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

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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 $

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