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T0mmy Group TitleBest ResponseYou've already chosen the best response.0
Do you want to find 'a'?
 one year ago

howdomath Group TitleBest ResponseYou've already chosen the best response.0
Being by isolating the variable, to do this we divide both sides of the equation by 5/4. Remember that dividing a fraction by another fraction is really multiplying the first fraction by the reciprocal of the dividing fraction. \[(a  8) = \frac{ 2 }{ 3 } \times \frac{ 4 }{ 5 }\] We get\[a  8 = \frac{ 8 }{ 15 }\] Now you add 8 to both sides to get a on it's own..\[a = \frac{ 8 }{ 15 } +8\] Now we need a common denominator to add these fractions, so..\[a = \frac{ 8 }{ 15 } +\frac{ 120 }{ 15 }\] Now we have \[a = \frac{ 128 }{ 15 }\]
 one year ago

horsealot203 Group TitleBest ResponseYou've already chosen the best response.0
okkk...
 one year ago

howdomath Group TitleBest ResponseYou've already chosen the best response.0
What doesn't make sense, I'll try to explain better.
 one year ago

horsealot203 Group TitleBest ResponseYou've already chosen the best response.0
so can you simplify the fraction?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
No, factors of 15 are 3,5, and 128 is not divisible by either one.
 one year ago

T0mmy Group TitleBest ResponseYou've already chosen the best response.0
No but you can create a mixed fraction.
 one year ago

howdomath Group TitleBest ResponseYou've already chosen the best response.0
Well look at the factors of 15, there's 1, 3, 5 and 15. 128 isn't divisible by any of those, so it can't be simplified. If you want it in mixed number it's:\[8\frac{ 8 }{ 15 }\]
 one year ago

horsealot203 Group TitleBest ResponseYou've already chosen the best response.0
The answers I can choose from are 6 14/15 7 1/15 8 8/15 8 2/3
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
You could write it as a mixed fraction = \[8\frac{ 8 }{ 15 }\]
 one year ago

horsealot203 Group TitleBest ResponseYou've already chosen the best response.0
ok, great. Thank you so much!
 one year ago
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