Question

Perform the indicated operations if

Answer 1

Answer:

$\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5} = \frac{3(3y + 2x)}{2x(y+5)}$  If $x \ne 0 \ \ or\ y \ne 0$

Step-by-step explanation:

Given

$\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5}$

Required

Solve

First, we change / to *

$\frac{x^2 - 9}{y^2 - 25} * \frac{3y^2 - 15y}{2x^2 - 6x} + \frac{3 - 1.5y}{y + 5}$

Apply difference of two squares

$\frac{(x- 3)(x+3)}{(y - 5)(y+5)} * \frac{3y^2 - 15y}{2x^2 - 6x} + \frac{3 - 1.5y}{y + 5}$

Factorize:

$\frac{(x- 3)(x+3)}{(y - 5)(y+5)} * \frac{3y(y - 5)}{2x(x - 3)} + \frac{3 - 1.5y}{y + 5}$

x - 3 cancels out:

$\frac{(x+3)}{(y - 5)(y+5)} * \frac{3y(y - 5)}{2x} + \frac{3 - 1.5y}{y + 5}$

y - 5 cancels out

$\frac{x+3}{y+5} * \frac{3y}{2x} + \frac{3 - 1.5y}{y + 5}$

$\frac{3y(x+3)}{2x(y+5)} + \frac{3 - 1.5y}{y + 5}$

Take L.C.M

$\frac{3y(x+3) + 2x(3 - 1.5y)}{2x(y+5)}$

Open brackets

$\frac{3xy+9y + 6x - 3xy}{2x(y+5)}$

Collect Like Terms

$\frac{3xy- 3xy+9y + 6x }{2x(y+5)}$

$\frac{9y + 6x}{2x(y+5)}$

Factorize:

$\frac{3(3y + 2x)}{2x(y+5)}$

Hence:

$\frac{x^2 - 9}{y^2 - 25} / \frac{2x^2 - 6x}{3y^2 - 15y} + \frac{3 - 1.5y}{y + 5} = \frac{3(3y + 2x)}{2x(y+5)}$   If $x \ne 0 \ \ or\ y \ne 0$