2/5x+3/7x=1-4/7x

Solution for 2/5x+3/7x=1-4/7x equation:

2/5x+3/7x=1-4/7x

We move all terms to the left:

2/5x+3/7x-(1-4/7x)=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



Domain of the equation: 7x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2/5x+3/7x-(-4/7x+1)=0

We get rid of parentheses

2/5x+3/7x+4/7x-1=0

We calculate fractions


14x/35x^2+(20x+3)/35x^2-1=0

We multiply all the terms by the denominator


14x+(20x+3)-1*35x^2=0

Wy multiply elements

-35x^2+14x+(20x+3)=0

We get rid of parentheses

-35x^2+14x+20x+3=0

We add all the numbers together, and all the variables

-35x^2+34x+3=0

a = -35; b = 34; c = +3;
Δ = b2-4ac
Δ = 342-4·(-35)·3
Δ = 1576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1576}=\sqrt{4*394}=\sqrt{4}*\sqrt{394}=2\sqrt{394}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{394}}{2*-35}=\frac{-34-2\sqrt{394}}{-70}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{394}}{2*-35}=\frac{-34+2\sqrt{394}}{-70}
$