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2/3x+5=5/6x+3
We move all terms to the left:
2/3x+5-(5/6x+3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 6x+3)!=0We get rid of parentheses
x∈R
2/3x-5/6x-3+5=0
We calculate fractions
12x/18x^2+(-15x)/18x^2-3+5=0
We add all the numbers together, and all the variables
12x/18x^2+(-15x)/18x^2+2=0
We multiply all the terms by the denominator
12x+(-15x)+2*18x^2=0
Wy multiply elements
36x^2+12x+(-15x)=0
We get rid of parentheses
36x^2+12x-15x=0
We add all the numbers together, and all the variables
36x^2-3x=0
a = 36; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·36·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*36}=\frac{0}{72} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*36}=\frac{6}{72} =1/12 $
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