X+y+z=15
xy+yz+zx=72
then the value of x is
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you have given two equations only and there are three variables.
Hence one cannot solve for x as a numerical value. But x can be found
in terms of y and z.
x + y + z = 15 --- 4
(x + z) y + z x = 72
(15 - y) y + z x = 72
x = ( y^2 - 15 y + 72 ) / z = 15 - y - z --- 3
then y and z satisfy : y^2 + z^2 - 15 (y+z) + y z +72 = 0 - -- 2
you can solve for y in terms of z.
Further,
(x+y+z)^2 - 2 (xy+yz+zx) = 15^2 - 2 * 72
x^2 + y^2 + z^2 = 81 ----- equation 5
81 - x^2 - 15 (y + z) + y z + 72 = 0
x^2 = 153 - 15 ( y + z ) + y z --- 1
substitute the value of y in terms of z, then x can be obtained in terms of z.
x + y + z = 15 --- 4
(x + z) y + z x = 72
(15 - y) y + z x = 72
x = ( y^2 - 15 y + 72 ) / z = 15 - y - z --- 3
then y and z satisfy : y^2 + z^2 - 15 (y+z) + y z +72 = 0 - -- 2
you can solve for y in terms of z.
Further,
(x+y+z)^2 - 2 (xy+yz+zx) = 15^2 - 2 * 72
x^2 + y^2 + z^2 = 81 ----- equation 5
81 - x^2 - 15 (y + z) + y z + 72 = 0
x^2 = 153 - 15 ( y + z ) + y z --- 1
substitute the value of y in terms of z, then x can be obtained in terms of z.
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