SOLUTION: [x/(4x^2-9)] - [(x+3)/(8x^2+6x-9)] = [1/(8x^2-18x+9)]

Question 65133: [x/(4x^2-9)] - [(x+3)/(8x^2+6x-9)] = [1/(8x^2-18x+9)]
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
[x/(4x^2-9)] - [(x+3)/(8x^2+6x-9)] = [1/(8x^2-18x+9)]

Factor the denominators to find the lowest common denominator and restriced values for x:

2x+3 cannot = 0 and 2x-3 cannot =0
x cannot =-3/2 and x cannot = 3/2

x cannot = 3/4 and x cannot= -3/2

x cannot=3/4 and x cannot=3/2
Your LCD would be (2x+3)(2x-3)(4x-3)
Your restricted values are: x cannot = -3/2, 3/4, or 3/2
Multiply both sides by the LCD and if you get one of the restricted values, reject it beacuse its extraneous.

Cancel the matching numerators and denominators:

x-3=0 or x-1=0
x-3+3=0+3 or x-1+1=0+1
x=3 or x=1
These aren't one of the restricted values, so they're valid. You can double check them, byt substitution to see if the equation balances to be double safe. I checked it on my calculator.
Happy Calculating!!!
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