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1/2x+16=2/3x-4
We move all terms to the left:
1/2x+16-(2/3x-4)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 3x-4)!=0We get rid of parentheses
x∈R
1/2x-2/3x+4+16=0
We calculate fractions
3x/6x^2+(-4x)/6x^2+4+16=0
We add all the numbers together, and all the variables
3x/6x^2+(-4x)/6x^2+20=0
We multiply all the terms by the denominator
3x+(-4x)+20*6x^2=0
Wy multiply elements
120x^2+3x+(-4x)=0
We get rid of parentheses
120x^2+3x-4x=0
We add all the numbers together, and all the variables
120x^2-1x=0
a = 120; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·120·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*120}=\frac{0}{240} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*120}=\frac{2}{240} =1/120 $
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