How do you add or subtract #2/(x^2+8x+15) + 1/(x^2+11x+30)#?

Question

1578 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T22:02:04

#3/((x+3)(x+6))#

factorise both denominators:

#x^2+8x+15 = (x+3)(x+5)#

#x^2+11x+30 = (x+6)(x+5)#

then the fractions can be written as #2/((x+3)(x+5))# and #1/((x+6)(x+5))#.

the common denominator between these is their lowest common multiple.

they have a common factor of #x+6#, and the lowest common multiple of #x+3# and #x+5# is #(x+3)(x+5)#.

therefore the lowest common multiple of #(x+3)(x+5)# and #(x+6)(x+5)# is #(x+3)(x+6)(x+5)#.

#2/((x+3)(x+5)) - (2(x+6))/((x+3)(x+6)(x+5))#

#= (2x+12)/((x+3)(x+6)(x+5))#

#1/((x+6)(x+5)) = (x+3)/((x+3)(x+6)(x+5))#

#2x + 12 + x + 3 = 3x + 15#

#3x + 15 = 3(x+5)#

#3(x+5)/((x+3)(x+6)(x+5)) = 3/((x+3)(x+6))#