# Thread: Solve for X

1. ## Solve for X

State Restrictions and solve for X.

$
\frac{x}{x-1}+3=\frac{1}{x-1}-1
$

Restrictions: x≠1
Answer: x = 1 = null

Correct? If not, suggest correct answer.

$\frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}$

Restrictions: x≠3
Answer: x = 3 = null

Correct? If not, suggest correct answer.

Thank you very much.

2. Originally Posted by larry21
State Restrictions and solve for X.

$
\frac{x}{x-1}+3=\frac{1}{x-1}-1
$

Restrictions: x≠1
Answer: x = 1 = null

Correct? If not, suggest correct answer.
Unfortunately you didn't post your work ...

If this is actually the correct question then this equation has no solution:

$\frac{x}{x-1}+3=\frac{1}{x-1}-1~\implies~\frac{x}{x-1}-\frac{1}{x-1} = -4~\implies~1\neq -4
$

$\frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}$

Restrictions: x≠3
Answer: x = 3 = null

Correct? If not, suggest correct answer.

Thank you very much.
The common dnominator is 2x-6:

$\frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}~\implies~\frac{x+2}{2x-6}-\frac{6}{2x-6}=\frac{3}{2}$

And now go ahead!

3. Hello,
I each case, you clearly state that the answer you give for x is exactly the value that it is not allowed to take. Isn't there a problem ?