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1/2(12x-16)=1/5(20x-15)
We move all terms to the left:
1/2(12x-16)-(1/5(20x-15))=0
Domain of the equation: 2(12x-16)!=0
x∈R
Domain of the equation: 5(20x-15))!=0We calculate fractions
x∈R
(5x2/(2(12x-16)*5(20x-15)))+(-2x1/(2(12x-16)*5(20x-15)))=0
We calculate terms in parentheses: +(5x2/(2(12x-16)*5(20x-15))), so:
5x2/(2(12x-16)*5(20x-15))
We multiply all the terms by the denominator
5x2
We add all the numbers together, and all the variables
5x^2
Back to the equation:
+(5x^2)
We calculate terms in parentheses: +(-2x1/(2(12x-16)*5(20x-15))), so:We get rid of parentheses
-2x1/(2(12x-16)*5(20x-15))
We multiply all the terms by the denominator
-2x1
We add all the numbers together, and all the variables
-2x
Back to the equation:
+(-2x)
5x^2-2x=0
a = 5; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·5·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*5}=\frac{0}{10} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*5}=\frac{4}{10} =2/5 $
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