Algebra 2

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I know the answer to this problem is -1. I found that by substituting for x and seeing if it worked. How should I do it the correct way?

(x/(3x+6))-(x+1/3x-6)=(1/(x^2)-4)

  • Algebra 2 -

    first, use enough parens to be correct:

    (x/(3x+6))-(x+1)/(3x-6)=(1/(x^2)-4)

    factor out the 1/3 on the left side, and factor the right side

    1/3 (x/(x+2) - (x+1)/(x-2)) = 1/((x+2)(x-2))

    now put LS over a common denominator

    1/3 (x(x-2) - (x+1)(x+2))/((x-2)(x+2)) = 1/((x+2)(x-2))

    multiply through by 3(x-2)(x+2)

    x(x-2) - (x+1)(x+2) = 3
    x^2 - 2x - x^2 - 3x - 2 = 3
    -5x - 2 = 3
    -5x = 5
    x = 1

  • Algebra 2 -

    Thank you for your help. I still need some more explaining. I understand factoring out the 1/3 on the left hand side. I also understand factoring the denominator on the right hand side. After that I have trouble. Could you please show it to me again, but this time with more steps shown?

  • Algebra 2 -

    If you are ok to here:

    1/3 (x/(x+2) - (x+1)/(x-2)) = 1/((x+2)(x-2))

    Think of it as something like

    1/3 (3/5 - 4/7) = 1/(35)

    To clear the fractions, multiply by the LCD of them all: 3*5*7 = 35

    In our case, 3(x+2)(x-2)

    1/3 x/(x+2) * 3*(x-2)(x+2) --> x(x-2)
    the other factors cancel out. Do that for all three terms of the equation and you get to the next step, which is much easier to read:

    x(x-2) - (x+1)(x+2) = 3
    and from there it should be clear. That help?

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