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Assuming no denominator equals zero, which expression is equivalent to the given expression?

[tex]
\frac{\frac{r-5}{9r}}{\frac{5-r}{3r^2}}
[/tex]

A. [tex]-\frac{3}{r}[/tex]
B. [tex]\frac{r}{3}[/tex]
C. [tex]-\frac{r}{3}[/tex]
D. [tex]\frac{r-5}{5-3r}[/tex]

Sagot :

To determine which expression is equivalent to the given one, we need to simplify the following fraction:
[tex]\[ \frac{\frac{r-5}{9 r}}{\frac{5-r}{3 r^2}} \][/tex]

First, let's rewrite the overall fraction:
[tex]\[ \frac{\frac{r-5}{9 r}}{\frac{5-r}{3 r^2}} = \left( \frac{r-5}{9 r} \right) \div \left( \frac{5-r}{3 r^2} \right) \][/tex]

Dividing by a fraction is the same as multiplying by its reciprocal. Hence, we get:
[tex]\[ \left( \frac{r-5}{9 r} \right) \times \left( \frac{3 r^2}{5-r} \right) \][/tex]

Now, let's perform the multiplication:
[tex]\[ \frac{(r-5) \cdot 3 r^2}{9 r \cdot (5-r)} \][/tex]

Next, we simplify the numerator and the denominator before multiplying:
[tex]\[ \frac{3 r^2 (r-5)}{9 r (5-r)} \][/tex]

Notice that [tex]\(5 - r = - (r - 5)\)[/tex]. So:
[tex]\[ \frac{3 r^2 (r-5)}{9 r (- (r-5))} = \frac{3 r^2 (r-5)}{-9 r (r-5)} \][/tex]

Both the numerator and the denominator have the factor [tex]\((r-5)\)[/tex], so we can cancel them out:
[tex]\[ \frac{3 r^2}{-9 r} \][/tex]

Further simplify by dividing both the numerator and the denominator by [tex]\(r\)[/tex]:
[tex]\[ \frac{3 r^2}{-9 r} = \frac{3 r}{-9} \][/tex]

Perform the final arithmetic simplification:
[tex]\[ \frac{3 r}{-9} = -\frac{r}{3} \][/tex]

Thus, the given expression simplifies to:
[tex]\[ -\frac{r}{3} \][/tex]

Therefore, the equivalent expression is:
[tex]\[ C. -\frac{r}{3} \][/tex]
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