Andrew M. answered • 10/06/18

Tutor

New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

It appears you have left off grouping parenthesis.... THESE ARE IMPORTANT!!

Note: Area = Length*width ... A=Lw ... w = A/L

A = (x

^{4}-y^{4})/(8xy) and L = (x^{2}-y^{2})/(2xy)W = [(x

^{4}-y^{4})/(8xy)]/[(x^{2}-y^{2})/(2xy)]w = [(x

^{4}-y^{4})/(8xy)][(2xy)/(x^{2}-y^{2})]Note: we can cancel the 2xy in numerator of 2nd term into the 8xy in denominator of

1st term, leaving 4 in denominator of 1st term.

w = [(x

^{4}-y^{4})/4][1/(x^{2}-y^{2})]Note that we can use the formula for the difference of two squares here:

a

^{2}- b^{2}= (a-b)(a+b) and thus x^{4}-y^{4}= (x^{2})^{2}- (y^{2})^{2}= (x^{2}-y^{2})(x^{2}+y^{2})w = [(x

^{2}-y^{2})(x^{2}+y^{2})/4][1/(x^{2}-y^{2})]We can now cancel the (x

^{2}-y^{2}) terms from the numerator and denominator terms**w = (x**

^{2}+y^{2})/4************************************

As for the 2nd part of the question.

Since area = length*width ... A = Lw

If we double the length and keep the width the same we have

A = 2Lw so all it does is double the area

Thus the new area is the old area multiplied by 2

A = 2[(x

^{4}-y^{4})/(8xy)] = (x^{4}-y^{4})/(4xy)