9x-7/5x-1=7x+3/5x+4

Solution for 9x-7/5x-1=7x+3/5x+4 equation:

9x-7/5x-1=7x+3/5x+4

We move all terms to the left:

9x-7/5x-1-(7x+3/5x+4)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 5x+4)!=0

x∈R


We get rid of parentheses

9x-7/5x-7x-3/5x-4-1=0

We multiply all the terms by the denominator


9x*5x-7x*5x-4*5x-1*5x-7-3=0

We add all the numbers together, and all the variables

9x*5x-7x*5x-4*5x-1*5x-10=0

Wy multiply elements

45x^2-35x^2-20x-5x-10=0

We add all the numbers together, and all the variables

10x^2-25x-10=0

a = 10; b = -25; c = -10;
Δ = b2-4ac
Δ = -252-4·10·(-10)
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{41}}{2*10}=\frac{25-5\sqrt{41}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{41}}{2*10}=\frac{25+5\sqrt{41}}{20}
$