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1/5x+4=3/10x
We move all terms to the left:
1/5x+4-(3/10x)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/5x-(+3/10x)+4=0
We get rid of parentheses
1/5x-3/10x+4=0
We calculate fractions
10x/50x^2+(-15x)/50x^2+4=0
We multiply all the terms by the denominator
10x+(-15x)+4*50x^2=0
Wy multiply elements
200x^2+10x+(-15x)=0
We get rid of parentheses
200x^2+10x-15x=0
We add all the numbers together, and all the variables
200x^2-5x=0
a = 200; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·200·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*200}=\frac{0}{400} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*200}=\frac{10}{400} =1/40 $
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