How do you solve #(2x)/1 = 1/(x-1)#?

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Answer 1 · 2017-08-17T07:30:59

#x=1/2(1+-sqrt3)#

#(2x)/1=1/(x-1)#

Cross multiply

#2x(x-1)=1#

#2x^2-2x-1=0#

For a quadratic equation of the form: #ax^2+bx+c#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In our case: #a=2, b=-2, c=-1#

#:. x=(+2+-sqrt(4+4*2*1))/(2*2)#

#=(+2+-sqrt(12))/(4)#

#=(2+-2sqrt(3))/(4)#

#= 1/2(1+-sqrt3)#

Answer 2 · 2017-08-17T07:33:28

Multiply through by the denominator of the right-hand side and then solve the resulting quadratic equation.

#x=1.366# or #-0.366#

#(2x)/1# is just #2x#, so

#2x=1/(x-1)#

Multiply both sides by #(x-1)#

#2x(x-1)=1(cancel(x-1))/cancel(x-1)#

#2x^2-2x=1#

#2x^2-2x-1=0#

Now we can use the quadratic formula (or any other method) to solve this quadratic equation.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#=(2+-sqrt(4+8))/(4)#

#x=1.366# or #-0.366#