How do you solve #x-7+14/x=2# and check for extraneous solutions?

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Answer 1 · 2016-12-22T14:57:06

#{7, 2}# with #x != 0#

Put on a common denominator #(x)#.

#(x^2 - 7x + 14)/x = 2#

#x^2 - 7x + 14 = 2x#

#x^2 - 9x + 14 = 0#

#(x - 7)(x - 2) = 0#

#x = 7 and 2#

CHECK

#7 - 7 + 14/7 =^? 2#

#0 + 2 = 2" "color(green)(√)#

#2 - 7 + 14/2 =^? 2#

#-5 + 7 = 2" "color(green)(√)#

Hopefully this helps!