Solve for X

May 2010
26
1
State Restrictions and solve for X.

\(\displaystyle
\frac{x}{x-1}+3=\frac{1}{x-1}-1
\)

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

Correct? If not, suggest correct answer.


\(\displaystyle \frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}\)

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

Correct? If not, suggest correct answer.

Thank you very much. (Nod)
 

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
State Restrictions and solve for X.

\(\displaystyle
\frac{x}{x-1}+3=\frac{1}{x-1}-1
\)

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:

\(\displaystyle \frac{x}{x-1}+3=\frac{1}{x-1}-1~\implies~\frac{x}{x-1}-\frac{1}{x-1} = -4~\implies~1\neq -4
\)
\(\displaystyle \frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}\)

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

Correct? If not, suggest correct answer.

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

\(\displaystyle \frac{x+2}{2x-6}-\frac{3}{x-3}=\frac{3}{2}~\implies~\frac{x+2}{2x-6}-\frac{6}{2x-6}=\frac{3}{2}\)

And now go ahead!
 
Nov 2009
927
260
Wellington
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 ?