Answer:
[tex]\frac{2x-3y}{2}[/tex]
Step-by-step explanation:
Dividing by a fraction means multiplying by the reciprocal. So we can write the problem as:
[tex]\frac{5}{2x+3y}*\frac{4x^2-9y^2}{10}[/tex]
We can simplify [tex]4x^2-9y^2[/tex] into [tex](2x+3y)(2x-3y)[/tex] by using the formula [tex]a^2-b^2=(a+b)(a-b)[/tex]
Now we can write:
[tex]\frac{5}{2x+3y}*\frac{(2x-3y)(2x+3y)}{10}[/tex]
We can cancel out (2x+3y) and also reduce "5" and 10". Thus we have:
[tex]\frac{5}{2x+3y}*\frac{(2x-3y)(2x+3y)}{10}\\=\frac{1}{1}*\frac{(2x-3y)}{2}\\=\frac{2x-3y}{2}[/tex]
This is the simplified form.
