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-5=1/3(Y+12)
We move all terms to the left:
-5-(1/3(Y+12))=0
Domain of the equation: 3(Y+12))!=0We multiply all the terms by the denominator
Y∈R
-(1-5*3(Y+12))=0
We calculate terms in parentheses: -(1-5*3(Y+12)), so:
1-5*3(Y+12)
determiningTheFunctionDomain -5*3(Y+12)+1
Wy multiply elements
-15Y(Y+1
Back to the equation:
-(-15Y(Y+1)
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| 3s+12=16 | | y+2y-5-5y=7 | | (2x^2)=(2x^2)=64 | | w-8/9=7 | | 19.50+2x+2.50=36.50 | | z=3=9/z-6=-2 | | 19.50+2x+2.50=36.59 | | 2a-12a=12-8a | | 2j+2j+3j-4j=9 | | (3x-4)+(3x-4)=40 | | 16-2b=5b+9 | | -4(x+4)+2x=-10 | | 1/3+q=2/3 | | -4x=9(x+4) | | y/6=5/6 | | 20-8k=8k+4 | | 60=16n | | (3x-1)+4=2x | | z-8=1/2 | | 7-2x=97 | | -4x=3(x-6) | | v^2−6v=−91 | | 3v+9v=60 | | q-7=1/2 | | 1/2y=-30 | | -7-1/3x=1/3x+3 | | n/5-6=24 | | -155h;h=-5 | | n-7/10=8 | | 2k-4=-29-3k | | 3.14r^2=64 | | -8v=2-9v |