4/5x-4/3=7/3x-3

Solution for 4/5x-4/3=7/3x-3 equation:

4/5x-4/3=7/3x-3

We move all terms to the left:

4/5x-4/3-(7/3x-3)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 3x-3)!=0

x∈R


We get rid of parentheses

4/5x-7/3x+3-4/3=0

We calculate fractions


108x/135x^2+(-35x)/135x^2+(-20x)/135x^2+3=0

We multiply all the terms by the denominator


108x+(-35x)+(-20x)+3*135x^2=0

Wy multiply elements

405x^2+108x+(-35x)+(-20x)=0

We get rid of parentheses

405x^2+108x-35x-20x=0

We add all the numbers together, and all the variables

405x^2+53x=0

a = 405; b = 53; c = 0;
Δ = b2-4ac
Δ = 532-4·405·0
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-53}{2*405}=\frac{-106}{810}
=-53/405
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+53}{2*405}=\frac{0}{810}
=0
$