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1/3x+x=32
We move all terms to the left:
1/3x+x-(32)=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-32=0
We multiply all the terms by the denominator
x*3x-32*3x+1=0
Wy multiply elements
3x^2-96x+1=0
a = 3; b = -96; c = +1;
Δ = b2-4ac
Δ = -962-4·3·1
Δ = 9204
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9204}=\sqrt{4*2301}=\sqrt{4}*\sqrt{2301}=2\sqrt{2301}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-2\sqrt{2301}}{2*3}=\frac{96-2\sqrt{2301}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+2\sqrt{2301}}{2*3}=\frac{96+2\sqrt{2301}}{6} $
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