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Math Help - Help with solving linear equation

  1. #1
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    Help with solving linear equation

    Problem: 1/3 + 1/9(k+5) - k/4 = 2

    Question... When you multiply by 36 (LCD) to kill the fractions...

    Why do you not multiply (k+5)?

    I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?
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  2. #2
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    Ok... when you multiply by 36, you multiply BOTH sides. It doesn't matter whether (x+5) would be affected or not, just as long as you multiply both sides by 36. Therefore, 1/3 + 1/9(k+5) - k/4 = 2 would become 12+4(k+5)-9k = 72, and then you would expand.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by erj145aviator View Post
    Problem: 1/3 + 1/9(k+5) - k/4 = 2

    Question... When you multiply by 36 (LCD) to kill the fractions...

    Why do you not multiply (k+5)?

    I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?
    \frac{1}{3} + \frac{1}{9}(k+5) - \frac{k}{4} = 2

    \frac{1}{3} + \frac{(k+5)}{9} - \frac{k}{4} = 2

    \frac{12}{36} + \frac{4(k+5)}{36} - \frac{9k}{36} = 2

    12 + 4(k+5) - 9k = 72

    12 + 4k + 20 - 9k = 72

    -5k + 32 = 72

    -5k = 40

    k = -8
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  4. #4
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    Quote Originally Posted by erj145aviator View Post
    Problem: 1/3 + 1/9(k+5) - k/4 = 2

    Question... When you multiply by 36 (LCD) to kill the fractions...

    Why do you not multiply (k+5)?

    I mean 36(1/3 + 1/9(k+5) - k/4) ... (k+5) is there... if you distribute 36 over... (k+5) would be affected. So why in the process of solving, would you not do that?
    As far as I know you don't multiply k+5 by 35 because of the distributive property. You are keeping proportion with the rest of the equation.
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