SOLUTION: Solve the equation. 30 over x^2-9 = 2 over x-3. I worked it out to: x=12 or x=3 x=3 is extraneous Am I correct?

Algebra ->  Equations -> SOLUTION: Solve the equation. 30 over x^2-9 = 2 over x-3. I worked it out to: x=12 or x=3 x=3 is extraneous Am I correct?      Log On


   

Question 189602This question is from textbook saxon algebra 2
: Solve the equation.
30 over x^2-9 = 2 over x-3.
I worked it out to: x=12 or x=3
x=3 is extraneous
Am I correct? This question is from textbook saxon algebra 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct. Here's the check



Checking x=12


30%2F%28x%5E2-9%29+=+2%2F%28x-3%29 Start with the given equation.


30%2F%2812%5E2-9%29+=+2%2F%2812-3%29 Plug in x=12


30%2F%28144-9%29+=+2%2F%2812-3%29 Square 12 to get 144


30%2F135+=+2%2F9 Subtract


2%2F9+=+2%2F9 Reduce. Since the equation is true, this verifies that x=12 is a solution.


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Checking x=3


30%2F%28x%5E2-9%29+=+2%2F%28x-3%29 Start with the given equation.


30%2F%283%5E2-9%29+=+2%2F%283-3%29 Plug in x=3


30%2F%289-9%29+=+2%2F%283-3%29 Square 3 to get 9


30%2F0+=+2%2F0 Subtract. Since division by zero is undefined, this means that x=3 is NOT part of the domain. So it is NOT a solution.