Solve for $x$ in $(a+b-x)/c+(a+c-x)/b+(c+b-x)/a+(4x)/(a+b+c)=1$ ?

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Answer 1 · 2018-07-20T03:41:57

$(a+b-x)/c+(a+c-x)/b+(c+b-x)/a+(4x)/(a+b+c)=1$

$=>(a+b-x)/c+1+(a+c-x)/b+1+(c+b-x)/a+1+(4x)/(a+b+c)-3-1=0$
$=>(a+b+c-x)/c+(a+c+b-x)/b+(c+b+a-x)/a-4(1-x/(a+b+c))=0$

$=>(a+b+c-x)(1/c+1/b+1/a)-4((a+b+c-x)/(a+b+c))=0$

$=>(a+b+c-x)(1/c+1/b+1/a-4/(a+b+c))=0$

So

$=>(a+b+c-x)=0$

For $(1/c+1/b+1/a-4/(a+b+c))!=0$

Hence $x=a+b+c$

Solve for xx in (a+b-x)/c+(a+c-x)/b+(c+b-x)/a+(4x)/(a+b+c)=1(a+b-x)/c+(a+c-x)/b+(c+b-x)/a+(4x)/(a+b+c)=1?