Answer:
2 is the extraneous solution
Step-by-step explanation:
Given equation is
[tex]\frac{2x}{x-2} -\frac{11}{x} =\frac{8}{x^2-2x}[/tex]
Factor the denominator
[tex]\frac{2x}{x-2} -\frac{11}{x} =\frac{8}{x(x-2)}[/tex]
LCD is x(x-2), multiply all the fractions by LCD
[tex]2x \cdot x-11(x-2)=8[/tex]
[tex]2x^2-11x+22= 8[/tex], subtract 8 from both sides
[tex]2x^2-11x+14=0[/tex]
factor the left hand side
[tex]2x^2-7x-4x+14= 0[/tex]
[tex]x(2x-7)-2(2x-7)=0[/tex]
[tex](x-2)(2x-7)=0[/tex]
x-2=0, so x=2
2x-7=0, [tex]x=\frac{7}{2}[/tex]
when x=2, then the denominator becomes 0 that is undefined
So 2 is the extraneous solution