How do you add the rational expressions #(2x-1)/((x-3)(x+2))+(x-4)/(x-3)#?

Question

4246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-30T14:30:29

#(x+3)/(x+2),x!=3#

Writing
#(2x-1)/((x-3)(x+2))+((x-4)(x+2))/((x-3)(x+2))=#

#(2*x-1+x^2-4x+2x-8)/((x-3)(x+2))=#

#(x^2-9)/((x-3)(x+2))=#

#((x-3)(x+3))/((x-3)(x+2))=#

#(x+3)/(x+2), x!=3#

Answer 2 · 2018-05-30T15:04:18

#(x+3)/(x+2),x!=3#

1st way:

#(2x-1)/((x-3)*(x+2))+(x-4)/(x-3)#

=#((2x-1)+(x-4)(x+2))/((x-3)(x+2))#

=#(2x-1+x^2-2x-8)/((x-3)(x+2))#

=#(x^2-9)/((x-3)(x+2))#

=#((x+3)(x-3))/((x-3)(x+2))#

=#(x+3)/(x+2), x!=3#

2nd way:

#(2x-1)/((x-3)*(x+2))+(x-4)/(x-3)#

=#((x-3)+(x+2))/((x-3)*(x+2))+(x-4)/(x-3)#

=#1/(x-3)+1/(x+2)+(x-4)/(x-3)#

=#1/(x+2)+(x-3)/(x-3)#

=#1/(x+2)+1,x!=3#

=#(x+3)/(x+2), x!=3#