solve the equation. check for extraneous solutions. will fan and medal

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- anonymous

solve the equation. check for extraneous solutions. will fan and medal

- katieb

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- anonymous

\[9\left| 9-8x \right|=2x+3\]

- SolomonZelman

well, you will get two results:
\(9\left| 9-8x \right|=2x+3\)
\(9(9-8x )=2x+3\) OR \(9(9-8x )=-2x-3\)

- SolomonZelman

solve each case individually.

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## More answers

- anonymous

i dont know where to start. will you walk me through it?

- SolomonZelman

well, expand the right side in case 1

- SolomonZelman

I mean the left side (not the right side)

- anonymous

would you move the x to the left side first?

- SolomonZelman

\(9(9-8x )=2x+3\)
\(9\cdot9-9\cdot8x =2x+3\)
\(81-72x =2x+3\)

- SolomonZelman

So in case two you would have the same, but the right side would be -2x-3

- SolomonZelman

Solve for each:
\(81-72x =2x+3\) \(81-72x =-2x-3\)

- anonymous

x would go on the left side?

- SolomonZelman

you can add 72x to both sides in case 2 and case 1

- anonymous

leaving us with 81=74x+3

- SolomonZelman

two cases:
\(81-72x =2x+3\)
\(81-72x+72x =2x+72x+3\)
\(81=74x+3\)
right!

- SolomonZelman

then subtract 3 from both sides.

- anonymous

78=74x?

- SolomonZelman

yes, so x=?

- anonymous

74/78?

- SolomonZelman

yes

- anonymous

what about in the second case?

- SolomonZelman

Now in case 2:
\(81-72x =-2x-3\)

- SolomonZelman

you also add 72x to both sides,
THEN, add 3 to both sides

- anonymous

x=70/84?

- SolomonZelman

oh the first one is incorrect, and this one too

- anonymous

noooo

- SolomonZelman

when in case one you have:
78=74x
you divide by 74 on both sides, and thus
x=78/74 (not 74/78)

- SolomonZelman

And in case two you made the same mistake

- anonymous

so it would be 84/70?

- SolomonZelman

Yes, for case 2.

- SolomonZelman

And they are not extraneous.

- SolomonZelman

let me say correctly what would have been extraneous:

- SolomonZelman

Any x-solution, in this case, that is equal to \(C\) that satisfies:
\(C<-2/3\)

- SolomonZelman

because if x is less than -2/3
then 2x+3 is negative, and absolute value can NOT e equal to negative
(since absolute value is really a distance)

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