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the-system-of-equation-a-c-2-1-16-b-2-1-16-b-a-2-1-25-c-2-1-25-c-b-2-1-36-a-2-1-36-given-that-a-b-c-are-real-number




Question Number 11831 by Peter last updated on 02/Apr/17
the system of equation    a − (√(c^2  −(1/(16)) ))= (√(b^2  − (1/(16))))  b − (√(a^2  − (1/(25))))= (√(c^2  − (1/(25))))  c − (√(b^2  − (1/(36))))= (√(a^2  − (1/(36))))    given that a, b, c are real numbers   that satisfy they system of equation ...  if a + b + c = (x/( (√(y )))) where x, y are   positive integers and y is square free  find the value of x + y
$$\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equation} \\ $$$$ \\ $$$$\mathrm{a}\:−\:\sqrt{\mathrm{c}^{\mathrm{2}} \:−\frac{\mathrm{1}}{\mathrm{16}}\:}=\:\sqrt{\mathrm{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$\mathrm{b}\:−\:\sqrt{\mathrm{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}}=\:\sqrt{\mathrm{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}} \\ $$$$\mathrm{c}\:−\:\sqrt{\mathrm{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}}=\:\sqrt{\mathrm{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}} \\ $$$$ \\ $$$$\mathrm{given}\:\mathrm{that}\:\mathrm{a},\:\mathrm{b},\:\mathrm{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\: \\ $$$$\mathrm{that}\:\mathrm{satisfy}\:\mathrm{they}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equation}\:… \\ $$$$\mathrm{if}\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:=\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{y}\:}}\:\mathrm{where}\:\mathrm{x},\:\mathrm{y}\:\mathrm{are}\: \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{and}\:\mathrm{y}\:\mathrm{is}\:\mathrm{square}\:\mathrm{free} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:+\:\mathrm{y}\: \\ $$$$ \\ $$

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