Answer:
D. [tex]-\dfrac{1}{2}[/tex]
Step-by-step explanation:
Given the fraction:
[tex]\dfrac{\dfrac{3}{4y}-\dfrac{2}{y}}{\dfrac{1}{y}+\dfrac{3}{2y}}[/tex]
Consider the numerator and the denominator separately:
Numerator:
[tex]\dfrac{3}{4y}-\dfrac{2}{y}=\dfrac{3}{4y}-\dfrac{8}{4y}=\dfrac{3-8}{4y}=-\dfrac{5}{4y}[/tex]
Denominator:
[tex]\dfrac{1}{y}+\dfrac{3}{2y}=\dfrac{2}{2y}+\dfrac{3}{2y}=\dfrac{2+3}{2y}=\dfrac{5}{2y}[/tex]
Now,
[tex]\dfrac{\dfrac{3}{4y}-\dfrac{2}{y}}{\dfrac{1}{y}+\dfrac{3}{2y}}=\dfrac{-\dfrac{5}{4y}}{\dfrac{5}{2y}}=-\dfrac{5}{4y}\div \dfrac{5}{2y}=-\dfrac{5}{4y}\times \dfrac{2y}{5}=-\dfrac{1}{2}[/tex]
