How do you solve #(2x)/(x-4)=8/(x-4)+3#?

2 Answers
May 14, 2017

Answer:

No solution!

Explanation:

First, note that 4 cannot be a solution (division by zero)

Then, multiply both sides by #(x-4)#, you get

#2x = 8+3*(x-4)#
#=> 3x-2x= 3*4-8#
#=> x = 4# which is impossible!
So there is no solution

May 14, 2017

Answer:

The equation is unsolvable.

Explanation:

We have:

#(2x)/(x-4) = 8/(x-4)+3#

Multiply all terms by #x-4#.

#2x=8+3(x-4)#

Expand the brackets.

#2x=8+3x-12=3x-4#

Add #4-2x# to both sides.

#4=x# or #x=4#

Unfortunately this leads to a problem that #x=4# is a singularity (mathematicians don't like infinities).

Reorganise the original equation by subtracting #8/(x-4)# from both sides.

#(2x-8)/(x-4)=3#

This gives:

#(2(x-4))/(x-4)=3#

This gives #2=3# the equation makes no sense unless #x=4# in which case both sides are infinite..