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srry cant help
28+7a=7(4+1)
Thanks anyway
=7(4+a)
It's supposed to be a final answer of a fraction
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a^2-16=(a+4)(a-4) so a=4 will get cancelled
2 fraction are in multiplication
from those first solve one on right side ..we will multiply them later
so on right we left with\[\frac{ 7 }{ a-4 }\]
now one on left
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okay,
3a^2-13a+4=3a^2-12a-a+4 3a(a-4)-1(a-4)=(a-4)(3a-1)
so now we have (7)/(a-4) and (a-4)/(3a-1)?
yeah multiply them
\[\frac{ a-4 }{ 3a-1 }*\frac{ 7 }{ a-4}\]
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okay
a-4 will get cancelled
so it becomes 7 over 3a-1?
yeppp
THANK YOU SO MUCH!
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welcm :)
Can you help me with my last problem also?
yep post new one
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