2/3x=2/5x+16

Solution for 2/3x=2/5x+16 equation:

2/3x=2/5x+16

We move all terms to the left:

2/3x-(2/5x+16)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 5x+16)!=0

x∈R


We get rid of parentheses

2/3x-2/5x-16=0

We calculate fractions


10x/15x^2+(-6x)/15x^2-16=0

We multiply all the terms by the denominator


10x+(-6x)-16*15x^2=0

Wy multiply elements

-240x^2+10x+(-6x)=0

We get rid of parentheses

-240x^2+10x-6x=0

We add all the numbers together, and all the variables

-240x^2+4x=0

a = -240; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-240)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-240}=\frac{-8}{-480}
=1/60
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-240}=\frac{0}{-480}
=0
$