Question

If, bc+1/a=ca+2/b=ab+7/c=1/(a+b+c). then, a+b+c=?

Answer 1

Given :   bc+1/a=ca+2/b=ab+7/c=1/(a+b+c)

To find : a+b+c=?

Solution:

bc+1/a=ca+2/b=ab+7/c=1/(a+b+c).  = k

=> bc+1/a= k

=> abc + 1  = ak   Eq1

 

ca+2/b = k

=> abc + 2 = bk    Eq2

ab+7/c=  k

=> abc + 7 = ck   Eq3

1/(a+b+c).  = k

=> (a + b + c)k = 1

Eq1 + Eq2 + Eq3

=>  abc + 1  + abc + 2 + abc + 7  = ak + bk + ck

=> 3abc  + 10 = (a + b + c)k

=> 3abc + 10 =  1

=> 3abc = -9

=> abc = -3

abc = -3

-3 + 1 = ak => ak = -2

-3 + 2 = bk => bk = -1

-3 + 7 = ck => ck = 4

multiplying

abck³ = 8

=> k³ = - 8/3

=> k = -2/∛3

a + b + c = 1/k

=> a + b + c = -∛3 / 2

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Answer 2

Given :$bc+\frac{1}{a}=ca+\frac{2}{b}=ab+\frac{7}{c}=\frac{1}{(a+b+c)}$

To find : a+b+c=?

Solution:

We are given that :

$bc+\frac{1}{a}=ca+\frac{2}{b}=ab+\frac{7}{c}=\frac{1}{(a+b+c)}$

Now we are supposed to find a+b+c

So,

$bc+\frac{1}{a}=ca+\frac{2}{b}=ab+\frac{7}{c}=\frac{1}{(a+b+c)}= k$

$\Rightarrow bc+\frac{1}{a}= k$

$\Rightarrow abc + 1= ak ------- 1$

$\Rightarrow ca+\frac{2}{b} = k$

$\Rightarrow abc + 2 = bk ----- 2$

$\Rightarrow ab+\frac{7}{c}=k$

$\Rightarrow abc + 7 = ck ------- 3$

$\Rightarrow \frac{1}{(a+b+c)}= k$

$\Rightarrow (a + b + c)k = 1 ----4$

Add 1 ,2 and 3

$\Rightarrow abc + 1 + abc + 2 + abc + 7= ak + bk + ck\\\Rightarrow 3abc+ 10 = (a + b + c)k$

Using 4

$\Rightarrow 3abc + 10 =1\\\Rightarrow 3abc = -9\\\Rightarrow abc = -3$

Substitute value in 1

$-3 + 1 = ak \\\Rightarrow ak = -2 ----5$

Substitute value in 2

$-3 + 2 = bk \\\Rightarrow bk = -1 ----6$

Substitute value in 3

$-3 + 7 = ck \\\Rightarrow ck = 4 ---7$

Now multiply 5,6 and 7

$abck^3 = 8\\\Rightarrow k^3= \frac{-8}{3}\\\Rightarrow k =\sqrt[3]{\frac{-8}{3}}=-\frac{2}{\sqrt[3]{3}}$

Now $a + b + c = \frac{1}{k}$

$\Rightarrow a + b + c =-\frac{\sqrt[3]{3}}{2}$

Hence $a + b + c =-\frac{\sqrt[3]{3}}{2}$