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1/5x-1/5=1/10x-2
We move all terms to the left:
1/5x-1/5-(1/10x-2)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x-2)!=0We get rid of parentheses
x∈R
1/5x-1/10x+2-1/5=0
We calculate fractions
10x/1250x^2+(-125x)/1250x^2+(-10x)/1250x^2+2=0
We multiply all the terms by the denominator
10x+(-125x)+(-10x)+2*1250x^2=0
Wy multiply elements
2500x^2+10x+(-125x)+(-10x)=0
We get rid of parentheses
2500x^2+10x-125x-10x=0
We add all the numbers together, and all the variables
2500x^2-125x=0
a = 2500; b = -125; c = 0;
Δ = b2-4ac
Δ = -1252-4·2500·0
Δ = 15625
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{15625}=125$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-125)-125}{2*2500}=\frac{0}{5000} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-125)+125}{2*2500}=\frac{250}{5000} =1/20 $
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