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2/3x+4=1/5x
We move all terms to the left:
2/3x+4-(1/5x)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 5x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2/3x-(+1/5x)+4=0
We get rid of parentheses
2/3x-1/5x+4=0
We calculate fractions
10x/15x^2+(-3x)/15x^2+4=0
We multiply all the terms by the denominator
10x+(-3x)+4*15x^2=0
Wy multiply elements
60x^2+10x+(-3x)=0
We get rid of parentheses
60x^2+10x-3x=0
We add all the numbers together, and all the variables
60x^2+7x=0
a = 60; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·60·0
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*60}=\frac{-14}{120} =-7/60 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*60}=\frac{0}{120} =0 $
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