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5/6x+4=3/4x+12
We move all terms to the left:
5/6x+4-(3/4x+12)=0
Domain of the equation: 6x!=0
x!=0/6
x!=0
x∈R
Domain of the equation: 4x+12)!=0We get rid of parentheses
x∈R
5/6x-3/4x-12+4=0
We calculate fractions
20x/24x^2+(-18x)/24x^2-12+4=0
We add all the numbers together, and all the variables
20x/24x^2+(-18x)/24x^2-8=0
We multiply all the terms by the denominator
20x+(-18x)-8*24x^2=0
Wy multiply elements
-192x^2+20x+(-18x)=0
We get rid of parentheses
-192x^2+20x-18x=0
We add all the numbers together, and all the variables
-192x^2+2x=0
a = -192; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-192)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-192}=\frac{-4}{-384} =1/96 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-192}=\frac{0}{-384} =0 $
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