Inequalities and linear programming - (8.1-8.3)

Solve compound linear inequalities in one unknown

  • Theory
    For any real number a and b

    1) If \(a > b\)
    Given \(x > a\) and \(x > b\)
    Then \(x > a\)
    for value of x greater than b may not be greater than a,
    the selected value range must satisfy both constraint connected by “and”)

    2) If \(a > b\)
    Given \(x > a\) or \(x > b\)
    Then \(x > b\)
    for value of x either greater than a or greater than b
    satisfies the constraint connected by “or” )
  • Examples
    Solve the following compound inequalities and represent the solutions graphically.
    \(a)\qquad\left\{ \begin{array}{a} 4(x - 1) < x + 5\\6(2x + 3) < 11(x - 2)\end{array} \right.\)
    \( b)\qquad x + 5 \le 2x + 7\) or \(7x - 5 < 3x - 4\)
  • Solutions

    \(a)\qquad 4(x - 1) < x + 5\) and \(6(2x + 3) < 11(x - 2)\)
    \(4x - 4 < x + 5\) and \(12x + 18 < 11x - 22\)
    \(3x < 9\) and \(x < - 40\)
    \(x < 3\) and \(x < - 40\)
    The solutions of the compound inequality are \(x < - 40\).
    Graphically,


    \( b)\qquad x + 5 \le 2x + 7\) or \(7x - 5 < 3x - 4\)
    \( - x \le 2\) or \(4x < 1\)
    \(x \ge - 2\) or \(x < \frac{1}{4}\)
    The solutions of the compound inequality are all real numbers.
    Graphically,

Solve quadratic inequalities in one unknown by the graphical method

  • Examples

  • Examples
    The figure shows the graph of\(\;y = \frac{4}{{25}}\left( {{x^2} - 14x + 24} \right)\) .
    Solve the quadratic equation \({x^2} - 14x + 24 < 0\) graphically.
  • Solutions

    From the graph \(y = \frac{4}{{25}}\left( {{x^2} - 14x + 12} \right),\;\)
    For \(2 < x < 12\) , \(y < 0\)
    \(\frac{4}{{25}}\left( {{x^2} - 14x + 12} \right)\left\langle {0\; = } \right\rangle \;{x^2} - 14x + 12 < 0\)
    \(\therefore \) the solution to \({x^2} - 14x + 12 < 0\) is \(2 < x < 12\)

Solve quadratic inequalities in one unknown by the algebraic method

  • Examples
    Solve the inequality \({x^2} - x - 56 \ge 0\) by the algebraic method.
  • Solutions

    \(\begin{array}{1}{x^2} - x - 56 \ge 0\\(x + 7)(x - 8) \ge 0\end{array}\)

    \(\left\{ \begin{array}{l}x + 7 \ge 0\\x - 8 \ge 0\end{array} \right.\) or \(\left\{ \begin{array}{l}x + 7 \le 0\\x - 8 \le 0\end{array} \right.\)

    \(\left\{ \begin{array}{l}x \ge - 7\\x \ge 8\end{array} \right.\) or \(\left\{ \begin{array}{l}x \le - 7\\x \le 8\end{array} \right.\)

    \(x \ge 8\) or \(x \le - 7\)
    Therefore,the solutions of the inequality are \(x \le - 7\) or \(x \ge 8\).

    Alternative method: \(\begin{array}{c}{x^2} - x - 56 \ge 0\\(x + 7)(x - 8) \ge 0\end{array}\)
    Construct the following table to show the signs of the expressions of
    \(x + 7\), \(x - 8\) and \((x + 7)(x - 8)\) in the relevant intervals of x.

     \(x < - 7\)\(x = - 7\)\( - 7 < x < 8\)\(x = 8\)\(x > 8\)
    \(x + 7\)-0+++
    \(x - 8\)---0+
    \((x + 7)(x - 8)\)+0-0+

    According to the table above, the solutions of the inequality are \(x \le - 7\) or \(x \ge 8\).