2/3x+16=4/5x

Solution for 2/3x+16=4/5x equation:

2/3x+16=4/5x

We move all terms to the left:

2/3x+16-(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

2/3x-(+4/5x)+16=0

We get rid of parentheses

2/3x-4/5x+16=0

We calculate fractions


10x/15x^2+(-12x)/15x^2+16=0

We multiply all the terms by the denominator


10x+(-12x)+16*15x^2=0

Wy multiply elements

240x^2+10x+(-12x)=0

We get rid of parentheses

240x^2+10x-12x=0

We add all the numbers together, and all the variables

240x^2-2x=0

a = 240; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·240·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*240}=\frac{0}{480}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*240}=\frac{4}{480}
=1/120
$