How do you solve #2(x-5)^-1 + 1/x = 0 #?

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Answer 1 · 2016-09-04T12:41:00

#x = 5/3#

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

Re-arrange the terms to have one fraction on each side.
(note that the negative index moves the bracket to the denominator)

#x!=+5 and x!=0#

#2/(x-5) = (-1)/x " "larr"cross multiply"#

#2x = -x+5#

#3x = 5#

#x= 5/3#

Answer 2 · 2016-09-04T13:48:46

#x = (5) / (3)#

We have: #2 (x - 5)^(-1) + (1) / (x) = 0#

The terms within the parentheses can be expressed as a fraction:

#=> 2 cdot (1) / (x - 5) + (1) / (x) = 0#

#=> (2) / (x - 5) + (1) / (x) = 0#

Let's combine the fractions:

#=> ((2) (x) + (1) (x - 5)) / ((x - 5) (x)) = 0#

#=> 2 x + x - 5 = 0#

#=> 3 x = 5#

#=> x = (5) / (3)#