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1/2k+3=1/3k-1
We move all terms to the left:
1/2k+3-(1/3k-1)=0
Domain of the equation: 2k!=0
k!=0/2
k!=0
k∈R
Domain of the equation: 3k-1)!=0We get rid of parentheses
k∈R
1/2k-1/3k+1+3=0
We calculate fractions
3k/6k^2+(-2k)/6k^2+1+3=0
We add all the numbers together, and all the variables
3k/6k^2+(-2k)/6k^2+4=0
We multiply all the terms by the denominator
3k+(-2k)+4*6k^2=0
Wy multiply elements
24k^2+3k+(-2k)=0
We get rid of parentheses
24k^2+3k-2k=0
We add all the numbers together, and all the variables
24k^2+k=0
a = 24; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·24·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*24}=\frac{-2}{48} =-1/24 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*24}=\frac{0}{48} =0 $
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