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2/5x-16=1/3x+10
We move all terms to the left:
2/5x-16-(1/3x+10)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 3x+10)!=0We get rid of parentheses
x∈R
2/5x-1/3x-10-16=0
We calculate fractions
6x/15x^2+(-5x)/15x^2-10-16=0
We add all the numbers together, and all the variables
6x/15x^2+(-5x)/15x^2-26=0
We multiply all the terms by the denominator
6x+(-5x)-26*15x^2=0
Wy multiply elements
-390x^2+6x+(-5x)=0
We get rid of parentheses
-390x^2+6x-5x=0
We add all the numbers together, and all the variables
-390x^2+x=0
a = -390; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-390)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-390}=\frac{-2}{-780} =1/390 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-390}=\frac{0}{-780} =0 $
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