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

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Answer 1 · 2017-05-14T14:37:57

No solution!

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

Answer 2 · 2017-05-14T14:39:46

The equation is unsolvable.

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..

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