1/5x+1/3=4+3/4x

Solution for 1/5x+1/3=4+3/4x equation:

1/5x+1/3=4+3/4x

We move all terms to the left:

1/5x+1/3-(4+3/4x)=0


Domain of the equation: 5x!=0

x!=0/5

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

1/5x-(3/4x+4)+1/3=0

We get rid of parentheses

1/5x-3/4x-4+1/3=0

We calculate fractions


80x^2/180x^2+36x/180x^2+(-135x)/180x^2-4=0

We multiply all the terms by the denominator


80x^2+36x+(-135x)-4*180x^2=0

Wy multiply elements

80x^2-720x^2+36x+(-135x)=0

We get rid of parentheses

80x^2-720x^2+36x-135x=0

We add all the numbers together, and all the variables

-640x^2-99x=0

a = -640; b = -99; c = 0;
Δ = b2-4ac
Δ = -992-4·(-640)·0
Δ = 9801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9801}=99$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-99)-99}{2*-640}=\frac{0}{-1280}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-99)+99}{2*-640}=\frac{198}{-1280}
=-99/640
$