(2/3*x)+16=3/4*x

Solution for (2/3*x)+16=3/4*x equation:

(2/3x)+16=3/4x

We move all terms to the left:

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


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


8x/12x^2+(-9x)/12x^2+16=0

We multiply all the terms by the denominator


8x+(-9x)+16*12x^2=0

Wy multiply elements

192x^2+8x+(-9x)=0

We get rid of parentheses

192x^2+8x-9x=0

We add all the numbers together, and all the variables

192x^2-1x=0

a = 192; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·192·0
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*192}=\frac{0}{384}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*192}=\frac{2}{384}
=1/192
$