3/5x-3=1/7x+9

Solution for 3/5x-3=1/7x+9 equation:

3/5x-3=1/7x+9

We move all terms to the left:

3/5x-3-(1/7x+9)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 7x+9)!=0

x∈R


We get rid of parentheses

3/5x-1/7x-9-3=0

We calculate fractions


21x/35x^2+(-5x)/35x^2-9-3=0

We add all the numbers together, and all the variables

21x/35x^2+(-5x)/35x^2-12=0

We multiply all the terms by the denominator


21x+(-5x)-12*35x^2=0

Wy multiply elements

-420x^2+21x+(-5x)=0

We get rid of parentheses

-420x^2+21x-5x=0

We add all the numbers together, and all the variables

-420x^2+16x=0

a = -420; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-420)·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*-420}=\frac{-32}{-840}
=4/105
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-420}=\frac{0}{-840}
=0
$