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x+3/3x+19=x/3x+4
We move all terms to the left:
x+3/3x+19-(x/3x+4)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 3x+4)!=0We get rid of parentheses
x∈R
x+3/3x-x/3x-4+19=0
We multiply all the terms by the denominator
x*3x-x-4*3x+19*3x+3=0
We add all the numbers together, and all the variables
-1x+x*3x-4*3x+19*3x+3=0
Wy multiply elements
3x^2-1x-12x+57x+3=0
We add all the numbers together, and all the variables
3x^2+44x+3=0
a = 3; b = 44; c = +3;
Δ = b2-4ac
Δ = 442-4·3·3
Δ = 1900
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-10\sqrt{19}}{2*3}=\frac{-44-10\sqrt{19}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+10\sqrt{19}}{2*3}=\frac{-44+10\sqrt{19}}{6} $
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