Question

x-a-b/c+x-b-c/a+x-c-a/b=3

Answer 1

Heya User,

We bring all the one's at the R.H.S. to have :->

[tex] \frac{x-a-b}{c} + \frac{x-a-c}{b} + \frac{x-b-c}{a} = 3 \\ \\ \frac{x-a-b}{c} - 1 + \frac{x-a-c}{b} -1 + \frac{x-b-c}{a} -1 = 0 \\ \\ \frac{x-a-b -c}{c} + \frac{x-a-b -c}{b} + \frac{x-a-b -c}{a} = 0[/tex]

Now, this implies that :->

[tex] [ x - a - b - c ][ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ] = 0 \\ \\ Implies \ \ x = a + b + c ; [/tex]