addition/subtraction of rational expressions

[juliag]

a / (5-a) - 2 /(a+3) + 2a[sup:2n0aeidw]2[/sup:2n0aeidw] - 2a / a[sup:2n0aeidw]2[/sup:2n0aeidw] - 2a - 15
a/ (5-a) - 2/(a+3) + 2a (a+1) / (a+3) (a-5)

I know that I have to factor a negative one out of the bolded parts of the equation I am just not sure when I should do this. Now or at the end of the equation. Could you please show me this equation fully worked out I am really struggling and cannot find a proper example in my textbook. Thank you in advance.

Julia

 

[Loren]

\(\displaystyle \frac{a}{5-a}=\frac{-a}{-(5-a)}=\frac{-a}{a-5}or -\frac{a}{a-5}\)
Does that help?

 

[juliag]

Yes I understand that part. I am just not sure as to when I should do this for ease of multiplying to get a LCD. Should I add a -1 in as part of the LCD?

 

[Denis]

a / (5-a) - 2 /(a+3) + 2a[sup:1x4gg5b0]2[/sup:1x4gg5b0] - 2a / a[sup:1x4gg5b0]2[/sup:1x4gg5b0] - 2a - 15

Is right portion correct: 2a^2 - 2a / a^2 - 2a - 15
or should it be: (2a^2 - 2a) / (a^2 - 2a - 15) ?
VERY important to bracket properly...

 

[Loren]

a/ (5-a) - 2/(a+3) + 2a (a+1) / ((a+3) (a-5))
\(\displaystyle \frac{-a}{a-5}-\frac{2}{a+3} + \frac{2a(a+1)}{(a+3)(a-5)}=\)

\(\displaystyle \frac{-a(a+3)}{(a+3)(a-5)}-\frac{2(a-5)}{(a+3)(a-5)} + \frac{2a(a+1)}{(a+3)(a-5)}=??\)

Now they all have the same denominator. Can you take it from here?

 

[Denis]

a / (5-a) - 2 /(a+3) + 2a[sup:2aj0imhj]2[/sup:2aj0imhj] - 2a / a[sup:2aj0imhj]2[/sup:2aj0imhj] - 2a - 15

Easier if you re-arrange after changing the 5-a to a-5:

[-a/(a-5)] - [2/(a+3)] + [2a(a-1) / ((a+3) (a-5))]

= [2a(a-1) / ((a+3) (a-5))] - [a/(a-5)] - [2/(a+3)] : having no minus term at beginning is "easier" (opinion!)
Apply LCM:
= [2a(a-1) - a/(a+3) - 2/(a-5)] / [(a+3) (a-5)]
Simplify numerator:
= (a^2 - 7a + 10) / [(a+3) (a-5)]
Factor numerator:
= [(a-2)(a-5)] / [(a+3) (a-5)]

= (a-2) / (a+3) ; finished!

By the way, you can check if your final expression is correct by substituting a value for a in the original expression;
if you let a = -4, the result of your original expression will be 6;
substitute same in (a-2) / (a+3) and result is 6 : so you know you're correct: get my drift?

 

[juliag]

Thanks for all the help i got it!