1/5x+15-3/7x+3=5

Solution for 1/5x+15-3/7x+3=5 equation:

1/5x+15-3/7x+3=5

We move all terms to the left:

1/5x+15-3/7x+3-(5)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R


We add all the numbers together, and all the variables

1/5x-3/7x+13=0

We calculate fractions


7x/35x^2+(-15x)/35x^2+13=0

We multiply all the terms by the denominator


7x+(-15x)+13*35x^2=0

Wy multiply elements

455x^2+7x+(-15x)=0

We get rid of parentheses

455x^2+7x-15x=0

We add all the numbers together, and all the variables

455x^2-8x=0

a = 455; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·455·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*455}=\frac{0}{910}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*455}=\frac{16}{910}
=8/455
$