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(1/5)x+(1/3)=1
We move all terms to the left:
(1/5)x+(1/3)-(1)=0
Domain of the equation: 5)x!=0determiningTheFunctionDomain (1/5)x-1+(1/3)=0
x!=0/1
x!=0
x∈R
We add all the numbers together, and all the variables
(+1/5)x-1+(+1/3)=0
We multiply parentheses
x^2-1+(+1/3)=0
We get rid of parentheses
x^2-1+1/3=0
We multiply all the terms by the denominator
x^2*3+1-1*3=0
We add all the numbers together, and all the variables
x^2*3-2=0
Wy multiply elements
3x^2-2=0
a = 3; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·3·(-2)
Δ = 24
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*3}=\frac{0-2\sqrt{6}}{6} =-\frac{2\sqrt{6}}{6} =-\frac{\sqrt{6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*3}=\frac{0+2\sqrt{6}}{6} =\frac{2\sqrt{6}}{6} =\frac{\sqrt{6}}{3} $
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