Question

Adding and Subtracting Rational Expressions:Help!!

Answer 1

\[\frac{ x^2+2x-35 }{ x-5 }+\frac{ x-4 }{ x+3 }\]

Answer 2

Try factoring the quadratic in the 1st equation

Answer 3

ok so i have done that im stuck on what to do next

Answer 4

what did you get?

Answer 5

\[\frac{ (x+7)(x-5) }{ x-5 }\]

Answer 6

notice that (x-5)/(x-5) is 1
anything divided by itself is 1 (except 0, so make a note: x≠5 or we get in trouble)

so now you have
x+7 + (x-4)/(x+3)

I would multiply the first term (x+7) by (x+3)/(x+3) so that you get a common denominator.
then add the tops

Answer 7

or you could do it this way
\[ x \cdot \frac{(x+3)}{(x+3)} + 7 \cdot \frac{(x+3)}{(x+3)} + \frac{x-4}{x+3}\]

Answer 8

but would we multiply \[\frac{ x-4 }{x+3 }*\frac{ x+3 }{ x+3 }\]

Answer 9

you could do that, but you really shouldn't. the denominator is already x+3
there is a rule: "If it ain't broken, don't fix it"

Answer 10

ooh i heard that lol so now would i cx like terms

Answer 11

yes

Answer 12

so it shoul look like\[\frac{ x^2+3x+7x+21+x-4 }{x+3 ?}\]

Answer 13

meaning\[\frac{ x^2+11x+17 }{ x+3}\]