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x x + 3 1
--------- - -------------- = --------------
4x² - 9 8x² + 6x - 9 8x²- 18x + 9
Factor all the denominators:
x x + 3 1
-------------- - -------------- = --------------
(2x-3)(2x+3) (2x+3)(4x-3) (4x-3)(2x-3)
LCD = (2x-3)(2x+3)(4x-3)
Put it over 1, like this
(2x-3)(2x+3)(4x-3)
--------------------
1
Now for each of the two fractions terms on the left,
and the one fraction term on the right, multiply
through by this fraction.
First fraction term on the left
x (2x-3)(2x+3)(4x-3)
-------------- · --------------------
(2x-3)(2x+3) 1
Cancel the (2x-3)'s and the (2x+3)'s
1 1
x (2x-3)(2x+3)(4x-3)
-------------- · --------------------
(2x-3)(2x+3) 1
1 1
x(4x - 3)
4x² - 4x
------------------
Second fraction term on the left
x + 3 (2x-3)(2x+3)(4x-3)
- -------------- · --------------------
(2x+3)(4x-3) 1
Cancel the (2x+3)'s and the (4x-3)'s
1 1
x + 3 (2x-3)(2x+3)(4x-3)
- -------------- · --------------------
(2x+3)(4x-3) 1
1 1
- (x + 3)(2x - 3)
-(2x² - 3x + 6x - 9)
-(2x² + 3x - 9)
-2x² - 3x + 9
Fraction term on right:
1 (2x-3)(2x+3)(4x-3)
-------------- · --------------------
(4x-3)(2x-3) 1
Cancel the (4x-3)'s and the (2x-3)'s
1 1
1 (2x-3)(2x+3)(4x-3)
-------------- · --------------------
(4x-3)(2x-3)
1 1
2x + 3
Now you have
4x² - 3x - 2x² - 3x + 9 = 2x + 3
2x² - 6x + 9 = 2x + 3
2x² - 8x + 6 = 0
Divide every term by 2
x² - 4x + 3 = 0
Factor
(x - 3)(x - 1) = 0
x - 3 = 0 gives solution x = 3
x - 1 = 0 gives solution x = 1
Neither of these cause any of the denominators
in the original to become 0, so they are
solutions.
Edwin
