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X+y+z+xy+yz+zx+xyz=384 Find x+y+z??? |
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Answer» Answer: Step-by-step explanation: x y z + XY + y z + x z + x + y + z = 384 x y ( z + 1) + z ( x + y) + (x + y) + z = 384 x y (z + 1) + ( x + y ) ( z + 1) + z = 384 add 1 to both the sides x y (z + 1) + (x + y)(z + 1) +( z + 1) = 384 +1 (z + 1) ( xy + x+y + 1) = 385 ( z + 1) {x(y+1) + 1( y+1)} = 385 ( z + 1) ( y + 1) ( x + 1) = 5 x 7 x 11 (x+1), (y+1), (z+1) hold either of these values. 5 or 7 or 11 x, y, z hold either of the values 4, 6 , or 10 Hence, x + y + z = 4 + 6 + 10 = 20 |
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