OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

$\large 5+\frac{x+3}{x-1} \div \frac{3x-1}{1-x^2}$

3 years ago
OpenStudy (anonymous):

Is that is?

3 years ago
OpenStudy (anonymous):

Yes

3 years ago
OpenStudy (anonymous):

$5+\frac{x+3}{x-1} * \frac{1-x^2}{3x-1}$

3 years ago
OpenStudy (anonymous):

$$1-x^2=-(x-1)(x+1)$$ ummm are you sure I wrote out the equation correctly??? Can you write it out?

3 years ago
OpenStudy (anonymous):

Yes I did, but the answer key may be wrong.

3 years ago
OpenStudy (anonymous):

5+(x+3/x-1)/(3x-1/1-x^2)

3 years ago
OpenStudy (bibby):

you're doing it right, yeah. the x-1's cancel

3 years ago
OpenStudy (shamil98):

you have to add 5 to the fraction first, then divide :P $\huge (5 + \frac{ x+3 }{ x-1 } )\div \frac{ 3x-1 }{ 1-x^2 }$ $\huge \frac{ 2(3x-1) }{ x-1 } * \frac{ -(x+1)(x-1) }{ 3x-1 }$ $\huge \frac{ 2\cancel{(3x-1)} }{ \cancel{x-1} } * \frac{ -(x+1)\cancel{(x-1)} }{ \cancel{3x-1} }$

3 years ago
OpenStudy (shamil98):

$\huge -2(x+1)$

3 years ago
OpenStudy (anonymous):

Thank you Shamil98

3 years ago