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