Question

Please help if u can!

Answer 1

9514 1404 393

Answer:

  C

Step-by-step explanation:

When you cross multiply by denominator expressions that share a zero, you are effectively making that zero a solution of the resulting equation. That zero cannot be part of the domain of the original equation, and so must be an extraneous solution.

This condition is seen in choice C.

  $\dfrac{8}{x^2-9}=\dfrac{5}{2x-6}\qquad\text{original equation. x\ne3 }\\\\8(2x-6)=5(x^2-9)\qquad\text{result of cross multiplying}\\\\16(x -3)=5(x+3)(x-3)\qquad\text{factored form}$

The solutions to this equation are x=1/5 and x=3. Of these, x=3 is extraneous.

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A better solution method is multiplying by the least common denominator:

  $2(8)=5(x+3)\qquad\text{result of multiplying by 2(x^2-9) }$