How do you solve the system #Y=x+7#, #x^2+y^2=25#?

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Answer 1 · 2016-01-01T17:21:29

Substitute #y = x+7# into #x^2+y^2 = 25# to get a quadratic in #x#, hence solutions:

#(x, y) = (-3, 4) or (-4, 3)#

Substitute #y = x+7# into #x^2+y^2 = 25# ...

#25= x^2+y^2#

#=x^2+(x+7)^2#

#=x^2+x^2+14x+49#

#=2x^2+14x+49#

Subtract #25# from both ends to get:

#2x^2+14x+24 = 0#

Divide through by #2# to get:

#x^2+7x+12 = 0#

We can factor this by finding a pair of factors of #12# whose sum is #7#. The pair #3, 4# works.

Hence:

#0 = x^2+7x+12 = (x+3)(x+4)#

This has roots #x=-3# and #x=-4#, corresponding to #y = 4# and #y=3# respectively.