Find x x in terms of a a , b b and c c

Algebra Level pending

Real numbers x x , y y , z z , a a , b b and c c are such that

y z + z x + x y = a 2 x 2 = b 2 y 2 = c 2 z 2 \large yz+zx+xy=a^2 - x^2=b^2 - y^2=c^2 - z^2

Solve for x x in terms of a a , b b and c c .

x = a 2 b 2 + c 2 a 2 c 2 b 2 2 a b c x=\frac {a^2b^2+c^2a^2-c^2b^2}{2abc} x = a + b c 2 x= \frac {a+b-c}2 x = a b c x=abc x = a 2 + b 2 c 2 2 a b c x= \frac {a^2+b^2-c^2}{2 abc}

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1 solution

Debanik Samanta
Apr 12, 2017

z(y+x) + xy=a^2 - x^2
=>a^2=z(x+y) + x^2 + xy
=>a^2=(z+x)(x+y)
Similarly,
b^2=(y+z)(x+y) ,and
c^2=(x+z)(y+z)
So, c^2=(a^2 b^2) / (x+y)^2
=>c=(ab) / (x+y) [LEAVING NEGATIVE VALUE]
=>x+y=(ab) / c...........(2)
a^2 - x^2=b^2 - y^2
=>x^2 - y^2=a^2 - b^2
=>(x+y)(x-y)=a^2 - b^2
=>(ab)(x-y)/c=a^2 - b^2
=>x-y=c(a^2 - b^2) / (ab).........(1)
Now, adding (1) and (2)
x +y=ab/c
+ x - y=c(a^2 - b^2) / (ab)
=>2x=(ab) / (c) + c(a^2 - b^2) / (ab)
=>x=(a^2
b^2 + c^2 a^2 - c^2 b^2) / (2 abc)

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