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Mathematics 13 Online
OpenStudy (thadds2003):

Help! Fan and medal.

OpenStudy (thadds2003):

OpenStudy (anonymous):

First, i'd move the 9 in the denominator. So, multiply both sides by 9.

OpenStudy (thadds2003):

Wait, how?

OpenStudy (faiqraees):

You have to make x alone. So first remove 9 then -2 then 8

OpenStudy (anonymous):

\[ 2*9 = \frac{8x-2}{9}*9\]

OpenStudy (anonymous):

\[2∗9=(8x−2)*\frac{9}{9}\]

OpenStudy (thadds2003):

so, \[ \frac{ 6_{x} }{ 9 } \times 9?\]

OpenStudy (anonymous):

I'm not sure where you got the 6x.

OpenStudy (thadds2003):

It says 8x - 2, witch is 6 x.

OpenStudy (anonymous):

nope, you can't combine the 8x and the -2. Because the -2 doesn't have an x.

OpenStudy (thadds2003):

Oh, so what is that then?

OpenStudy (anonymous):

You want to get the stuff with x's on one side and the other stuff on the other

OpenStudy (thadds2003):

So I have to isolate the x's and the regulars?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Does this make sense so far? \[2∗9=(8x−2)∗\frac{9}{9}\]

OpenStudy (thadds2003):

Nope.

OpenStudy (anonymous):

ok, I took the whole equation, and multiplied both sides by 9.

OpenStudy (anonymous):

\[\frac{2}{1} = \frac{8x-2}{9}\] becomes \[\frac{2}{1}*9 = \frac{8x-2}{9}*9\]. Which is allowed because we did it to both sides

OpenStudy (anonymous):

gtg, so \[18=(8x-2)*\frac{9}{9} \iff 18=8x-2 \iff 18+2 = 8x-2+2\]

OpenStudy (thadds2003):

Dang. thats a long equation.

OpenStudy (anonymous):

sorry, it's three equations. the <==> just means one is the same as the other

OpenStudy (anonymous):

\[x=20/8\]

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