Answer:
x ≈ 8.37819064728
Step-by-step explanation:
Maybe you want to solve for x.
[tex]\dfrac{x-4}{x-3}=\dfrac{2x^2-6}{x^2+2x-3}-\dfrac{x-1}{x+1}=\dfrac{2(x^2-3)}{(x-1)(x+3)}-\dfrac{x-1}{x+1}\\\\\dfrac{x-4}{x-3}=\dfrac{2(x^2-3)(x+1)}{(x-1)(x+3)(x+1)}-\dfrac{(x-1)(x-1)(x+3)}{(x+1)(x-1)(x+3)}\\\\\dfrac{x-4}{x-3}=\dfrac{(2x^3+2x^2-6x-6)-(x^3+x^2-5x+3)}{(x^2-1)(x+3)}\\\\\dfrac{(x^2-1)(x+3)(x-4)}{(x^2-1)(x+3)(x-3)}=\dfrac{(x^3+x^2-x-9)(x-3)}{(x^2-1)(x+3)(x-3)}\\\\\dfrac{x^4-x^3-13x^2+x+12}{(x^2-1)(x^2-9)}=\dfrac{x^4-2x^3-4x^2-6x+27}{(x^2-1)(x^2-9)}\\\\\dfrac{x^3-9x^2+7x-15}{(x^2-1)(x^2-9)}=0[/tex]
A graphing calculator shows the numerator cubic to have one real irrational zero near x ≈ 8.37819064728.
