1/9x+38=5/7x

Solution for 1/9x+38=5/7x equation:

1/9x+38=5/7x

We move all terms to the left:

1/9x+38-(5/7x)=0


Domain of the equation: 9x!=0

x!=0/9

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

1/9x-(+5/7x)+38=0

We get rid of parentheses

1/9x-5/7x+38=0

We calculate fractions


7x/63x^2+(-45x)/63x^2+38=0

We multiply all the terms by the denominator


7x+(-45x)+38*63x^2=0

Wy multiply elements

2394x^2+7x+(-45x)=0

We get rid of parentheses

2394x^2+7x-45x=0

We add all the numbers together, and all the variables

2394x^2-38x=0

a = 2394; b = -38; c = 0;
Δ = b2-4ac
Δ = -382-4·2394·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-38}{2*2394}=\frac{0}{4788}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+38}{2*2394}=\frac{76}{4788}
=1/63
$