(a)

\begin{align} 3x + 4 & < - 2 \\ 3x & < - 2 - 4 \\ 3x & < - 6 \\ x & < {-6 \over 3} \\ x & < -2 \end{align}

(b)

\begin{align} 2x + 7 & > 6x + 3 \\ 2x - 6x & > 3 - 7 \\ -4x & > -4 \\ x & < {-4 \over -4} \\ x & < 1 \end{align}

(c)

\begin{align} -5x + 16 & \le 3x \\ -5x - 3x & \le -16 \\ -8x & \le -16 \\ x & \ge {-16 \over -8} \\ x & \ge 2 \end{align}

(d)

\begin{align} 6x - 1 & \ge 8x + 3 \\ 6x - 8x & \ge 3 + 1 \\ -2x & \ge 4 \\ x & \le {4 \over -2} \\ x & \le -2 \end{align}

(a)

\begin{align} {x - 5 \over 4} & \le {-3 \over 1} \\ {x - 5 \over 4} & \le {-12 \over 4} \\ x - 5 & \le -12 \\ x & \le -12 + 5 \\ x & \le -7 \end{align}

(b)

\begin{align} {3x + 1 \over 5} & < {3 - x \over 2} \\ {2(3x + 1) \over 10} & < {5(3 - x) \over 10} \\ 2(3x + 1) & < 5(3 - x) \\ 6x + 2 & < 15 - 5x \\ 6x + 5x & < 15 - 2 \\ 11x & < 13 \\ x & < {13 \over 11} \end{align}

(c)

\begin{align} {x + 3 \over 3} & \ge {2x + 1 \over 2} \\ {2(x + 3) \over 6} & \ge {3(2x + 1) \over 6} \\ 2(x + 3) & \ge 3(2x + 1) \\ 2x + 6 & \ge 6x + 3 \\ 2x - 6x & \ge 3 - 6 \\ -4x & \ge -3 \\ x & \le {-3 \over -4} \\ x & \le {3 \over 4} \end{align}

(d)

\begin{align} {2 - 3x \over 4} & > {x + 5 \over 6} \\ {3(2 - 3x) \over 12} & > {2(x + 5) \over 12} \\ 3(2 - 3x) & > 2(x + 5) \\ 6 - 9x & > 2x + 10 \\ -9x - 2x & > 10 - 6 \\ -11x & > 4 \\ x & < {4 \over -11} \\ x & < -{4 \over 11} \end{align}

(a)

\begin{align} -2x + 10 & > -{4x \over 3} \\ {-2x + 10 \over 1} & > {-4x \over 3} \\ {3(-2x + 10) \over 3} & > {-4x \over 3} \\ 3(-2x + 10) & > -4x \\ -6x + 30 & > -4x \\ -6x + 4x & > -30 \\ -2x & > -30 \\ x & < {-30 \over -2} \\ x & < 15 \\ \\ \\ \therefore x & = 14 \end{align}

(b) A prime number is a positive integer with only two factors - one and itself

$$ x = 13 $$

(c) Perfect squares: 1², 2², 3², 4², ...

$$ x = 3^2 = 9 $$

(a)

\begin{align} {6 - 4x \over 5} + {x + 3 \over 2} & < 0 \\ {2(6 - 4x) \over 10} + {5(x + 3) \over 10} & < {0 \over 10} \\ {2(6 - 4x) + 5(x + 3) \over 10} & < {0 \over 10} \\ 2(6 - 4x) + 5(x + 3) & < 0 \\ 12 - 8x + 5x + 15 & < 0 \\ 27 - 3x & < 0 \\ -3x & < -27 \\ x & > {-27 \over -3} \\ x & > 9 \end{align}

(b)

\begin{align} {6 - 4x \over 5} + {x + 3 \over 2} & \le -3 \\ {2(6 - 4x) \over 10} + {5(x + 3) \over 10} & \le {-30 \over 10} \\ {2(6 - 4x) + 5(x + 3) \over 10} & \le {-30 \over 10} \\ 2(6 - 4x) + 5(x + 3) & \le -30 \\ 12 - 8x + 5x + 15 & \le -30 \\ 27 - 3x & \le -30 \\ -3x & \le -30 -27 \\ -3x & \le -57 \\ x & \ge {-57 \over -3} \\ x & \ge 19 \end{align}

(a)

\begin{align} x - y & = \underbrace{3}_\text{Largest} - \underbrace{(-2)}_\text{Smallest} \\ & = 5 \end{align}

(b) The square of a non-zero number will be a positive number, i.e. (-3)² = 9

\begin{align} x^2 - y & = \underbrace{(-7)^2}_\text{Largest} - \underbrace{(-2)}_\text{Smallest} \\ & = 51 \end{align}

(c)

\begin{align} y^2 - x & = \underbrace{(0)^2}_\text{Smallest} - \underbrace{3}_\text{Largest} \\ & = -3 \end{align}

(d)

\begin{align} xy & = (-7)(-2) \\ & = 14 \end{align}

(a) The square of a non-zero number will be a positive number, i.e. (-3)² = 9. Thus we should make x - y as close to 0 as possible

\begin{align} (x - y)^2 & = (0 - 0)^2 \\ & = 0 \end{align}

(b)

\begin{align} y^3 - x^2 & = (-4)^3 - \underbrace{ (-3)^2}_\text{Largest} \\ & = -73 \end{align}

(c) Note: The product of two negative numbers is positive!

\begin{align} x^3 y & = (-3)^3 (-4) \\ & = (-27)(-4) \\ & = 108 \end{align}

(d)

\begin{align} {y \over z^2} & = {2 \over (-2)^2} \\ & = {1 \over 2} \end{align}

(a)

$$ -3 \le x \le 2 $$

(b)(i)

(b)(ii)

\begin{align} y - x & = \underbrace{ 6 }_\text{Largest} - \underbrace{ (-3) }_\text{Smallest} \\ & = 9 \end{align}

(a)

\begin{align} {4x - 7 \over 3} - {2 \over 1} & \ge {1 \over 1} + {10x - 7 \over 9} \\ {3(4x - 7) \over 9} - {18 \over 9} & \ge {9 \over 9} + {10x - 7 \over 9} \\ {3(4x - 7) - 18 \over 9} & \ge {9 + 10x - 7 \over 9} \\ 3(4x - 7) - 18 & \ge 9 + 10x - 7 \\ 12x - 21 - 18 & \ge 2 + 10x \\ 12x - 39 & \ge 2 + 10x \\ 12x - 10x & \ge 2 + 39 \\ 2x & \ge 41 \\ x & \ge {41 \over 2} \\ x & \ge 20{1 \over 2} \end{align}

(b)

\begin{align} 4x^2 - 12x - 40 & = 4(x^2 - 3x - 10) \\ & = 4(x + 2)(x - 5) \\ \\ \text{Let } & x = 20{1 \over 2}, \\ 4\left(20{1 \over 2} + 2\right) \left(20{1 \over 2} - 5\right) & = 1395 \end{align}

\begin{align} \text{Largest possible (before rounding)} & = 174 \\ \\ \text{Smallest possible (before rounding)} & = 165 \\ \\ \therefore 165 \le x \le 174 & \end{align}