Question

Q. Solve for 'x' and 'y':

x + xy + y = 11,
x²y + xy² = 30

Answer 1

Hey !!

Here is your answer.. ⬇⬇

$x + xy + y = 17 \\ \\ x + y = 17 - xy \: \: .......(1) \\ \\ {x}^{2} y + {y}^{2} x = 30 \\ \\ take \: ( xy )\: as \: \: common. \\ \\ xy(x + y) = 30 \\ \\ xy(11 - xy) = 30 \: \: ......(from \: eq.(1)) \\ \\ 11xy - {(xy)}^{2} = 30 \\ \\ 30 - 11xy + {(xy) }^{2} \\ \\ {(xy)}^{2} - 11xy + 30 = 0 \\ \\ {(xy)}^{2} - 6xy - 5xy + 30 = 0 \\ \\ xy(xy - 6) - 5(xy - 6) \\ \\ (xy - 5)(xy - 6) = 0 \\ \\ xy = 5 \\ xy = 6 \\ = = = = = = = = = = = = = = = = = = = = = \\ \\ taking \: \: xy = 5.... \: \: (y = \frac{5}{x} ) \\ \\ from \: \: eq. \: \: (1) \\ \\ x + y = 11 - 5 \\ \\ x + y = 6 \\ \\ x + \frac{5}{x} = 6 \\ \\ {x}^{2} + 5 = 6x \\ \\ {x}^{2} - 6x + 5 = 0 \\ \\ {x}^{2} - x - 5x + 5 = 0 \\ \\ x(x - 1) - 5(x - 1) = 0 \\ \\ (x - 1)(x - 5) = 0 \\ \\ x = 1 \\ x = 5 \\ \\ y = \frac{5}{x} \\ \\ y = \frac{5}{5} = 1 \\ \\ y = \frac{5}{1} = 5 \\ = = = = = = = = = = = = = = = = = = = \\ \\ now \: taking \: \: xy = 6 \: \: .....(y = \frac{6}{x}) \\ \\ from \: \: eq. \: \: (1) \\ \\ x + y = 11 - 6 \\ \\ x + y = 5 \\ \\ x + \frac{6}{x} = 5 \\ \\ {x}^{2} - 5x + 6 = 0 \\ \\ {x}^{2} - 3x - 2x + 6 = 0 \\ \\ x(x - 3) - 2(x - 3) \\ \\ (x - 2)(x - 3) = 0 \\ \\ x = 2 \\ x = 3 \\ \\ y = \frac{6}{x} \\ \\ y = \frac{6}{2} = 3 \\ \\ y = \frac{6}{3} = 2 \\ \\ y = 2 \\ y = 3 \\ \\ = = = = = = = = = = = = = = = = = =$



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