Question

Addition and Subtraction:
9x+2/3x^2-2x-8 + 7/3x^2+x-4

Answer 1

can you add some brackets, its really hard to read

Answer 2

\(\large9x+(2/3x^{2})-2x-8+(7/3x^{2})+x-4\)

right?

Answer 3

something tells me its: \[\large {{9x+2}\over{3x^2-2x-8}}+{{7}\over{3x^2+x-4}}\]

Answer 4

yes @yummydum

Answer 5

please help me

Answer 6

K, gimme a minute

Answer 7

\[\large {{9x+2}\over{3x^2-2x-8}}+{{7}\over{3x^2+x-4}}\]
factor the denominator: \[\large {{9x+2}\over{(x-2)(3x+4)}}+{{7}\over{(x-1)(3x+4)}}\]
find a common denominator:\[\large {{(9x+2)(x-1)}\over{(x-2)(3x+4)(x-1)}}+{{(7)}(x-2)\over{(x-1)(3x+4)(x-2)}}\]
multiply out again:\[\large {{9x^2-7x-2}\over{3 x^3-5 x^2-6 x+8}}+{{7x-14}\over{3 x^3-5 x^2-6 x+8}}\]
now add:\[\large {{9x^2-16}\over{3 x^3-5 x^2-6 x+8}}\]
factor out once more:\[\large{(3x-4)(3x+4)}\over\large{(x-1)(x-2)(3x+4)}\] cancel out and multiply, you get:\[\large{3x-4}\over\large{x^2-3x+2}\]