Question

Simplify: 1/a-b + 4/b-a - 8/a+b - 11a-5b/b^2-a^2

Answer 1

Okayy, when the denominators are all messed up like here, you need to look for something that you can change them all to. If it were normal fractions you could just multiply them together right and just change them. You can also do that here. The key thing to notice is that\[b^2 - a^2 = (b+a)(b-a)\] The difference of two squares. Knowing this makes it all a bit easier. Now we need to change the other ones slightly so that they look like either (b+a) or (b-a). \[\frac{ 1 }{ a-b } + \frac{ 4 }{ b-a } - \frac{ 8 }{ a+b } + \frac{ 11a-5b }{ b^2-a^2 }\] Now take a factor of (-1)/(-1) out of the first term so that it is b-a not a-b on the bottom. The second one is fine, and the third one is okay aswell (as a+b = b+a).\[\frac{ -1 }{ b-a } + \frac{ 4 }{ b-a } - \frac{ 8 }{ b+a } + \frac{ 11a-5b }{ b^2-a^2 }\] Now the first term we can multiply by (b+a)/(b+a), the second by the same, and the third by (b-a)/(b-a) to get.\[\frac{ -1(b+a) }{ b^2-a^2 } + \frac{ 4(b+a) }{ b^2-a^2 } - \frac{ 8(b-a) }{ b^2-a^2 } + \frac{ 11a-5b }{ b^2-a^2 }\] Since (b-a)(b+a) is (b^2 -a^2) remember. So now we can put it all on the same denominator to get \[\frac{ -1(b+a)+4(b+a)-8(b-a)+ 11a-5b }{ b^2-a^2 } \] Now expand and simplify.\[\frac{ -b-a+4b+4a-8b+8a+ 11a-5b }{ b^2-a^2 }\]\[\frac{ 22a-10b }{ b^2-a^2 }\] It could go a tiny bit further but it doesn't look as nice. Hope that this helped :)

Answer 2

thanks