Answer
$\frac{x-y}{2x-y}$ and $x\ne-3y$.
Work Step by Step
$\frac{x^2+2xy-3y^2}{2x^2+5xy-3y^2}=\frac{x^2+3xy-xy-3y^2}{2x^2+6xy-xy-3y^2}=\frac{x(x+3y)-y(x+3y)}{2x(x+3y)-y(x+3y)}=\frac{(x+3y)(x-y)}{(x+3y)(2x-y)}=\frac{x-y}{2x-y}$ and $x\ne-3y$.
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 6 - Section 6.1 - Rational Expressions and Functions; Multiplying and Dividing - Exercise Set - Page 414: 46
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