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Mathematics 7 Online
OpenStudy (anonymous):

Classify the solutions of 3 over x plus 5, plus one fifth, equals 2 over x plus 5 as extraneous or non-extraneous.

OpenStudy (anonymous):

@Algorithmic @ganeshie8 @CaptainLlama49

OpenStudy (anonymous):

is the answer extraneous?

OpenStudy (mertsj):

I have typed this twice and lost it both times. I will try one more time.

OpenStudy (anonymous):

ok:)

OpenStudy (mertsj):

\[\frac{3}{x+5}+\frac{1}{5}=\frac{2}{x+5}\]

OpenStudy (mertsj):

Common denominator is 5(x+5). Multiply each term by that.

OpenStudy (anonymous):

so you get 15/5(x+5)+x+5/5(x+5)=10/5(x+5)

OpenStudy (mertsj):

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

OpenStudy (mertsj):

Cancel all the denominators.

OpenStudy (mertsj):

\[15 +x+5=10\]

OpenStudy (anonymous):

yup! :)

OpenStudy (mertsj):

The 3 on the right side is supposed to be a 2

OpenStudy (mertsj):

\[x=-10\]

OpenStudy (mertsj):

Now. Is that extraneous or not? Does it cause any denominator to be 0? Does it make the original equation true?

OpenStudy (anonymous):

extraneous?

OpenStudy (mertsj):

Why is it extraneous?

OpenStudy (anonymous):

3/ -5 +1/5 = 2/-5

OpenStudy (anonymous):

I don't know. When i plugged it in, that's the equation I got above,

OpenStudy (mertsj):

\[-\frac{3}{5}+\frac{1}{5}=-\frac{2}{5}\]

OpenStudy (mertsj):

Is that true or not?

OpenStudy (mertsj):

Is any denominator equal to 0 ?

OpenStudy (anonymous):

no

OpenStudy (mertsj):

Then it is a perfectly fine solution and is not extraneous.

OpenStudy (anonymous):

Oh!

OpenStudy (anonymous):

Thank you:)

OpenStudy (anonymous):

Can I ask one more?

OpenStudy (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.

OpenStudy (mertsj):

Yes

OpenStudy (anonymous):

Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus 5 over 10.

OpenStudy (anonymous):

Okay! I understand now:)

OpenStudy (anonymous):

For this one, I got x=-3,2

OpenStudy (mertsj):

\[\frac{2}{5}+\frac{3}{5x}=\frac{x+5}{10}\]

OpenStudy (mertsj):

Is that it?

OpenStudy (anonymous):

yeah:)

OpenStudy (mertsj):

\[10x(\frac{2}{5})+10x(\frac{3}{5x})=10x(\frac{x+5}{10})\]

OpenStudy (mertsj):

\[4x+6=x^2+5x\]

OpenStudy (anonymous):

-1x^2+5x+36

OpenStudy (mertsj):

\[x^2+x-6=0\]

OpenStudy (mertsj):

\[(x+3)(x-2)=0\]

OpenStudy (anonymous):

So, -3 and 2 ?

OpenStudy (mertsj):

\[x=-3,2\]

OpenStudy (anonymous):

Last one:)?

OpenStudy (mertsj):

\[\frac{2}{5}+\frac{3}{5(-3)}=\frac{-3+5}{10}\]

OpenStudy (mertsj):

\[\frac{2}{5}-\frac{3}{15}=\frac{2}{10}\]

OpenStudy (mertsj):

\[\frac{2}{5}-\frac{1}{5}=\frac{1}{5}\]

OpenStudy (mertsj):

That seems to check. Try 2. See if it checks.

OpenStudy (anonymous):

Okay, that makes sense!! :D

OpenStudy (anonymous):

Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or non-extraneous.

OpenStudy (anonymous):

So, I got -4

OpenStudy (mertsj):

\[\frac{1}{x+4}+\frac{1}{2}=\frac{1}{x+4}\]

OpenStudy (mertsj):

And -4 causes the denominator to be 0 so it is extraneous.

OpenStudy (anonymous):

yes, I agree

OpenStudy (mertsj):

Good job!!

OpenStudy (anonymous):

Solve for x: 1 over 2 plus 4 over 2 x equals quantity x plus 4 over 10.

OpenStudy (mertsj):

Me too. Did you check them?

OpenStudy (mertsj):

They both check.

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