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(1)/(2)x+4=(2)/(3)x
We move all terms to the left:
(1)/(2)x+4-((2)/(3)x)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/2x-(+2/3x)+4=0
We get rid of parentheses
1/2x-2/3x+4=0
We calculate fractions
3x/6x^2+(-4x)/6x^2+4=0
We multiply all the terms by the denominator
3x+(-4x)+4*6x^2=0
Wy multiply elements
24x^2+3x+(-4x)=0
We get rid of parentheses
24x^2+3x-4x=0
We add all the numbers together, and all the variables
24x^2-1x=0
a = 24; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·24·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*24}=\frac{0}{48} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*24}=\frac{2}{48} =1/24 $
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