1/9(2m-1.6)=1/3(2m+4)

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Solution for 1/9(2m-1.6)=1/3(2m+4) equation:



1/9(2m-1.6)=1/3(2m+4)
We move all terms to the left:
1/9(2m-1.6)-(1/3(2m+4))=0
Domain of the equation: 9(2m-1.6)!=0
m∈R
Domain of the equation: 3(2m+4))!=0
m∈R
We calculate fractions
(3m2/(9(2m-1.6)*3(2m+4)))+(-9m2/(9(2m-1.6)*3(2m+4)))=0
We calculate terms in parentheses: +(3m2/(9(2m-1.6)*3(2m+4))), so:
3m2/(9(2m-1.6)*3(2m+4))
We multiply all the terms by the denominator
3m2
We add all the numbers together, and all the variables
3m^2
Back to the equation:
+(3m^2)
We calculate terms in parentheses: +(-9m2/(9(2m-1.6)*3(2m+4))), so:
-9m2/(9(2m-1.6)*3(2m+4))
We multiply all the terms by the denominator
-9m2
We add all the numbers together, and all the variables
-9m^2
Back to the equation:
+(-9m^2)
We add all the numbers together, and all the variables
3m^2+(-9m^2)=0
We get rid of parentheses
3m^2-9m^2=0
We add all the numbers together, and all the variables
-6m^2=0
a = -6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-6)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{-12}=0$

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