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How do you simplify # (y²-25) / (3y-7)#?

1 Answer

Nov 7, 2015

#(y^2-25)/(3y-7)=((y+5)(y-5))/(3y-7)#

Explanation:

#(y^2-25)/(3y-7)#

Rewrite #25# as #5^2#.

#(y^2-5^2)/(3y-7)#

#(y^2-5^2)# is an example of the difference of squares, #a^2-b^2=(a+b)(a-b)#, where #a=y and b=5#.

#((y+5)(y-5))/(3y-7)#

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