1/3x+140=4/5x

Solution for 1/3x+140=4/5x equation:

1/3x+140=4/5x

We move all terms to the left:

1/3x+140-(4/5x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x-(+4/5x)+140=0

We get rid of parentheses

1/3x-4/5x+140=0

We calculate fractions


5x/15x^2+(-12x)/15x^2+140=0

We multiply all the terms by the denominator


5x+(-12x)+140*15x^2=0

Wy multiply elements

2100x^2+5x+(-12x)=0

We get rid of parentheses

2100x^2+5x-12x=0

We add all the numbers together, and all the variables

2100x^2-7x=0

a = 2100; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2100·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2100}=\frac{0}{4200}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2100}=\frac{14}{4200}
=1/300
$