2x/3x+(4x+5x)=5

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

2x/3x+(4x+5x)=5

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


2x+9x*3x-5*3x=0

Wy multiply elements

27x^2+2x-15x=0

We add all the numbers together, and all the variables

27x^2-13x=0

a = 27; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·27·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*27}=\frac{0}{54}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*27}=\frac{26}{54}
=13/27
$