What is the integral value of x which satisfies the given equation?

$\frac{2x}{2x^2-5x+3}+\frac{13x}{2x^2+x+3}=6.$

We look for a way to make a substitution. Since clearly $x\ne 0$ , we can divide by 𝑥, and create a common term of$2x+\frac{3}{x}$ in each fraction:

$\frac{2}{2x+\frac{3}{x}-5}+\frac{13}{2x+\frac{3}{x}+1}=6.$

$\text{Then, the substitution}y=2x+\frac{3}{x}\text{yields the equation}$

Now, we can clear the denominators to obtain a quadratic equation. We obtain

$2(y+1)+13(y-5)=6(y-5)(y+1),$

and moving all terms to one side gives

$6y^2-39y+33=3(y-1)(2y-11)=0,$

$\text{implying}y=1\text{and}y=\frac{11}{2}\text{. We now solve for}x\text{. When}y=1\text{, we have }$$2x+\frac{3}{x}=1\text{, which is equivalent to}$

$2x^2-x+3=0\text{.}$

$\text{The discriminant of this quadratic is}\mathrm{\Delta }=1^2-4\cdot 2\cdot 3<0\text{, so this equation}$

$\text{has no real solutions. When}y=\frac{11}{2}\text{, we have}2x+\frac{3}{x}=\frac{11}{2}\text{, which reduces to}$

$4x^2-11x+6=(x-2)(4x-\frac{3}{4})=0.$

$\text{We obtain the solutions}x=2\text{and}x=\frac{3}{4}\text{.}$