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

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: [x/(4x^2-9)] - [(x+3)/(8x^2+6x-9)] = [1/(8x^2-18x+9)]       Log On


   

Question 65133: [x/(4x^2-9)] - [(x+3)/(8x^2+6x-9)] = [1/(8x^2-18x+9)]
Answer by funmath(2933) About Me  (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)]
x%2F%284x%5E2-9%29-%28x%2B3%29%2F%288x%5E2%2B6x-9%29=1%2F%288x%5E2-18x%2B9%29
Factor the denominators to find the lowest common denominator and restriced values for x:
4x%5E2-9=%282x%2B3%29%282x-3%29
2x+3 cannot = 0 and 2x-3 cannot =0
x cannot =-3/2 and x cannot = 3/2
8x%5E2%2B6x-9
8x%5E2%2B12x-6x-9
%288x%5E2%2B12x%29%2B%28-6x-9%29
4x%282x%2B3%29-3%282x%2B3%29
%284x-3%29%282x%2B3%29
x cannot = 3/4 and x cannot= -3/2
8x%5E2-18x%2B9
8x%5E2-12x-6x%2B9
%288x%5E2-12x%29%2B%28-6x%2B9%29
4x%282x-3%29-3%282x-3%29
%284x-3%29%282x-3%29
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%284x-3%29-%28x%2B3%29%282x-3%29=%282x%2B3%29
4x%5E2-3x-%282x%5E2-3x%2B6x-9%29=2x%2B3
4x%5E2-3x-%282x%5E2%2B3x-9%29=2x%2B3
4x%5E2-3x-2x%5E2-3x%2B9=2x%2B3
2x%5E2-6x%2B9=2x%2B3
2x%5E2-6x-2x%2B9-3=2x-2x%2B3-3
2x%5E2-8x%2B6=0
2%28x%5E2-4x%2B3%29=0
2%28x-3%29%28x-1%29=0
2%28x-3%29%28x-1%29%2F2=0%2F2
%28x-3%29%28x-1%29=0
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!!!