How do you simplify these two expressions? (3 and 4)
thanks!!~!
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thanks!!~!
http://img842.imageshack.us/img842/1866/imag0144tp.jpg
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3. Complex fraction
First we simply the numerator.
1 / (3 + x) - 1 / 3
= 3 / 3(3 + x) - (3 + x) / 3(3 + x)
= (3 - 3 - x) / (3(3 + x))
= -x / 3(3 + x)
So,
[1 / (3 + x) - 1 / 3] / x
= [-x / 3(3 + x)] / x
= -x / 3(3 + x) divided by x
= -x / 3(3 + x) * 1 / x [cancel out the x]
= -1 / 3(x + x)
4.
x^2 - 6x + 9 = (x - 3)(x - 3) and x^2 - 2x - 3 = (x - 3)(x + 1) so,
LCD = (x - 3)(x - 3)(x + 1)
2x / (x^2 - 6x + 9) - 1 / (x + 1) - 8 / (x^2 - 2x - 3)
= 2x(x + 1) / LCD - (x - 3)(x - 3) / LCD - 8(x - 3) / LCD
= [2x(x + 1) - (x - 3)(x - 3) - 8(x - 3)] / LCD
= (2x^2 - 2x - x^2 + 6x - 9 - 8x + 24) / LCD
= (x^2 - 4x + 15) / LCD
= (x^2 - 4x + 15) / (x - 3)(x - 3)(x + 1)
First we simply the numerator.
1 / (3 + x) - 1 / 3
= 3 / 3(3 + x) - (3 + x) / 3(3 + x)
= (3 - 3 - x) / (3(3 + x))
= -x / 3(3 + x)
So,
[1 / (3 + x) - 1 / 3] / x
= [-x / 3(3 + x)] / x
= -x / 3(3 + x) divided by x
= -x / 3(3 + x) * 1 / x [cancel out the x]
= -1 / 3(x + x)
4.
x^2 - 6x + 9 = (x - 3)(x - 3) and x^2 - 2x - 3 = (x - 3)(x + 1) so,
LCD = (x - 3)(x - 3)(x + 1)
2x / (x^2 - 6x + 9) - 1 / (x + 1) - 8 / (x^2 - 2x - 3)
= 2x(x + 1) / LCD - (x - 3)(x - 3) / LCD - 8(x - 3) / LCD
= [2x(x + 1) - (x - 3)(x - 3) - 8(x - 3)] / LCD
= (2x^2 - 2x - x^2 + 6x - 9 - 8x + 24) / LCD
= (x^2 - 4x + 15) / LCD
= (x^2 - 4x + 15) / (x - 3)(x - 3)(x + 1)