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