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1/2j-16=2/3j-14
We move all terms to the left:
1/2j-16-(2/3j-14)=0
Domain of the equation: 2j!=0
j!=0/2
j!=0
j∈R
Domain of the equation: 3j-14)!=0We get rid of parentheses
j∈R
1/2j-2/3j+14-16=0
We calculate fractions
3j/6j^2+(-4j)/6j^2+14-16=0
We add all the numbers together, and all the variables
3j/6j^2+(-4j)/6j^2-2=0
We multiply all the terms by the denominator
3j+(-4j)-2*6j^2=0
Wy multiply elements
-12j^2+3j+(-4j)=0
We get rid of parentheses
-12j^2+3j-4j=0
We add all the numbers together, and all the variables
-12j^2-1j=0
a = -12; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-12)·0
Δ = 1
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}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-12}=\frac{0}{-24} =0 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-12}=\frac{2}{-24} =-1/12 $
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