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

Hey there!

Given that, a, b, c are positive numbers.

Also, a+b+ab= 8,
b+c+bc=15 ,
c+a+ca = 35,

Often making the Values perfect square helps and we see that all the RHS are one less than a perfect square. So, We add 1 to all equations .


a + b + ab + 1 = 9
b + c + bc + 1 = 16
c + a + ac + 1 = 36

We can factorise the equations as follows,

a ( b + 1 ) + 1 ( b + 1 ) = 9
( a + 1 ) ( b + 1 ) = 9 ----> Equation 1

Other equations will get factorized as

(b+1) ( c + 1 ) = 16 --> Equation 2

(c + 1 )( a + 1 ) = 36 --> Equation 3

From the second equation,
( b + 1 ) ( c + 1 ) = 16

Factors of 16 are 1 , 16 , 2 , 4 , 8

If we take b + 1 = 1, c+1 = 16 other equations doesn't get satisfied. If you take b + 1 = 4 , c + 1 = 4 then also the other equations won't have possible solutions.

So, We are left with 2 , 8

( b + 1 )( c + 1 ) = 2 * 8
b + 1 = 2 , c + 1 = 8
b = 1 , c = 7

We haven't take b = 7 , c = 1 as they won't yield positive rational values of a .

Now, Substitute value of c in equation 3 or value of b in equation 1

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

a + 1 = 36/8

a + 1 = 9/2

a = 7/2 = 3.5

Now,

a + b + c + abc = 3.5 + 1 + 7 + 3.5 × 1 × 7 = 36

$\boxed {\boxed{ \textbf{Therefore, The required value of a + b + c + abc is 36 }}}$

Hope helped!