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2/3x+1/10x=19
We move all terms to the left:
2/3x+1/10x-(19)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 10x!=0We calculate fractions
x!=0/10
x!=0
x∈R
20x/30x^2+3x/30x^2-19=0
We multiply all the terms by the denominator
20x+3x-19*30x^2=0
We add all the numbers together, and all the variables
23x-19*30x^2=0
Wy multiply elements
-570x^2+23x=0
a = -570; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·(-570)·0
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*-570}=\frac{-46}{-1140} =23/570 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*-570}=\frac{0}{-1140} =0 $
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