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1/3z-3=5-3/2z
We move all terms to the left:
1/3z-3-(5-3/2z)=0
Domain of the equation: 3z!=0
z!=0/3
z!=0
z∈R
Domain of the equation: 2z)!=0We add all the numbers together, and all the variables
z!=0/1
z!=0
z∈R
1/3z-(-3/2z+5)-3=0
We get rid of parentheses
1/3z+3/2z-5-3=0
We calculate fractions
2z/6z^2+9z/6z^2-5-3=0
We add all the numbers together, and all the variables
2z/6z^2+9z/6z^2-8=0
We multiply all the terms by the denominator
2z+9z-8*6z^2=0
We add all the numbers together, and all the variables
11z-8*6z^2=0
Wy multiply elements
-48z^2+11z=0
a = -48; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·(-48)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*-48}=\frac{-22}{-96} =11/48 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*-48}=\frac{0}{-96} =0 $
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