A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or nonextraneous.
anonymous
 one year ago
Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or nonextraneous.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you draw the equation, it is difficult to understand .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439570678077:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ x+4 }+\frac{ 1 }{ 2 }=\frac{ 1 }{ x+4 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you subtract 1/(x+4) from both sides, you get 1/2 = 0, which is a contradiction. There are no solutions to this equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x = 4; extraneous x = 4; nonextraneous x = 8; extraneous x = 8; nonextraneous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these are the answers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you solve the equation $$ \frac{1}{(x+4)} + \frac 1 2 = \frac{ 1}{(x+4)} \\ \ \\ \frac{2}{2(x+4)} + \frac {(x+4)}{2(x+4)} = \frac{ 1}{(x+4)} \\ \ \\ \frac{2+ (x+4) }{2(x+4)} = \frac{ 1}{(x+4)} \\ \ \\ \frac{x+6 }{2(x+4)} = \frac{ 1}{(x+4)} \\ \ \\ x + 6 = \frac{ 1}{(x+4)} \cdot 2(x+4) \\ \ \\ x + 6 = 2 \\ \\ x = 4 $$ But 4 does not satisfy original equation. Hence we say 4 is 'extraneous'
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.