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1/2q+3/5=2/3q
We move all terms to the left:
1/2q+3/5-(2/3q)=0
Domain of the equation: 2q!=0
q!=0/2
q!=0
q∈R
Domain of the equation: 3q)!=0We add all the numbers together, and all the variables
q!=0/1
q!=0
q∈R
1/2q-(+2/3q)+3/5=0
We get rid of parentheses
1/2q-2/3q+3/5=0
We calculate fractions
54q^2/150q^2+75q/150q^2+(-100q)/150q^2=0
We multiply all the terms by the denominator
54q^2+75q+(-100q)=0
We get rid of parentheses
54q^2+75q-100q=0
We add all the numbers together, and all the variables
54q^2-25q=0
a = 54; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·54·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*54}=\frac{0}{108} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*54}=\frac{50}{108} =25/54 $
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