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What is the common denominator of [tex]y+\frac{y-3}{3}[/tex] in the complex fraction [tex]\frac{y+\frac{y-3}{3}}{\frac{5}{9}+\frac{2}{3y}}[/tex]?

A. [tex]3y(y-3)[/tex]
B. [tex]y(y-3)[/tex]
C. [tex]3y[/tex]
D. [tex]3[/tex]


Sagot :

To solve the given complex fraction, let's start by examining the components of the numerator and the denominator in detail and identifying their least common denominators (LCD).

First, consider the numerator:
[tex]\[ y + \frac{y - 3}{3} \][/tex]

To combine the terms, we need to find a common denominator. The term [tex]\( y \)[/tex] can be written as [tex]\( \frac{3y}{3} \)[/tex] so that it has the same denominator as [tex]\(\frac{y-3}{3}\)[/tex].

Thus, the numerator becomes:
[tex]\[ \frac{3y}{3} + \frac{y - 3}{3} \][/tex]

Now, combine the fractions:
[tex]\[ \frac{3y + (y - 3)}{3} = \frac{3y + y - 3}{3} = \frac{4y - 3}{3} \][/tex]

Next, consider the denominator:
[tex]\[ \frac{5}{9} + \frac{2}{3y} \][/tex]

For these fractions, the common denominator is the product of the individual denominators, which is [tex]\( 9 \cdot 3y = 27y \)[/tex]. Rewrite each fraction with this common denominator:

The first term:
[tex]\[ \frac{5}{9} = \frac{5 \cdot 3y}{9 \cdot 3y} = \frac{15y}{27y} \][/tex]

The second term:
[tex]\[ \frac{2}{3y} = \frac{2 \cdot 9}{3y \cdot 9} = \frac{18}{27y} \][/tex]

Combine the fractions:
[tex]\[ \frac{15y}{27y} + \frac{18}{27y} = \frac{15y + 18}{27y} \][/tex]

Putting it all together, we can rewrite the original complex fraction:
[tex]\[ \frac{\frac{4y - 3}{3}}{\frac{15y + 18}{27y}} \][/tex]

To simplify this complex fraction, multiply the numerator by the reciprocal of the denominator:
[tex]\[ \frac{4y - 3}{3} \times \frac{27y}{15y + 18} = \frac{(4y - 3) \cdot 27y}{3 (15y + 18)} \][/tex]

Simplify the expression:
[tex]\[ \frac{27y (4y - 3)}{3 (15y + 18)} = \frac{27y (4y - 3)}{3 \cdot 15y + 3 \cdot 18} = \frac{27y (4y - 3)}{45y + 54} \][/tex]

Simplifying further by dividing the numerator and the denominator by 3:
[tex]\[ \frac{9y (4y - 3)}{15y + 18} \][/tex]

After this step, consider the common denominator adjustments:
[tex]\[ \frac{y (4y - 3)}{\frac{15y + 18}{9}} \][/tex]

Which simplifies to:
[tex]\[ \frac{y (4y - 3)}{5y + 2} \][/tex]

Thus, the simplified final expression is:
[tex]\[ y \cdot \frac{4y-3}{5/9 + 2/(3y)} \][/tex]

From this equation, the common denominator for the original question is:
[tex]\[ \boxed{3y} \][/tex]
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