1/7x=5x=35

Solution for 1/7x=5x=35 equation:

1/7x=5x=35

We move all terms to the left:

1/7x-(5x)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R


We add all the numbers together, and all the variables

-5x+1/7x=0

We multiply all the terms by the denominator


-5x*7x+1=0

Wy multiply elements

-35x^2+1=0

a = -35; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-35)·1
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*-35}=\frac{0-2\sqrt{35}}{-70}
=-\frac{2\sqrt{35}}{-70}
=-\frac{\sqrt{35}}{-35}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*-35}=\frac{0+2\sqrt{35}}{-70}
=\frac{2\sqrt{35}}{-70}
=\frac{\sqrt{35}}{-35}
$