Math, asked by lakshyashukl, 11 months ago

if x^2+xy+xz= 141,y^2+yz+xy=363 and z^2+xz+yz=72,then find the value of (X+y-z).

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Answered by Anonymous

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$(x + y+ z) {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} + 2(xy+ yz + zx) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =( x {}^{2} + xy + xz) + (y {}^{2} + yz + xy) + (z {}^{2} + xz + yz) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 141 + 363 + 72 = 576 \\ (x + y + z) = \sqrt{576} \\ (x + y + z) = 24 \: \: \: (answer)$

Answered by Anonymous

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if x^2+xy+xz= 141,y^2+yz+xy=363 and z^2+xz+yz=72,then find the value of (X+y-z).​

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if x^2+xy+xz= 141,y^2+yz+xy=363 and z^2+xz+yz=72,then find the value of (X+y-z).