9. 10. To solve 2x x 2 11 x = 8 x 2 2x , Ren multiplied both sides by the least common denominator. Which statement is true? A. 2 is an extraneous solution. B. 7 2 is an extraneous solution. C. 0 and 2 are extraneous solutions. D. This equation does not contain any extraneou

Respuesta :

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

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