1/5x+14=1/7x+3

Solution for 1/5x+14=1/7x+3 equation:

1/5x+14=1/7x+3

We move all terms to the left:

1/5x+14-(1/7x+3)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 7x+3)!=0

x∈R


We get rid of parentheses

1/5x-1/7x-3+14=0

We calculate fractions


7x/35x^2+(-5x)/35x^2-3+14=0

We add all the numbers together, and all the variables

7x/35x^2+(-5x)/35x^2+11=0

We multiply all the terms by the denominator


7x+(-5x)+11*35x^2=0

Wy multiply elements

385x^2+7x+(-5x)=0

We get rid of parentheses

385x^2+7x-5x=0

We add all the numbers together, and all the variables

385x^2+2x=0

a = 385; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·385·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*385}=\frac{-4}{770}
=-2/385
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*385}=\frac{0}{770}
=0
$