The equivalent fraction to the given complex fraction is [tex]\frac{2y-1}{3y+1}[/tex].
Given expression,
[tex]\dfrac{2-\frac{1}{y} }{3+\frac{1}{y} } } \\[/tex]
We have to find the equivalent fraction of [tex]\frac{2-\frac{1}{y} }{3+\frac{1}{y} } } \\[/tex] .
On solving,
[tex]\dfrac{2-\frac{1}{y} }{3+\frac{1}{y} } } =\dfrac{\frac{2y-1}{y} }{\frac{3y+1}{y} }[/tex]
Eliminating y from numerator and denominator we get,
[tex]\dfrac{2-\frac{1}{y} }{3+\frac{1}{y} } } =\dfrac{2y-1} {3y+1} }[/tex]
Hence the required expression is [tex]\frac{2y-1}{3y+1}[/tex].
For more details follow the link:
https://brainly.com/question/17912
Answer:
The answer is D :)
Step-by-step explanation:
2y-1/3y+1
