5/7x+1/5x=96

Solution for 5/7x+1/5x=96 equation:

5/7x+1/5x=96

We move all terms to the left:

5/7x+1/5x-(96)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We calculate fractions


25x/35x^2+7x/35x^2-96=0

We multiply all the terms by the denominator


25x+7x-96*35x^2=0

We add all the numbers together, and all the variables

32x-96*35x^2=0

Wy multiply elements

-3360x^2+32x=0

a = -3360; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·(-3360)·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*-3360}=\frac{-64}{-6720}
=1/105
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*-3360}=\frac{0}{-6720}
=0
$