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1/3x-2/5x=10
We move all terms to the left:
1/3x-2/5x-(10)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 5x!=0We calculate fractions
x!=0/5
x!=0
x∈R
5x/15x^2+(-6x)/15x^2-10=0
We multiply all the terms by the denominator
5x+(-6x)-10*15x^2=0
Wy multiply elements
-150x^2+5x+(-6x)=0
We get rid of parentheses
-150x^2+5x-6x=0
We add all the numbers together, and all the variables
-150x^2-1x=0
a = -150; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-150)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-150}=\frac{0}{-300} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-150}=\frac{2}{-300} =-1/150 $
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