How do you solve the following system algebraically: #x^2+y^2=25#, #x-2y+5=0#?

1 Answer
Lucy
May 10, 2018

When #y=0#, #x=-5#
When #y=0#, #x=3#

Explanation:

#x^2+y^2=25# --- (1)
#x-2y+5=0# --- (2)

From (2),
#x-2y+5=0#
#x=2y-5# --- (3)

Sub (3) into (1)
#(2y-5)^2+y^2=25#
#4y^2-20y+25+y^2=25#
#5y^2-20y=0#
#y^2-4y=0#
#y(y-4)=0#
#y=0,4#

If #y=0#,
#x^2+0^2=25#
#x=+-5#
But, if you sub #y=0# into (3), only #x=-5# works

If #y=4#
#x^2+4^2=25#
#x^2=9#
#x=+-3#
But if you sub #y=4# into (3), only #x=3# works

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