quick fraction question?

Question

x/((x+1)(x+3))+(x+12)/(x^2-9). EXpress it as a single fraction in its simplest form. show all working

Answer 1

first term has denominator (x+1)(x+3)
second term has denominator (x-3)(x+3)
the common denominator will have to be
(x+1)(x-3)(x+3)
so the first term needs to be multiplied by
(x-3)/(x-3)
and the second term needs to be multiplied by
(x+1)/(x+1)
now the numerator will be
x(x-3) + (x+12)(x+1) over the common
denominator of (x+1)(x-3)(x+3)
simplify the numerator to
x² - 3x + x² +13x + 12
to 2x² + 10x + 12 and
factor to 2(x² + 5x + 6)
to 2(x + 2)(x + 3)
now cancel the factor (x+3)from top and bottom
answer is
[2(x+2)] / [(x-3)(x+1)]

~~

Answer 2

1) like any fraction, you cannot add the two fractions together without a common denominator.

2) (x+1) (x+3) (x-3) is the common denominator. x^2-9 changes to (x+3)(x-3).

3) multiply both fractions by the common denominator. Do some canceling and see what you have.

Answer 3

2(x+2)/[(x-3)*(x+1)]

Answer 4

x/((x+1)(x+3))+(x+12)/(x^2-9)
x/((x+1)(x+3)) +(x+12)/(x+3)(x-3)


x(x+3)(x-3) + (x+12)(x+3)(x+1)
________________________
(x+1)(x+3)(x-3)


x(x-3) + (x+12)(x+1)
_________________
(x+1)(x-3)


x² -3x + x² +12x +x +12
_________________
(x² +x -3x -3)


2x² +10x +12
___________
(x² -2x -3)



2(x² +5x +6)
_________
(x² -2x -3)

OR


2(x-1)(x+6)
_________
(x+1)(x-3)

Answer 5

x/((x+1)(x+3))+(x+12)/(x^2-9)=
x/((x+1)(x+3))+(x+12)/((x-3)(x+3))=
1/((x+1)(x+3)(x-3))[x(x-3)+(x+12)(x+1)]=
1/((x+1)(x+3)(x-3))[(x^2-3x)+(x^2+13x+12)]=
1/((x+1)(x+3)(x-3))[2x^2+10x+12]=
2/((x+1)(x+3)(x-3))[x^2+5x+6]=
2/((x+1)(x+3)(x-3))[(x+3)(x+2)]=
2/((x+1)(x-3))[(x+2)]=
=2(x+2)/((x+1)(x-3))