Answer:
[tex]\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5} = \frac{3(3y + 2x)}{2x(y+5)}[/tex] If [tex]x \ne 0 \ \ or\ y \ne 0[/tex]
Step-by-step explanation:
Given
[tex]\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5}[/tex]
Required
Solve
First, we change / to *
[tex]\frac{x^2 - 9}{y^2 - 25} * \frac{3y^2 - 15y}{2x^2 - 6x} + \frac{3 - 1.5y}{y + 5}[/tex]
Apply difference of two squares
[tex]\frac{(x- 3)(x+3)}{(y - 5)(y+5)} * \frac{3y^2 - 15y}{2x^2 - 6x} + \frac{3 - 1.5y}{y + 5}[/tex]
Factorize:
[tex]\frac{(x- 3)(x+3)}{(y - 5)(y+5)} * \frac{3y(y - 5)}{2x(x - 3)} + \frac{3 - 1.5y}{y + 5}[/tex]
x - 3 cancels out:
[tex]\frac{(x+3)}{(y - 5)(y+5)} * \frac{3y(y - 5)}{2x} + \frac{3 - 1.5y}{y + 5}[/tex]
y - 5 cancels out
[tex]\frac{x+3}{y+5} * \frac{3y}{2x} + \frac{3 - 1.5y}{y + 5}[/tex]
[tex]\frac{3y(x+3)}{2x(y+5)} + \frac{3 - 1.5y}{y + 5}[/tex]
Take L.C.M
[tex]\frac{3y(x+3) + 2x(3 - 1.5y)}{2x(y+5)}[/tex]
Open brackets
[tex]\frac{3xy+9y + 6x - 3xy}{2x(y+5)}[/tex]
Collect Like Terms
[tex]\frac{3xy- 3xy+9y + 6x }{2x(y+5)}[/tex]
[tex]\frac{9y + 6x}{2x(y+5)}[/tex]
Factorize:
[tex]\frac{3(3y + 2x)}{2x(y+5)}[/tex]
Hence:
[tex]\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5} = \frac{3(3y + 2x)}{2x(y+5)}[/tex] If [tex]x \ne 0 \ \ or\ y \ne 0[/tex]
