2-1/3z=1/8z+9

Solution for 2-1/3z=1/8z+9 equation:

2-1/3z=1/8z+9

We move all terms to the left:

2-1/3z-(1/8z+9)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 8z+9)!=0

z∈R


We get rid of parentheses

-1/3z-1/8z-9+2=0

We calculate fractions


(-8z)/24z^2+(-3z)/24z^2-9+2=0

We add all the numbers together, and all the variables

(-8z)/24z^2+(-3z)/24z^2-7=0

We multiply all the terms by the denominator


(-8z)+(-3z)-7*24z^2=0

Wy multiply elements

-168z^2+(-8z)+(-3z)=0

We get rid of parentheses

-168z^2-8z-3z=0

We add all the numbers together, and all the variables

-168z^2-11z=0

a = -168; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·(-168)·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*-168}=\frac{0}{-336}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*-168}=\frac{22}{-336}
=-11/168
$