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2/3x+1=4/5x-3
We move all terms to the left:
2/3x+1-(4/5x-3)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 5x-3)!=0We get rid of parentheses
x∈R
2/3x-4/5x+3+1=0
We calculate fractions
10x/15x^2+(-12x)/15x^2+3+1=0
We add all the numbers together, and all the variables
10x/15x^2+(-12x)/15x^2+4=0
We multiply all the terms by the denominator
10x+(-12x)+4*15x^2=0
Wy multiply elements
60x^2+10x+(-12x)=0
We get rid of parentheses
60x^2+10x-12x=0
We add all the numbers together, and all the variables
60x^2-2x=0
a = 60; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·60·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*60}=\frac{0}{120} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*60}=\frac{4}{120} =1/30 $
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