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2/3x+3=1/2x+5
We move all terms to the left:
2/3x+3-(1/2x+5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x+5)!=0We get rid of parentheses
x∈R
2/3x-1/2x-5+3=0
We calculate fractions
4x/6x^2+(-3x)/6x^2-5+3=0
We add all the numbers together, and all the variables
4x/6x^2+(-3x)/6x^2-2=0
We multiply all the terms by the denominator
4x+(-3x)-2*6x^2=0
Wy multiply elements
-12x^2+4x+(-3x)=0
We get rid of parentheses
-12x^2+4x-3x=0
We add all the numbers together, and all the variables
-12x^2+x=0
a = -12; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-12)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-12}=\frac{-2}{-24} =1/12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-12}=\frac{0}{-24} =0 $
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