How do you solve the system #x^2+y^2-4x-4y=26# and #x^2+y^2-4x=54# and #y=3x-8#?

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Answer 1 · 2016-09-08T09:26:55

#x = 5, y = 7#

#x^2+y^2-4x-4y=26#
#x^2+y^2-4x=54#
#y=3x-8#

Subtracting the first from the second we get

Solving

#{(4 y = 28), (y = 3 x - 8):}#

we obtain

#x = 5, y = 7#

Substituting those results in the first

#25+49-20-28=26#

Substituting now in the second

#25+49-20 = 54#

so the system has the solution #x = 5, y = 7#