algebra question: solve (2x)/3-14=4

haleofwi

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Nov 1, 2006
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I'm not sure how to work this one. Can anyone help?

. . .(2x)/3 - 14 = 4

This is what I did:

. . .2x/3 - 14 = 4
. . .2x/-11 = 4
. . .2x/-11 + 11 = 4 + 11
. . .2x = 15
. . .2x/2 = 15/2
. . .x = 15/2

Is this right?
 

Denis

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Feb 17, 2004
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Re: algebra question: (2x)/3-14=4

haleofwi said:
I'm not sure how to work this one. Can anyone help?
(2x)/3-14=4
This is what I did: 2x/3-14=4
2x/-11=4
2x/-11+11=4+11
2x=15
2x/2=15/2
x=15/2 ?
is this right???
No. You need to start by adding 14 to both sides:
(2x) / 3 = 18
 

stapel

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Feb 4, 2004
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haleofwi said:
(2x)/3 - 14 = 4
Does this mean the following?

. . . . .\(\displaystyle \L \frac{2x}{3}\, -\,14\, =\, 4\)

Or this?

. . . . .\(\displaystyle \L \frac{2x}{3\, -\, 14}\, =\, 4\)

If the former, then you cannot combine the 3 and the 14 into the denominator, since the 14 isn't in there anyway.

haleofwi said:
2x/-11 + 11 = 4 + 11
Since the "11" is in the denominator, how could adding an 11 possibly "cancel" this off?


. . . . .\(\displaystyle \L \frac{2x}{-11}\, + \, 11\)

And even if it did somehow cancel the denominator, you would end up with a zero in the denominator, which isn't allowed.

Have you not studied fractions previously...?

Eliz.
 
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