How do you solve # x^2 + y^2 = 41# and #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)) :}#