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1+1/2x=1/5x
We move all terms to the left:
1+1/2x-(1/5x)=0
Domain of the equation: 2x!=0
x!=0/2
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
1/2x-(+1/5x)+1=0
We get rid of parentheses
1/2x-1/5x+1=0
We calculate fractions
5x/10x^2+(-2x)/10x^2+1=0
We multiply all the terms by the denominator
5x+(-2x)+1*10x^2=0
Wy multiply elements
10x^2+5x+(-2x)=0
We get rid of parentheses
10x^2+5x-2x=0
We add all the numbers together, and all the variables
10x^2+3x=0
a = 10; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·10·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*10}=\frac{-6}{20} =-3/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*10}=\frac{0}{20} =0 $
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