How do you solve #1/(2(x-3))+3/(2-x)=5 x#?

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Answer 1 · 2017-11-02T09:21:35

# x = 1.436#
and
#x = 3.064#

#1/(2(x-3))+3/(2-x)=5 x#

Make the denominators equal:

#1/(2(x-3)) xx(2-x)/(2-x) +3/(2-x) xx (2(x-3))/(2(x-3))=5 x#

#(2-x)/(2(x-3)(2-x)) + (3xx2(x-3))/(2(x-3)(2-x) =5#

Now we can add the numerators:

#=>( (2-x) + 6(x-3))/(2(x-3)(2-x)) =5#

# (2-x +6x-18)/(2(x-3)(2-x)) =5#

Transposition :

# => (2-x +6x-18) = 5xx2(x-3)(2-x)#

#=> 5x -16 = 10(x(2-x) -3(2-x)#

#=> 5x -16 = 10(2x-x^2 -6+ 3x)#

#=> 5x -16 = 10(5x-x^2 -6)#

#=> 5x -16 = 50x-10x^2 -60#

#=> 50x-5x-10x^2 -60+16 = 0#

#=> -10x^2 +45x -44 =0#

#= 10x^2 -45x+ 44 = 0#

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

Here# a= 10, b= -45 and c= 44#

#b^2 -4ac = 2025-1760 = 265#

We get :
# x = 1.436# or #x = 3.064#