(5/7)x+(1/7)=3

Solution for (5/7)x+(1/7)=3 equation:

(5/7)x+(1/7)=3

We move all terms to the left:

(5/7)x+(1/7)-(3)=0


Domain of the equation: 7)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(5/7)x-3+(1/7)=0

We add all the numbers together, and all the variables

(+5/7)x-3+(+1/7)=0

We multiply parentheses

5x^2-3+(+1/7)=0

We get rid of parentheses

5x^2-3+1/7=0

We multiply all the terms by the denominator


5x^2*7+1-3*7=0

We add all the numbers together, and all the variables

5x^2*7-20=0

Wy multiply elements

35x^2-20=0

a = 35; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·35·(-20)
Δ = 2800
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:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{7}}{2*35}=\frac{0-20\sqrt{7}}{70}
=-\frac{20\sqrt{7}}{70}
=-\frac{2\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{7}}{2*35}=\frac{0+20\sqrt{7}}{70}
=\frac{20\sqrt{7}}{70}
=\frac{2\sqrt{7}}{7}
$