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1/5x+2=1/4x
We move all terms to the left:
1/5x+2-(1/4x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 4x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/5x-(+1/4x)+2=0
We get rid of parentheses
1/5x-1/4x+2=0
We calculate fractions
4x/20x^2+(-5x)/20x^2+2=0
We multiply all the terms by the denominator
4x+(-5x)+2*20x^2=0
Wy multiply elements
40x^2+4x+(-5x)=0
We get rid of parentheses
40x^2+4x-5x=0
We add all the numbers together, and all the variables
40x^2-1x=0
a = 40; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·40·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*40}=\frac{0}{80} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*40}=\frac{2}{80} =1/40 $
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