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  2. Elementary algebra - Wikipedia

    en.wikipedia.org/wiki/Elementary_algebra

    If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0. Then we can substitute again, letting x = b and y = c, to show that if bc = 0 then b = 0 or c = 0. Therefore, if abc = 0, then a = 0 or (b = 0 or c = 0), so abc = 0 implies ...

  3. Algebra - Wikipedia

    en.wikipedia.org/wiki/Algebra

    Algebra is the branch of mathematics that studies certain abstract systems, known as algebraic structures, and the manipulation of expressions within those systems. It is a generalization of arithmetic that introduces variables and algebraic operations other than the standard arithmetic operations, such as addition and multiplication.

  4. FOIL method - Wikipedia

    en.wikipedia.org/wiki/FOIL_method

    In elementary algebra, FOIL is a mnemonic for the standard method of multiplying two binomials [1] —hence the method may be referred to as the FOIL method.The word FOIL is an acronym for the four terms of the product:

  5. Algebraic geometry - Wikipedia

    en.wikipedia.org/wiki/Algebraic_geometry

    To see how this might come about, consider the variety V(y − x 2). If we draw it, we get a parabola. As x goes to positive infinity, the slope of the line from the origin to the point (x, x 2) also goes to positive infinity. As x goes to negative infinity, the slope of the same line goes to negative infinity. Compare this to the variety V(y ...

  6. Outline of algebra - Wikipedia

    en.wikipedia.org/wiki/Outline_of_algebra

    Algebra is one of the main branches of mathematics, covering the study of structure, relation and quantity. Algebra studies the effects of adding and multiplying numbers , variables , and polynomials , along with their factorization and determining their roots .

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  8. Equation solving - Wikipedia

    en.wikipedia.org/wiki/Equation_solving

    For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.

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