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

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

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Answer 1 · 2017-05-14T14:37:57

No solution!

Explanation:

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

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

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

Answer 2 · 2017-05-14T14:39:46

The equation is unsolvable.

Explanation:

We have:

$\frac{2 x}{x - 4} = \frac{8}{x - 4} + 3$ $(2x)/(x-4) = 8/(x-4)+3$

Multiply all terms by $x - 4$ $x-4$ .

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

Expand the brackets.

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

Add $4 - 2 x$ $4-2x$ to both sides.

$4 = x$ $4=x$ or $x = 4$ $x=4$

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

Reorganise the original equation by subtracting $\frac{8}{x - 4}$ $8/(x-4)$ from both sides.

$\frac{2 x - 8}{x - 4} = 3$ $(2x-8)/(x-4)=3$

This gives:

$\frac{2 (x - 4)}{x - 4} = 3$ $(2(x-4))/(x-4)=3$

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

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

2 Answers

Explanation:

Explanation:

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