anonymous
  • anonymous
Classify the solutions of 3 over x plus 5, plus one fifth, equals 2 over x plus 5 as extraneous or non-extraneous.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Algorithmic @ganeshie8 @CaptainLlama49
anonymous
  • anonymous
is the answer extraneous?
Mertsj
  • Mertsj
I have typed this twice and lost it both times. I will try one more time.

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More answers

anonymous
  • anonymous
ok:)
Mertsj
  • Mertsj
\[\frac{3}{x+5}+\frac{1}{5}=\frac{2}{x+5}\]
Mertsj
  • Mertsj
Common denominator is 5(x+5). Multiply each term by that.
anonymous
  • anonymous
so you get 15/5(x+5)+x+5/5(x+5)=10/5(x+5)
Mertsj
  • Mertsj
\[5(x+5)(\frac{3}{x+5})+5(x+5)(\frac{1}{5})=5(x+5)(\frac{3}{x+5})\]
Mertsj
  • Mertsj
Cancel all the denominators.
Mertsj
  • Mertsj
\[15 +x+5=10\]
anonymous
  • anonymous
yup! :)
Mertsj
  • Mertsj
The 3 on the right side is supposed to be a 2
Mertsj
  • Mertsj
\[x=-10\]
Mertsj
  • Mertsj
Now. Is that extraneous or not? Does it cause any denominator to be 0? Does it make the original equation true?
anonymous
  • anonymous
extraneous?
Mertsj
  • Mertsj
Why is it extraneous?
anonymous
  • anonymous
3/ -5 +1/5 = 2/-5
anonymous
  • anonymous
I don't know. When i plugged it in, that's the equation I got above,
Mertsj
  • Mertsj
\[-\frac{3}{5}+\frac{1}{5}=-\frac{2}{5}\]
Mertsj
  • Mertsj
Is that true or not?
Mertsj
  • Mertsj
Is any denominator equal to 0 ?
anonymous
  • anonymous
no
Mertsj
  • Mertsj
Then it is a perfectly fine solution and is not extraneous.
anonymous
  • anonymous
Oh!
anonymous
  • anonymous
Thank you:)
anonymous
  • anonymous
Can I ask one more?
Mertsj
  • Mertsj
An extraneous solution is a number that appears to be a solution but doesn't hold up to scrutiny because it either doesn't check or causes a denominator to be 0.
Mertsj
  • Mertsj
Yes
anonymous
  • anonymous
Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus 5 over 10.
anonymous
  • anonymous
Okay! I understand now:)
anonymous
  • anonymous
For this one, I got x=-3,2
Mertsj
  • Mertsj
\[\frac{2}{5}+\frac{3}{5x}=\frac{x+5}{10}\]
Mertsj
  • Mertsj
Is that it?
anonymous
  • anonymous
yeah:)
Mertsj
  • Mertsj
\[10x(\frac{2}{5})+10x(\frac{3}{5x})=10x(\frac{x+5}{10})\]
Mertsj
  • Mertsj
\[4x+6=x^2+5x\]
anonymous
  • anonymous
-1x^2+5x+36
Mertsj
  • Mertsj
\[x^2+x-6=0\]
Mertsj
  • Mertsj
\[(x+3)(x-2)=0\]
anonymous
  • anonymous
So, -3 and 2 ?
Mertsj
  • Mertsj
\[x=-3,2\]
anonymous
  • anonymous
Last one:)?
Mertsj
  • Mertsj
\[\frac{2}{5}+\frac{3}{5(-3)}=\frac{-3+5}{10}\]
Mertsj
  • Mertsj
\[\frac{2}{5}-\frac{3}{15}=\frac{2}{10}\]
Mertsj
  • Mertsj
\[\frac{2}{5}-\frac{1}{5}=\frac{1}{5}\]
Mertsj
  • Mertsj
That seems to check. Try 2. See if it checks.
anonymous
  • anonymous
Okay, that makes sense!! :D
anonymous
  • anonymous
Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or non-extraneous.
anonymous
  • anonymous
So, I got -4
Mertsj
  • Mertsj
\[\frac{1}{x+4}+\frac{1}{2}=\frac{1}{x+4}\]
Mertsj
  • Mertsj
And -4 causes the denominator to be 0 so it is extraneous.
anonymous
  • anonymous
yes, I agree
Mertsj
  • Mertsj
Good job!!
anonymous
  • anonymous
Solve for x: 1 over 2 plus 4 over 2 x equals quantity x plus 4 over 10.
Mertsj
  • Mertsj
Me too. Did you check them?
Mertsj
  • Mertsj
They both check.

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