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4/5x-3=3/10x+5
We move all terms to the left:
4/5x-3-(3/10x+5)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 10x+5)!=0We get rid of parentheses
x∈R
4/5x-3/10x-5-3=0
We calculate fractions
40x/50x^2+(-15x)/50x^2-5-3=0
We add all the numbers together, and all the variables
40x/50x^2+(-15x)/50x^2-8=0
We multiply all the terms by the denominator
40x+(-15x)-8*50x^2=0
Wy multiply elements
-400x^2+40x+(-15x)=0
We get rid of parentheses
-400x^2+40x-15x=0
We add all the numbers together, and all the variables
-400x^2+25x=0
a = -400; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-400)·0
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-400}=\frac{-50}{-800} =1/16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-400}=\frac{0}{-800} =0 $
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