1+4/5x=2+3/4x

Solution for 1+4/5x=2+3/4x equation:

1+4/5x=2+3/4x

We move all terms to the left:

1+4/5x-(2+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

4/5x-(3/4x+2)+1=0

We get rid of parentheses

4/5x-3/4x-2+1=0

We calculate fractions


16x/20x^2+(-15x)/20x^2-2+1=0

We add all the numbers together, and all the variables

16x/20x^2+(-15x)/20x^2-1=0

We multiply all the terms by the denominator


16x+(-15x)-1*20x^2=0

Wy multiply elements

-20x^2+16x+(-15x)=0

We get rid of parentheses

-20x^2+16x-15x=0

We add all the numbers together, and all the variables

-20x^2+x=0

a = -20; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-20)·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*-20}=\frac{-2}{-40}
=1/20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-20}=\frac{0}{-40}
=0
$