Question

please help..schools almost over. 3/x-5 + 5/x^2 -25

Answer 1

Can you factor x^2 -25 at all?

Answer 2

x= -5 x= 5?

Answer 3

You're close, it's really (x-5)(x+5)

Answer 4

Ok.. i still dont know the anwser :[

Answer 5

In order to add the two fractions, what must the denominators be?

Answer 6

Any ideas?

Answer 7

no

Answer 8

Alright, here's how you solve this problem

Answer 9

k thank you!

Answer 10

\[\Large \frac{3}{x-5}+\frac{5}{x^{2}-25} \]

\[\Large \frac{3}{x-5}+\frac{5}{\left(x-5\right)\left(x+5\right)} \]

\[\Large \frac{3\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\frac{5}{\left(x-5\right)\left(x+5\right)} \]

\[\Large \frac{3x+15}{\left(x-5\right)\left(x+5\right)}+\frac{5}{\left(x-5\right)\left(x+5\right)} \]

\[\Large \frac{3x+15+5}{\left(x-5\right)\left(x+5\right)} \]

\[\Large \frac{3x+20}{\left(x-5\right)\left(x+5\right)} \]

\[\Large \frac{3x+20}{x^{2}-25} \]


So
\[\Large \frac{3}{x-5}+\frac{5}{x^{2}-25} \]
simplifies to
\[\Large \frac{3x+20}{x^{2}-25} \]

Answer 11

I'm basically doing the following

Step 1) Get all the denominators equal

Step 2) Combine the numerators over the common denominator

Answer 12

then the anwser is 3x+20/x-5 x+5?

Answer 13

It's either \[\Large \frac{3x+20}{\left(x-5\right)\left(x+5\right)} \] or it is \[\Large \frac{3x+20}{x^{2}-25} \]


Either one works since the two are equivalent. The book might like one answer over the other though.

Answer 14

Does that make sense?

Answer 15

Yup

Answer 16

that's great