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2/3x+3=4/5x-5
We move all terms to the left:
2/3x+3-(4/5x-5)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 5x-5)!=0We get rid of parentheses
x∈R
2/3x-4/5x+5+3=0
We calculate fractions
10x/15x^2+(-12x)/15x^2+5+3=0
We add all the numbers together, and all the variables
10x/15x^2+(-12x)/15x^2+8=0
We multiply all the terms by the denominator
10x+(-12x)+8*15x^2=0
Wy multiply elements
120x^2+10x+(-12x)=0
We get rid of parentheses
120x^2+10x-12x=0
We add all the numbers together, and all the variables
120x^2-2x=0
a = 120; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·120·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*120}=\frac{0}{240} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*120}=\frac{4}{240} =1/60 $
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