Division of rational expressions
Divide and simplify
y^2-81/25y+225 ÷ y-9/45
It does not come with brakets in my homework....Sorry :/
+118724 y^2-81/25y+225 ÷ y-9/45
(y^2-81)/(25y+225) ÷ (y-9)/45
if this is what you mean it would look like this - The fraction line replaces the brackets.
\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\)
Is that what your question looks like?
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ok I will assume that it is.
\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\\ =\frac{y^2-81}{25y+225} \times \frac{45}{y-9}\\ =\frac{(y-9)(y+9)}{25(y+9)} \times \frac{45}{y-9}\\ =\frac{(y+9)}{25(y+9)} \times \frac{45}{1}\\ =\frac{1}{25} \times \frac{45}{1}\\ =\frac{1}{5} \times \frac{9}{1}\\ =\frac{9}{5} \\ =1\frac{4}{5} \\\)
+118724 y^2-81/25y+225 ÷ y-9/45
(y^2-81)/(25y+225) ÷ (y-9)/45
if this is what you mean it would look like this - The fraction line replaces the brackets.
\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\)
Is that what your question looks like?
--------------------
ok I will assume that it is.
\(\frac{y^2-81}{25y+225} ÷ \frac{y-9}{45}\\ =\frac{y^2-81}{25y+225} \times \frac{45}{y-9}\\ =\frac{(y-9)(y+9)}{25(y+9)} \times \frac{45}{y-9}\\ =\frac{(y+9)}{25(y+9)} \times \frac{45}{1}\\ =\frac{1}{25} \times \frac{45}{1}\\ =\frac{1}{5} \times \frac{9}{1}\\ =\frac{9}{5} \\ =1\frac{4}{5} \\\)
