Algebra Notes: Prime Factorization, Inequalities, and Interval Techniques

Prime Factorization and Denominators

• Review the idea of factoring denominators into primes and canceling common factors when possible.
• If a denominator is prime, leave it as is; if not, factor it into primes.
• Example flow mentioned:
  • Consider two numbers where a common factor appears (e.g., 7 and 35).
  • 35 factors as
    $35 = 5 \cdot 7$.
  • Cancel the common factor 7 from numerator and denominator to simplify. In a simple cancellation, you'd have something like
    $\frac{7}{35} = \frac{7}{7\cdot 5} = \frac{1}{5}.$
• The idea of cancelling is to remove the common prime factor present in both numerator and denominator; you’re left with the remaining factors.
• The transcript’s phrasing about “left with a one” likely reflects canceling a common factor in a broader setup (e.g., canceling a 7 factor from a ratio like (\frac{7}{7\cdot 5}) leaving (\frac{1}{5})). The key concept is: cancel common prime factors to simplify fractions.
• Takeaway: always inspect denominators (and numerators) for prime factors; factor if needed, then cancel common primes to simplify.

Inequalities: Reading and Endpoints

• From the sum 10 + 21, the result is 31, so an inequality is formed:
  $31 \ge x.$
• This is equivalent to
  $x \le 31.$
• Reading direction and endpoints:
  • You can read from right to left by turning the inequality; when you switch sides, you’re effectively turning the expression around.
  • The choice of brackets vs parentheses indicates whether endpoints are included:
  • If you have a (\ge) or (\le), endpoints are included (closed brackets): use [ and ].
  • If you have a strict inequality ( < or > ), endpoints are not included (open): use ( and ).
• Example interpretation on a number line:
  • Values less than or equal to 31 lie on the left side of the line, including 31.
  • For the inequality (x \le 31), the end at 31 is included.
• Fractions in inequalities and the least common denominator (LCD):
  • If the inequality contains fractions, identify the LCD to clear denominators.
  • In the scenario discussed, the denominator mentioned is 6, so the LCD is
    $\text{LCD} = 6.$
• Example problem discussed in the session:
  • Consider the fraction component (\frac{5}{6}) with an inequality involving x on both sides. The approach described:
  • Multiply both sides by the LCD (6) to clear the fraction. This yields terms like $5x\;\text{and}\;6x.$
  • To get all x-terms on the left, subtract 5x (i.e., subtract the term that has 5x) from both sides.
  • If we start with something like
    $\frac{5}{6}x \le x,$
    multiplying by 6 gives
    $5x \le 6x.$
    Subtract 5x from both sides: $0 \le x,$ which simplifies to
    $x \ge 0.$
• Interval notation and graphing the solution:
  • For the inequality (x \ge 0), the interval notation is
    $[0, \infty).$
  • When representing on a number line, shade to the right of 0, including 0.
• Key takeaway about fractions in inequalities:
  • Clear denominators with the LCD, gather like terms, and solve for x; remember that multiplying or dividing by a positive number does not flip the inequality sign, while multiplying or dividing by a negative number does.

Intervals, Intersections, and Visual Strategies

• Intersection of two sets (two inequalities): the set of x that satisfy both inequalities.
• Example visualization:
  • Suppose one bound starts at 5 and the other ends at 18, and both ends are included (i.e., closed bounds):
    $[5, 18].$
  • This represents the intersection of the two conditions where x must lie between 5 and 18, inclusive.
• Three-region idea to interpret a compound inequality:
  • The real line can be seen as three regions relative to a target interval: left of the lower bound, inside the interval, and right of the upper bound.
  • If you “add 11 in the middle” as a time-saving trick, you’re shifting all three regions by the same amount:
  • Example: shifting the interval [5, 18] by +11 yields [16, 29].
  • Important rule: adding (or subtracting) the same constant to all parts of a compound inequality does not change the direction of the inequalities.
• Does adding a constant affect the inequality direction?