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1/5x+2/3x=13
We move all terms to the left:
1/5x+2/3x-(13)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 3x!=0We calculate fractions
x!=0/3
x!=0
x∈R
3x/15x^2+10x/15x^2-13=0
We multiply all the terms by the denominator
3x+10x-13*15x^2=0
We add all the numbers together, and all the variables
13x-13*15x^2=0
Wy multiply elements
-195x^2+13x=0
a = -195; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-195)·0
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-195}=\frac{-26}{-390} =1/15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-195}=\frac{0}{-390} =0 $
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