1/3z-3=3-3/4z

Solution for 1/3z-3=3-3/4z equation:

1/3z-3=3-3/4z

We move all terms to the left:

1/3z-3-(3-3/4z)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 4z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

1/3z-(-3/4z+3)-3=0

We get rid of parentheses

1/3z+3/4z-3-3=0

We calculate fractions


4z/12z^2+9z/12z^2-3-3=0

We add all the numbers together, and all the variables

4z/12z^2+9z/12z^2-6=0

We multiply all the terms by the denominator


4z+9z-6*12z^2=0

We add all the numbers together, and all the variables

13z-6*12z^2=0

Wy multiply elements

-72z^2+13z=0

a = -72; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-72)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-72}=\frac{-26}{-144}
=13/72
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-72}=\frac{0}{-144}
=0
$