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2/3x+4=1/6x+-5
We move all terms to the left:
2/3x+4-(1/6x+-5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 6x+-5)!=0We add all the numbers together, and all the variables
x∈R
2/3x-(1/6x-5)+4=0
We get rid of parentheses
2/3x-1/6x+5+4=0
We calculate fractions
12x/18x^2+(-3x)/18x^2+5+4=0
We add all the numbers together, and all the variables
12x/18x^2+(-3x)/18x^2+9=0
We multiply all the terms by the denominator
12x+(-3x)+9*18x^2=0
Wy multiply elements
162x^2+12x+(-3x)=0
We get rid of parentheses
162x^2+12x-3x=0
We add all the numbers together, and all the variables
162x^2+9x=0
a = 162; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·162·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*162}=\frac{-18}{324} =-1/18 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*162}=\frac{0}{324} =0 $
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