How do you simplify #(28n^2-175) / (2n^2-7n+5) div(14n^2+31n-10 )/ (n^2-6n+5)#?

Question

1563 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-12T18:35:49

#(7n-35)/(7n-2)#

Once, I simplified fractions separately,

#(28n^2-175)/(2n^2-7n+5)#

=#(7*(4n^2-25))/(2n^2-7n+5)#

=#[7(2n+5)(2n-5)]/[(2n-5)(n-1)]#

=#[7*(2n+5)]/(n-1)#

#(14n^2+31n-10)/(n^2-6n+5)#

#[14n^2+35n-4n-10]/[(n-1)*(n-5)]#

=#[7n*(2n+5)-2*(2n+5)]/[(n-1)*(n-5)#

=#[(7n-2)(2n+5)]/[(n-1)(n-5)]#

Thus,

#[(28n^2-175)/(2n^2-7n+5)]/[(14n^2+31n-10)/(n^2-6n+5)]#

=#(28n^2-175)/(2n^2-7n+5)*(n^2-6n+5)/(14n^2+31n-10)#

=#(7*(2n+5))/(n-1)#*#((n-1)(n-5))/[(7n-2)(2n+5)]#

=#(7*(n-5))/(7n-2)#

=#(7n-35)/(7n-2)#