1/3x+80=1/2x+120

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Solution for 1/3x+80=1/2x+120 equation:



1/3x+80=1/2x+120
We move all terms to the left:
1/3x+80-(1/2x+120)=0
Domain of the equation: 3x!=0
x!=0/3
x!=0
x∈R
Domain of the equation: 2x+120)!=0
x∈R
We get rid of parentheses
1/3x-1/2x-120+80=0
We calculate fractions
2x/6x^2+(-3x)/6x^2-120+80=0
We add all the numbers together, and all the variables
2x/6x^2+(-3x)/6x^2-40=0
We multiply all the terms by the denominator
2x+(-3x)-40*6x^2=0
Wy multiply elements
-240x^2+2x+(-3x)=0
We get rid of parentheses
-240x^2+2x-3x=0
We add all the numbers together, and all the variables
-240x^2-1x=0
a = -240; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-240)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-240}=\frac{0}{-480} =0 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-240}=\frac{2}{-480} =-1/240 $

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