3/5x=1/7x+16

Solution for 3/5x=1/7x+16 equation:

3/5x=1/7x+16

We move all terms to the left:

3/5x-(1/7x+16)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 7x+16)!=0

x∈R


We get rid of parentheses

3/5x-1/7x-16=0

We calculate fractions


21x/35x^2+(-5x)/35x^2-16=0

We multiply all the terms by the denominator


21x+(-5x)-16*35x^2=0

Wy multiply elements

-560x^2+21x+(-5x)=0

We get rid of parentheses

-560x^2+21x-5x=0

We add all the numbers together, and all the variables

-560x^2+16x=0

a = -560; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-560)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-560}=\frac{-32}{-1120}
=1/35
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-560}=\frac{0}{-1120}
=0
$