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1/3k+80=1/2k+120
We move all terms to the left:
1/3k+80-(1/2k+120)=0
Domain of the equation: 3k!=0
k!=0/3
k!=0
k∈R
Domain of the equation: 2k+120)!=0We get rid of parentheses
k∈R
1/3k-1/2k-120+80=0
We calculate fractions
2k/6k^2+(-3k)/6k^2-120+80=0
We add all the numbers together, and all the variables
2k/6k^2+(-3k)/6k^2-40=0
We multiply all the terms by the denominator
2k+(-3k)-40*6k^2=0
Wy multiply elements
-240k^2+2k+(-3k)=0
We get rid of parentheses
-240k^2+2k-3k=0
We add all the numbers together, and all the variables
-240k^2-1k=0
a = -240; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-240)·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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-240}=\frac{0}{-480} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-240}=\frac{2}{-480} =-1/240 $
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