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What is true about the solution of x^2/2x-6=9/6x-18?


x= +_ sqrt 3 and they are actual solutions.

x= +_ sqrt 3 but they are extraneous solutions.

x = 3, and it is an actual solution.

x = 3, but it is an extraneous solution.

Respuesta :

gmany
[tex]\text{The domain:}\\2x-6\neq0\ \wedge\ 6x-18\neq0\\D:x\neq3\\\\\dfrac{x^2}{2x-6}=\dfrac{9}{6x-18}\\\\\dfrac{x^2}{2x-6}=\dfrac{9}{3(2x-6)}\\\\\dfrac{x^2}{2x-6}=\dfrac{3}{2x-6}\iff x^2=3\to x=\pm\sqrt3[/tex]
Answer:
x= +-√3 and they are actual solutions.

1236benjamin

Answer:

Its A    on the forbidden edge of tears and no hope

Step-by-step explanation:

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