Question 145462
The denominators are the same, so just add the numerators;
3 will cancel into 9:
1.{{{(5n)/3}}} + {{{(4n)/3}}} = {{{(9n)/3}}} = 3n
:
2.{{{x/((x+1)(x+8))}}} + {{{8/((x^2+9x+8))}}}
Factor the 2nd denominator and you can see that you can add the numerators:
{{{x/((x+1)(x+8))}}} + {{{8/((x+1)(x+8))}}} = {{{((x+8))/((x+1)(x+8))}}}
The (x+8)'s will cancel, leaving:
{{{1/((x+1))}}}
:
Same rules apply here:
3.{{{a/((a^2-4))}}} + {{{(-2)/((a^2-4))}}} = {{{((a-2))/((a^2-4))}}}
However, the denominator can be factored (the difference of squares)
{{{((a-2))/((a-2)(a+2))}}}
You can see he (a-2)'s will cancel leaving us with:
{{{1/((a+2))}}}
;
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