Question

What is the solution to the rational equation (5/(x^2+3x+2))+(1/(x+2))=(1/(3x+3))

Answer 1

x = -8

Answer 2

manuel1234, you have given an answer without explanation. That is against CoC.

http://openstudy.com/code-of-conduct

Answer 3

I have to report you

Answer 4

nooooo dont

Answer 5

The purpose of this site is to help users. Giving answers does not help anyone learn anything. Giving answers is a form of cheating and cheating is not what this site is about.

Answer 6

...

Answer 7

@militarygirl96 , you should want explanations, not answers.

Answer 8

i want both but this is one that is trial and error so that would be alot of work to show

Answer 9

It's not trial and error. The steps are simple

Answer 10

You just have to make sure the denominators are factored first

Answer 11

\[\huge\color{salmon}{\text{Zepp}}\]

Answer 12

@zepp

Answer 13

o.o

Answer 14

\[\large \frac{5}{(x^2+3x+2}+\frac{1}{x+2}=\frac{1}{(3x+3)}\]

Answer 15

\( \humongous \color{orange} {\text{ZEPP}} \)
\( \giant \color{orange} {\text{ZEPP}} \)
\( \enormous \color{orange} {\text{ZEPP}} \)
\( \gigantic \color{orange} {\text{ZEPP}} \)
\( \xlarge \color{orange} {\text{ZEPP}} \)
\( \enlarged \color{orange} {\text{ZEPP}} \)

none of these work

Answer 16

\[\big\color{red}{\text{ZEPP}}\]

Answer 17

I don't know why that isn't working

Answer 18

First step would be to get the same denominator\[\large \begin{align}
\frac{5}{x^2+3x+2}+\frac{1}{x+2}&=\frac{1}{(3x+3)}\\
\\
\frac{5}{(x+1)(x+2)}+\frac{1}{x+2}&=\frac{1}{(3x+3)}\\
\\
\frac{5}{(x+1)(x+2)}+\frac{1(x+1)}{(x+2)(x+1)}&=\frac{1}{(3x+3)}\\
\\
\\
\frac{5+1(x+1)}{(x+2)(x+1)}&=\frac{1}{3*(x+1)}\\
\\
\\
\frac{5+x+1}{(x+2)(x+1)}&=\frac{1}{3*(x+1)}\\
\\
\frac{x+6}{(x+2)(x+1)}&=\frac{1}{3*(x+1)}\\
\\
\\
(x+2)(x+1)&=(x+6)*3*(x+1)\\
\\
\frac{(x+2)(x+1)}{(x+1)}&=3(x+6)\\
\\
\\
x+2&=3x+18\\
\\
x&=3x+18-2\\
\\
x&=3x+16\\
\\
x-3x&=16\\
\\
-2x&=16\\
\\
x&=\frac{16}{-2}\\
\\
x&=-8

\end{align}\]

Answer 19

My steps would be better

Answer 20

What's with all these adjective in front of my name, @Hero? o.O

Answer 21

Fails at getting bigger font size

Answer 22

Post your steps please? :D

Answer 23

I skipped the first factorization thing, otherwise it would be longer :S

Answer 24

Nah, that would be a complete waste of time

Answer 25

I can show you in vyew though

Answer 26

Hero you lazy ;C

Answer 27

Nah it's fine, I'm watching a lecture, thanks for the proposition though

Answer 28

Basically I can do these without using LCD