What is #(w-1)/4 -: (w^2 + 2w - 3)/4#?

1 Answer
mason m
Mar 2, 2016

#1/(w+3)#

Explanation:

First, note that dividing a fraction is the same as multiplying by its reciprocal.

Thus, instead of dividing by #(w^2+2w-3)/4#, can multiply by #4/(w^2+2w-3)#.

#=(w-1)/4xx4/(w^2+2w-3)#

Factor the quadratic term.

#=(w-1)/4xx4/((w+3)(w-1))#

Any terms found in both a numerator and denominator can be cancelled.

#=color(red)(cancel(color(black)((w-1))))/color(blue)(cancel(color(black)(4)))xxcolor(blue)(cancel(color(black)(4)))/((w+3)color(red)(cancel(color(black)((w-1)))))#

#=1/(w+3)#

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