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j-6=1/j+3
We move all terms to the left:
j-6-(1/j+3)=0
Domain of the equation: j+3)!=0We get rid of parentheses
j∈R
j-1/j-3-6=0
We multiply all the terms by the denominator
j*j-3*j-6*j-1=0
We add all the numbers together, and all the variables
-9j+j*j-1=0
Wy multiply elements
j^2-9j-1=0
a = 1; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·1·(-1)
Δ = 85
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{85}}{2*1}=\frac{9-\sqrt{85}}{2} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{85}}{2*1}=\frac{9+\sqrt{85}}{2} $
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