Answer:
- {-1/4, 1}
- {3/4, 6}
Step-by-step explanation:
1. We can clear fractions and solve the resulting quadratic. We clear fractions by multiplying the equation by the product of the denominators.
[tex]\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{3}\\\\3((2x+1)^2-(2x-1)^2)=8(2x-1)(2x+1)\\\\3(8x) = 8(4x^2 -1)\\\\4x^2 -3x -1 = 0\qquad\text{factor out 8, subtract 3x}\\\\x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(4)(-1)}}{2(4)}=\dfrac{3\pm\sqrt{25}}{8}\\\\x=\dfrac{3\pm5}{8}=\left\{-\dfrac{1}{4},1\right\}[/tex]
__
2. Using the same idea here, we get ...
[tex]\dfrac{2}{x-2}+\dfrac{3}{x}=\dfrac{9}{x+3}\\\\2(x)(x+3)+3(x-2)(x+3)=9(x-2)(x)\\\\2x^2+6x+3(x^2+x-6)=9x^2-18x\\\\4x^2-27x+18=0\\\\x=\dfrac{-(-27)\pm\sqrt{(-27)^2-4(4)(18)}}{2(4)}=\dfrac{27\pm\sqrt{441}}{8}\\\\x=\dfrac{27\pm21}{8}=\left\{\dfrac{3}{4},6\right\}[/tex]
