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(1)/(5)x-4=(1)/(10)x
We move all terms to the left:
(1)/(5)x-4-((1)/(10)x)=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-(+1/10x)-4=0
We get rid of parentheses
1/5x-1/10x-4=0
We calculate fractions
10x/50x^2+(-5x)/50x^2-4=0
We multiply all the terms by the denominator
10x+(-5x)-4*50x^2=0
Wy multiply elements
-200x^2+10x+(-5x)=0
We get rid of parentheses
-200x^2+10x-5x=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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-200}=\frac{-10}{-400} =1/40 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-200}=\frac{0}{-400} =0 $
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