How do you solve $x^{2} + y^{2} = 41$ $x^2 + y^2 = 41$ and $y - x = - 1$ $y - x = -1$ ?

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Answer 1 · 2016-04-16T22:38:09

${ ((x, y) = (5, 4)), ((x, y) = (-4, -5)) :}$

From the second equation, we can deduce: $y = x - 1$ .

Substitute this in the first equation to get:

$41 = x^2+(x-1)^2 = 2x^2-2x+1$

Subtract $41$ from both ends to get:

$2x^2-2x-40 = 0$

Divide through by $2$ to get:

$x^2-x-20 = 0$

Note that $5*4=20$ and $5-4=1$ , hence:

$0 = x^2-x-20 = (x-5)(x+4)$

So $x = 5$ or $x = -4$

Hence solutions:

${ ((x, y) = (5, 4)), ((x, y) = (-4, -5)) :}$

How do you solve x^2 + y^2 = 41x2+y2=41 x^2 + y^2 = 41 and y - x = -1y−x=−1y - x = -1?