Question

Find the value of of x, y, if x^5+y^5=275 and x+y = 5....

Answer 1

$x {}^{5} + {y}^{5} = 275 \\ \\ {x}^{5} = 275 - {y}^{5} \\ \\ x = \sqrt[5]{275 - {y}^{5} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - ( \: 1 \: )$




Now we have got the value of x in terms of y, then putting the value of x in terms of y in [ x + y = 5 ]


$x + y = 5 \\ \\ \sqrt[5]{275 - {y}^{5} } + y = 5 \\ \\ \sqrt[5]{275 + {y}^{5} } = 5 - y$



Making 5 as the power on both sides,


$( \sqrt[5]{275 - {y}^{5} } ) {}^{5} = (5 - y) {}^{5} \\ \\ ( \sqrt{275 - {y}^{5} } ) {}^{ \frac{1}{5} \times 5 } = (5 - y) {}^{2} (5 - y) {}^{3} \\ \\ 275 - {y}^{5} = (25 + {y}^{2} - 10y)( - {y}^{3} + 15 {y}^{2} - 75y + 125) \\ \\ 275 - {y}^{5} = 3125 - 1250y + 125y^{2} - 1875y + 750y^{2}- 75y^{3}+ 375y^{2}- 150y^{3} + 15y^{4} - 25x^{3} + 10x^{4} - x^{5}$


combining and adding like terms,


$275 - y^{5}=3125 - 3125y + 1250y^{2} - 250y^{3} + 25y^{4} - y^{5}$


275 - 3125 + 3125y - 1250y² + 250y³ - 25y⁴ = 0

- 2850 + 3125y - 1250y² + 250y³ - 25y⁴ = 0

- 25[ 114 - 125y + 50y² - 10y³ + y⁴ ] = 0

y⁴ - 10y³ + 50y² - 125y + 114 = 0

y⁴ - 2y³ - 8y³ + 16y² + 34y² - 125y + 114 = 0

y³( y - 2 ) - 8y²( y - 2 ) + 34y² - 68y - 57y + 114 = 0

y³( y - 2 ) - 8y²( y - 2 ) + 34y( y - 2 ) - 57( y - 2 ) = 0

( y - 2 )( y³ - 8y² + 34y - 57 ) = 0

( y - 2 )( y - 3 )( y² - 5x + 19 ) = 0




Thus,
value of y is either 2 or 3 .



Putting the value(s) of y in ( 1 ) ,

$x = \sqrt[5]{275 - ( {2)}^{2} } \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x = \sqrt[5]{275 - {(3)}^{2} } \\ \\ \\ x = \sqrt[5]{271} \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: x = \sqrt[5]{266}$

x » 3.06 or x » 3.05





Therefore,
value of x is 3.06 or 3.05
value of y is 2 because if we add 3.05 with 3 ,we will the result which will be < 5 ,so :-



$\textbf{Value of x is 3.05} \\\textbf{ Value of y is 2 }$

Answer 2

Step-by-step explanation:

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