3/4x=19+7/3x

Solution for 3/4x=19+7/3x equation:

3/4x=19+7/3x

We move all terms to the left:

3/4x-(19+7/3x)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

3/4x-(7/3x+19)=0

We get rid of parentheses

3/4x-7/3x-19=0

We calculate fractions


9x/12x^2+(-28x)/12x^2-19=0

We multiply all the terms by the denominator


9x+(-28x)-19*12x^2=0

Wy multiply elements

-228x^2+9x+(-28x)=0

We get rid of parentheses

-228x^2+9x-28x=0

We add all the numbers together, and all the variables

-228x^2-19x=0

a = -228; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·(-228)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*-228}=\frac{0}{-456}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*-228}=\frac{38}{-456}
=-1/12
$