Question

Will give brainliest!
We know that a + b + c = 7 and $\frac{1}{a+b} + \frac{1}{b+c} +\frac{1}{a+c} =\frac{7}{10}$. What is the value of $\frac{10a}{b+c} +\frac{10b}{a+c} +\frac{10c}{a+b}$?

Answer 1

Answer:

19

Step-by-step explanation:

Note that

$\dfrac{10a}{b+c} + \dfrac{10b}{a+c} + \dfrac{10c}{a+b} = 10\left(\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} \right)$

and

$\dfrac{7-b-c}{b+c}+\dfrac{7-a-c}{a+c}+\dfrac{7-a-b}{a+b}=\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b}$

from $a+b+c=7 \implies a=7-bc; b = 7-a-c; c = 7-a-b$

and

$\dfrac{7-b-c}{b+c}+\dfrac{7-a-c}{a+c}+\dfrac{7-a-b}{a+b} = 7\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)-3$

because it is not difficult to note that

$\sum \dfrac a{b+c}=\dfrac a{b+c}+\dfrac b{c+a}+\dfrac c{a+b}$

Therefore,

$7\left(\dfrac{7}{10}\right)-3 = \dfrac{49}{10}-3 =\boxed{ \dfrac{19}{10}} = \dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b}$

Finally,

$10\left(\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} \right) = 10 \cdot \dfrac{19}{10} = 19$