(1/7)*x+(1/7)=3

Solution for (1/7)*x+(1/7)=3 equation:

(1/7)*x+(1/7)=3

We move all terms to the left:

(1/7)*x+(1/7)-(3)=0


Domain of the equation: 7)*x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(1/7)*x-3+(1/7)=0

We add all the numbers together, and all the variables

(+1/7)*x-3+(+1/7)=0

We multiply parentheses

x^2-3+(+1/7)=0

We get rid of parentheses

x^2-3+1/7=0

We multiply all the terms by the denominator


x^2*7+1-3*7=0

We add all the numbers together, and all the variables

x^2*7-20=0

Wy multiply elements

7x^2-20=0

a = 7; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·7·(-20)
Δ = 560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{35}}{2*7}=\frac{0-4\sqrt{35}}{14}
=-\frac{4\sqrt{35}}{14}
=-\frac{2\sqrt{35}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{35}}{2*7}=\frac{0+4\sqrt{35}}{14}
=\frac{4\sqrt{35}}{14}
=\frac{2\sqrt{35}}{7}
$