2 + 49/2}



\Rightarrow  (23/2 + 49/2)



\Rightarrow (23 + 49)/2



\Rightarrow 72/2



\Rightarrow 36



HENCE, the value of (a + b + c + abc) = 36



I HOPE ITS HELP YOU DEAR,



THANKS

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