1/3x+40=1/2x-50

Solution for 1/3x+40=1/2x-50 equation:

1/3x+40=1/2x-50

We move all terms to the left:

1/3x+40-(1/2x-50)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 2x-50)!=0

x∈R


We get rid of parentheses

1/3x-1/2x+50+40=0

We calculate fractions


2x/6x^2+(-3x)/6x^2+50+40=0

We add all the numbers together, and all the variables

2x/6x^2+(-3x)/6x^2+90=0

We multiply all the terms by the denominator


2x+(-3x)+90*6x^2=0

Wy multiply elements

540x^2+2x+(-3x)=0

We get rid of parentheses

540x^2+2x-3x=0

We add all the numbers together, and all the variables

540x^2-1x=0

a = 540; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·540·0
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*540}=\frac{0}{1080}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*540}=\frac{2}{1080}
=1/540
$