(1/7t)+(1/5t)=1

Solution for (1/7t)+(1/5t)=1 equation:

(1/7t)+(1/5t)=1

We move all terms to the left:

(1/7t)+(1/5t)-(1)=0


Domain of the equation: 7t)!=0

t!=0/1

t!=0

t∈R



Domain of the equation: 5t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+1/7t)+(+1/5t)-1=0

We get rid of parentheses

1/7t+1/5t-1=0

We calculate fractions


5t/35t^2+7t/35t^2-1=0

We multiply all the terms by the denominator


5t+7t-1*35t^2=0

We add all the numbers together, and all the variables

12t-1*35t^2=0

Wy multiply elements

-35t^2+12t=0

a = -35; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-35)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-35}=\frac{-24}{-70}
=12/35
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-35}=\frac{0}{-70}
=0
$