Question

6)
If x-y = 1 and x2 + y2 = 41, find the value of xty.​

Answer 1

Answer:

y= 4, -5

x=-4, 5

Step-by-step explanation:

$x - y = 1 \\ x = 1 + y \: \: \: ......(1) \\ put \: x \: in \: other \: equation \\ {x}^{2} + {y}^{2} = 41 \\ {(1 + y)}^{2} + {y}^{2} - 41 = 0 \\ 1 + {y}^{2} + 2y + y {}^{2} - 41 \\ 2y {}^{2} + 2y - 40 \\ devide \: whole \: equation \: by \: 2 \\ y {}^{2} + y - 20y \\ y {}^{2} - 4y + 5y - 20y \\ y(y - 4) + 5(y - 4) \\ (y + 5)(y - 4) \\ \\ y = 4 \: and \: \: - 5 \\ \\ put \: both \: values \: of \: y \: in \: \\ first \: equation \\ x = 1 + y \\ x = 1 + 4 = 5 \\ \\ x = 1 + y \\ x = 1 + ( - 5) = 1 - 5 = - 4$

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Answer 2

x-y=1.........(1)
x=y+1

2x+2y=41..........(2)
2(y+1)+2y=41
2y+2+2y=41
4y+2=41
4y=41-2
y=39/4

put in (1)

x-39/4=1
x=39/4+1
x=(39+4)/4
x=43/4

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