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anonymous

  • 4 years ago

0 = 5/6x + 2/7x + 1 My teacher says to multiply by 42 to get rid of the fractions. ...It's not exactly working for me. How should I do it?

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  1. JamesJ
    • 4 years ago
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    is that \( \frac{5}{6x} \) or \( \frac{5}{6}x \)?

  2. anonymous
    • 4 years ago
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    I can't seem to be able to type it out using the Equation system, but it's the second one. Same with 5/7x, they're both fractions.

  3. y2o2
    • 4 years ago
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    just use brackets

  4. JamesJ
    • 4 years ago
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    So you have \[ 0 = \frac{5}{6}x + \frac{2}{7}x + 1 \] Now multiply both sides by 42 and you have \[ 42(0) = 0 = 42\frac{5}{6}x + 42\frac{2}{7}x + 1 \] Now simplify that expression.

  5. JamesJ
    • 4 years ago
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    *correction: \[ 0 = 42\frac{5}{6}x + 42\frac{2}{7}x + 42 \]

  6. anonymous
    • 4 years ago
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    Okay, I now have: \[0 = \frac{257}{6}x +\frac{296}{7}x + 42\] \[0 =\frac{3575}{42}x + 42\] or \[0 = 85\frac{5}{42}x + 42\] Do I divide by 42 to get rid of that fraction?

  7. phi
    • 4 years ago
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    If I were you, I would cancel first. for example: \[ \frac{42\cdot 5}{6}= 7\cdot5= 35\] do that for both fractions before taking the next step.

  8. phi
    • 4 years ago
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    That is why you multiply by 42, to get rid of the 6 and the 7 in the denominators

  9. anonymous
    • 4 years ago
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    I think I got it! 0 = 5/6x + 2/7x + 1 (42)0 = 42(5/6x) + 42(2/7x) + 42(1) 0 = 35x + 12x + 42 0 = 47x + 42 -42 = 47x -42/47 = x

  10. phi
    • 4 years ago
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    looks good. It is easier working with smaller numbers, isn't it?

  11. anonymous
    • 4 years ago
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    Definitely! Thanks to the both of you. :D

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