Operation on rational expressions
Step 1: division of fractions
The division of two fractions is the same as multiplying the first by the inverted second fraction:
Then, in this case:
[tex]\frac{24y^2}{5x^2}\div\frac{6y^3}{25x^2}=\frac{24y^2}{5x^2}\times\frac{25x^2}{6y^3}[/tex]
Step 2: multiplication of two fractions
We multiply two fractions by multiplying the numerators and the denominators:
[tex]\frac{24y^2}{5x^2}\times\frac{25x^2}{6y^3}=\frac{24y^2\times25x^2}{5x^2\times6y^3}[/tex]
Step 3: simplifying the numbers of the fraction
We know that
[tex]\frac{25}{5}=5\text{ and }\frac{24}{6}=4[/tex]
Then, we can use this in our fraction:
[tex]\begin{gathered} \frac{24y^2\times25x^2}{5x^2\times6y^3}=5\cdot4\frac{y^2x^2}{x^2y^3} \\ \downarrow\text{ since 5}\cdot4=20 \\ 5\cdot4\frac{y^2x^2}{x^2y^3}=20\frac{y^2x^2}{x^2y^3} \end{gathered}[/tex]
Step 4: exponents of the result
We know that if we have a division of same base expressions (same letters), the exponent is just a substraction:
[tex]\begin{gathered} \frac{y^2}{y^3}=y^{2-3}=y^{-1} \\ \frac{x^2}{x^2}=x^{2-2}=x^0=1 \end{gathered}[/tex]
Then,
[tex]20\frac{y^2x^2}{x^2y^3}=20y^{-1}\cdot1=20y^{-1}[/tex]
Since negative exponents correspond to a division, then we can express the answer in two different ways:
[tex]20y^{-1}=\frac{20}{y}[/tex]
Answer:
[tex]20y^{-1}=\frac{20}{y}[/tex]
