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(2a+1)/3)+(a-2)/4)=9
We move all terms to the left:
(2a+1)/3)+(a-2)/4)-(9)=0
Domain of the equation: 3)+(a!=0We calculate fractions
a∈R
((2a+1)*4)-9)/4a^2+(-2)*3)+a/4a^2=0
We calculate terms in parentheses: +((2a+1)*4), so:We get rid of parentheses
(2a+1)*4
We multiply parentheses
8a+4
Back to the equation:
+(8a+4)
8a-9)/4a^2+(-2)*3)+a/4a^2+4=0
We calculate fractions
8a+(-9)*4a^2+4)/(4a^2+(*4a^2+4)+(-2)*3)+a*4a^2+()/(4a^2+(*4a^2+4)=0
We get rid of parentheses
8a+(-9)*4a^2+4)/(4a^2+*4a^2+(-2)*3)+a*4a^2+()/(4a^2+*4a^2+4+4=0
We multiply all the terms by the denominator
8a*(4a^2+((-9)*4a^2)*(4a^2+4)+(*4a^2)*(4a^2+(-2)*3)+a*4a^2+()+(*4a^2)*(4a^2+4*(4a^2+4*(4a^2=0
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