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  2. Clearing denominators - Wikipedia

    en.wikipedia.org/wiki/Clearing_denominators

    The smallest common multiple of the two denominators 6 and 15z is 30z, so one multiplies both sides by 30z: 5 x z + 2 y = 30 z . {\displaystyle 5xz+2y=30z.\,} The result is an equation with no fractions.

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  4. Rhind Mathematical Papyrus 2/n table - Wikipedia

    en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus...

    The table consisted of 26 unit fraction series of the form 1/n written as sums of other rational numbers. [9] The Akhmim wooden tablet wrote difficult fractions of the form 1/n (specifically, 1/3, 1/7, 1/10, 1/11 and 1/13) in terms of Eye of Horus fractions which were fractions of the form ⁠ 1 / 2 k ⁠ and remainders expressed in terms of a ...

  5. Cross-multiplication - Wikipedia

    en.wikipedia.org/wiki/Cross-multiplication

    are solved using cross-multiplication, since the missing b term is implicitly equal to 1: =. Any equation containing fractions or rational expressions can be simplified by multiplying both sides by the least common denominator. This step is called clearing fractions.

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  7. Mediant (mathematics) - Wikipedia

    en.wikipedia.org/wiki/Mediant_(mathematics)

    However, if the fraction 1/1 is replaced by the fraction 2/2, which is an equivalent fraction denoting the same rational number 1, the mediant of the fractions 2/2 and 1/2 is 3/4. For a stronger connection to rational numbers the fractions may be required to be reduced to lowest terms , thereby selecting unique representatives from the ...

  8. Hunter Sallis has 27 points, 12 rebounds, 5 assists as Wake ...

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    WINSTON-SALEM, N.C. (AP) — Hunter Sallis scored 17 of his 27 points in the second half and finished with 12 rebounds and five assists on Tuesday night to help Wake Forest pull away for a 75-58 ...

  9. Verbal arithmetic - Wikipedia

    en.wikipedia.org/wiki/Verbal_arithmetic

    But O cannot be equal to M, so O is less than M. Therefore O = 0. Since O is 1 less than M, S is either 8 or 9 depending on whether there is a carry in column 4. But if there were a carry in column 4 (generated by the addition of column 3), N would be less than or equal to O. This is impossible since O = 0.

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