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1x+1/3x=16
We move all terms to the left:
1x+1/3x-(16)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
x+1/3x-16=0
We multiply all the terms by the denominator
x*3x-16*3x+1=0
Wy multiply elements
3x^2-48x+1=0
a = 3; b = -48; c = +1;
Δ = b2-4ac
Δ = -482-4·3·1
Δ = 2292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2292}=\sqrt{4*573}=\sqrt{4}*\sqrt{573}=2\sqrt{573}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-2\sqrt{573}}{2*3}=\frac{48-2\sqrt{573}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+2\sqrt{573}}{2*3}=\frac{48+2\sqrt{573}}{6} $
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