2/3x+9=9/4x-5

Solution for 2/3x+9=9/4x-5 equation:

2/3x+9=9/4x-5

We move all terms to the left:

2/3x+9-(9/4x-5)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 4x-5)!=0

x∈R


We get rid of parentheses

2/3x-9/4x+5+9=0

We calculate fractions


8x/12x^2+(-27x)/12x^2+5+9=0

We add all the numbers together, and all the variables

8x/12x^2+(-27x)/12x^2+14=0

We multiply all the terms by the denominator


8x+(-27x)+14*12x^2=0

Wy multiply elements

168x^2+8x+(-27x)=0

We get rid of parentheses

168x^2+8x-27x=0

We add all the numbers together, and all the variables

168x^2-19x=0

a = 168; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·168·0
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*168}=\frac{0}{336}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*168}=\frac{38}{336}
=19/168
$