4/3x-5=3/4x-4

Solution for 4/3x-5=3/4x-4 equation:

4/3x-5=3/4x-4

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


16x/12x^2+(-9x)/12x^2+4-5=0

We add all the numbers together, and all the variables

16x/12x^2+(-9x)/12x^2-1=0

We multiply all the terms by the denominator


16x+(-9x)-1*12x^2=0

Wy multiply elements

-12x^2+16x+(-9x)=0

We get rid of parentheses

-12x^2+16x-9x=0

We add all the numbers together, and all the variables

-12x^2+7x=0

a = -12; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-12)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-12}=\frac{-14}{-24}
=7/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-12}=\frac{0}{-24}
=0
$