1/t-1+t/3t-2=1/3

Solution for 1/t-1+t/3t-2=1/3 equation:

1/t-1+t/3t-2=1/3

We move all terms to the left:

1/t-1+t/3t-2-(1/3)=0


Domain of the equation: t!=0

t∈R



Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We add all the numbers together, and all the variables

1/t+t/3t-1-2-(+1/3)=0

We add all the numbers together, and all the variables

1/t+t/3t-3-(+1/3)=0

We get rid of parentheses

1/t+t/3t-3-1/3=0

We calculate fractions


t^2/27t^2+27t/27t^2+(-t)/27t^2-3=0

We add all the numbers together, and all the variables

t^2/27t^2+27t/27t^2+(-1t)/27t^2-3=0

We multiply all the terms by the denominator


t^2+27t+(-1t)-3*27t^2=0

Wy multiply elements

t^2-81t^2+27t+(-1t)=0

We get rid of parentheses

t^2-81t^2+27t-1t=0

We add all the numbers together, and all the variables

-80t^2+26t=0

a = -80; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·(-80)·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*-80}=\frac{-52}{-160}
=13/40
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*-80}=\frac{0}{-160}
=0
$