Question

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?

Answer 1

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

Answer 2

Question: A,b,c are positive numbers such that a+b+ab=8,b+c+bc=15 and c+a+ca=35 what is tha value of a+b+c+abc=?

Solution:

Acc to the question, there are 3 equations given:

a+b+ab= 8-----(i)

b+c+bc=15------(ii)

c+a+ca = 35---------(iii)

Now adding 1 to both the sides of equ (i) we get,

a+b+ab+1= 8+1

or, (a+1)(b+1)=9.

Similarly we get, (b+1)(c+1)=16 and (a+1)(c+1)=36.

now starting from the second equation,

We know that 16 has many factors but we have to brute force for satisfying the quation.

We know, 16 = 2 * 8 = (1+1) * (7+1) now, comparing it with equ(ii) we get, b=1,c=7.

Now putting the value of c in any two of the remaining equation we get,

(a+1)*(c+1)=36

or, (a+1)*(7+1)=36

or,(a+1) = 36/8 = 4.5

or, a = 4.5 - 1 = 3.5.

Now, a+b+c+abc = 3.5 + 1 + 7 + 3.5*1*7 = 36. [Ans]