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1/5x+2=1/2x-4
We move all terms to the left:
1/5x+2-(1/2x-4)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x-4)!=0We get rid of parentheses
x∈R
1/5x-1/2x+4+2=0
We calculate fractions
2x/10x^2+(-5x)/10x^2+4+2=0
We add all the numbers together, and all the variables
2x/10x^2+(-5x)/10x^2+6=0
We multiply all the terms by the denominator
2x+(-5x)+6*10x^2=0
Wy multiply elements
60x^2+2x+(-5x)=0
We get rid of parentheses
60x^2+2x-5x=0
We add all the numbers together, and all the variables
60x^2-3x=0
a = 60; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·60·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*60}=\frac{0}{120} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*60}=\frac{6}{120} =1/20 $
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