Mathematics
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OpenStudy (thadds2003):
Help! Fan and medal.
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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):
\[ 2*9 = \frac{8x-2}{9}*9\]
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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.
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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?
\[2∗9=(8x−2)∗\frac{9}{9}\]
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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.
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OpenStudy (anonymous):
sorry, it's three equations. the <==> just means one is the same as the other
OpenStudy (anonymous):
\[x=20/8\]