How do you multiply #(2x+2y)/(x+y) *( x^2-y^2)/(2x-2y)#?

Question

Kennedy Adams · Accepted Answer

#x+y#

#(2x+2y)/(x+y)*(x^2-y^2)/(2x-2y)#

Factoring:

#(2(x+y))/(x+y)*(x^2-y^2)/(2(x-y))#

Cancelling:

#(cancel(2)(x+y))/(x+y)*(x^2-y^2)/(cancel(2)(x-y))#

#(x+y)/(x+y)*(x^2-y^2)/(x-y)#

Further factoring:

#(x+y)/(x+y)*((x+y)(x-y))/(x-y)#

Cancelling again:

#cancel((x+y))/cancel((x+y))*((x+y)cancel((x-y)))/cancel((x-y))#

#1*(x+y)=x+y#

Kennedy Adams · Answer

undefined To multiply the given expression, you can follow these steps:

1. Simplify both fractions individually, if possible.
2. Multiply the numerators together and the denominators together.
3. Simplify the resulting expression, if possible.

Let's apply these steps to the given expression:

1. Simplify the fractions:
   - The first fraction, (2x+2y)/(x+y), cannot be simplified further.
   - The second fraction, (x^2-y^2)/(2x-2y), can be factored as (x+y)(x-y)/(2(x-y)).
   
2. Multiply the numerators and denominators:
   - Numerator: (2x+2y) * (x+y)(x-y)
   - Denominator: (x+y) * 2(x-y)
   
3. Simplify the resulting expression:
   - The (x+y) terms in the numerator and denominator cancel out.
   - The (x-y) terms in the numerator and denominator cancel out.
   - The 2 in the denominator cancels out with one of the 2's in the numerator.
   
   Therefore, the simplified expression is: (2x+2y)(x-y)/(2(x-y)).

Note: It is important to mention that (x-y) cannot be equal to zero, as it would result in division by zero, which is undefined.