I hope you consider giving me brainliest.
Answer: [tex]\frac{-4x+7}{2(x-2)}[/tex]
Step-by-step explanation:
[tex]\frac{\frac{3}{x-1}-4 }{2-\frac{2}{x-1} }[/tex]
Let's solve the numerator first.
[tex]\frac{3}{x-1} -4=?\\[/tex]
[tex]\frac{3-[(4)(x-1)]}{x-1} =\frac{3-(4x-4)}{x-1} =\frac{3-4x+4}{x-1} =\frac{-4x+7}{x-1}[/tex]
Now let's solve the denominator.
[tex]2-\frac{2}{x-1}[/tex]
[tex]\frac{2(x-1)-2}{x-1}=\frac{2x-2-2}{x-1}=\frac{2x-4}{x-1}[/tex]
So now we have this:
[tex]\frac{\frac{-4x+7}{x-1} }{\frac{2x-4}{x-1} }[/tex]
Multiply by the inverse operation of the denominator in order to get rid of the fraction.
[tex]\frac{\frac{-4x+7}{x-1} }{\frac{2x-4}{x-1} }*\frac{x-1}{2x-4}[/tex]
Solve,
[tex]\frac{-4x+7}{x-1} *\frac{x-1}{2x-4}[/tex]
Since we're multiplying, we can cancel out equal values in opposite positions eg: numerator/denominator. In this case, we can eliminate x-1 and we have;
[tex]\frac{-4x+7}{2x-4}[/tex]
2x-4 is factorable like: [tex]2(x-2)[/tex] leaving the fraction like:
[tex]\frac{-4x+7}{2(x-2)}[/tex]
So, the correct answer is b.
