(a)

\begin{align} 3(9 - 4x) & < -5x + 8 \\ 27 - 12x & < -5x + 8 \\ - 12x + 5x & < 8 - 27 \\ -7x & < -19 \\ x & > {-19 \over -7} \\ x & > 2{5 \over 7} \end{align}

Note: The textbook answer x > 3 is a misprint. The correct answer is x > 19/7 (i.e. x > 2 5/7).

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

\begin{align} x & = 2^2 = 4 \end{align}

(a)

\begin{align} {3x + 1 \over 4} & \ge {5x - 2 \over 6} \\ {3(3x + 1) \over 12} & \ge {2(5x - 2) \over 12} \\ 3(3x + 1) & \ge 2(5x - 2) \\ 9x + 3 & \ge 10x - 4 \\ 9x - 10x & \ge -4 - 3 \\ -x & \ge -7 \\ x & \le 7 \end{align}

(b)

$$ x = 7 $$

(a) A rational number is a number that can be expressed as a fraction a/b, where a and b are integers and b ≠ 0

\begin{align} {x + 2 \over 2} & \le {4x - 3 \over 5} - 1 \\ {5(x + 2) \over 10} & \le {2(4x - 3) \over 10} - {10 \over 10} \\ {5(x + 2) \over 10} & \le {2(4x - 3) - 10 \over 10} \\ 5(x + 2) & \le 2(4x - 3) - 10 \\ 5x + 10 & \le 8x - 6 - 10 \\ 5x - 8x & \le -6 - 10 - 10 \\ -3x & \le -26 \\ x & \ge {-26 \over -3} \\ x & \ge {26 \over 3} \\ \\ \\ x & = {26 \over 3} \end{align}

(b) Odd numbers: ..., 7, 9, 11, ...

\begin{align} x & \ge {26 \over 3} \\ x & \ge 8{2 \over 3} \\ \\ \\ x & = 9 \end{align}

(c) Perfect cubes: 1³, 2³, 3³, 4³, ...

\begin{align} x & \ge 8{2 \over 3} \\ \\ \\ x & = 3^3 = 27 \end{align}

(a)

$$ -5 \le x \le 1 $$

(b)

(c)(i)

\begin{align} x - y & = \underbrace{-5}_\text{Smallest} - \underbrace{3}_\text{Largest} \\ & = -8 \end{align}

(c)(ii) Note: The product of two negative numbers is positive! Compare (1)(3) = 3, so the greatest is (-5)(-2) = 10.

\begin{align} xy & = (-5)(-2) \\ & = 10 \end{align}

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

\begin{align} y^3 - x^2 & = \underbrace{(3)^3}_\text{Largest} - \underbrace{(0)^2}_\text{Smallest} \\ & = 27 \end{align}

(a)

\begin{align} 6x + 9 & \le 2x + 13 &&& 2x - 8 & < 7x + 12 \\ 6x - 2x & \le 13 - 9 &&& 2x - 7x & < 12 + 8 \\ 4x & \le 4 &&& -5x & < 20 \\ x & \le {4 \over 4} &&& x & > {20 \over -5} \\ x & \le 1 &&& x & > -4 \end{align}

$$ -4 < x \le 1 $$

(b)

$$ \text{Integer values of } x: -3, -2, -1, 0, 1 $$

(a)

\begin{align} 4 - x & < 7 - 4x &&& 7 - 4x & \le 2x - 13 \\ -x + 4x & < 7 - 4 &&& -4x - 2x & \le -13 - 7 \\ 3x & < 3 &&& -6x & \le -20 \\ x & < {3 \over 3} &&& x & \ge {-20 \over -6} \\ x & < 1 &&& x & \ge 3{1 \over 3} \end{align}

$$ \text{No solutions} $$

(b)

\begin{align} 6x - 1 & \le 3(x + 2) &&& 3(x + 2) & < 2 + 5(x + 2) \\ 6x - 1 & \le 3x + 6 &&& 3x + 6 & < 2 + 5x + 10 \\ 6x - 3x & \le 6 + 1 &&& 3x - 5x & < 2 + 10 - 6 \\ 3x & \le 7 &&& -2x & < 6 \\ x & \le {7 \over 3} &&& x & > {6 \over -2} \\ x & \le 2{1 \over 3} &&& x & > -3 \end{align}
\begin{align} -3 < & \phantom{.} x \le 2{1 \over 3} \\ \\ \text{Integer values of } x & = -2, -1, 0, 1, 2 \end{align}

(a)

\begin{align} {2x - 5 \over 3} & \le x &&& x & < {x - 1 \over 2} \\ {2x - 5 \over 3} & \le {3x \over 3} &&& {2x \over 2} & < {x - 1 \over 2} \\ 2x - 5 & \le 3x &&& 2x & < x - 1 \\ 2x - 3x & \le 5 &&& 2x - x & < -1 \\ -x & \le 5 &&& x & < -1 \\ x & \ge -5 \end{align}

$$ -5 \le x < -1 $$

(b)

$$ \text{Integers: } -5, -4, -3, -2 $$

(a)

\begin{align} {-5y \over 4} & < {1 - y \over 1} &&& {1 - y \over 1} & \le {4 - 7y \over 9} \\ {-5y \over 4} & < {4(1 - y) \over 4} &&& {9(1 - y) \over 9} & \le {4 - 7y \over 9} \\ -5y & < 4(1 - y) &&& 9(1 - y) & \le 4 - 7y \\ -5y & < 4 - 4y &&& 9 - 9y & \le 4 - 7y \\ -5y + 4y & < 4 &&& -9y + 7y & \le 4 - 9 \\ -y & < 4 &&& -2y & \le -5 \\ y & > - 4 &&& y & \ge {-5 \over -2} \\ & &&& y & \ge 2{1 \over 2} \end{align}

$$ y \ge 2{1 \over 2} $$

(b) Perfect cubes: 1³, 2³, 3³, 4³, ...

$$ y = 2^3 = 8 $$

\begin{align} \text{Let } y \text{ represent Susan's} & \text{ score in the 3rd component} \\ \\ \text{Susan's mean score} & = {57 + 69 + y \over 3} \\ & = {126 + y \over 3} \\ \\ \text{To obtain } C \text{ grade: } 60 & \le {126 + y \over 3} < 80 \end{align} \begin{align} 60 & \le {126 + y \over 3} &&& {126 + y \over 3} & < 80 \\ {180 \over 3} & \le {126 + y \over 3} &&& {126 + y \over 3} & < {240 \over 3} \\ 180 & \le 126 + y &&& 126 + y & < 240 \\ 180 - 126 & \le y &&& y & < 240 - 126 \\ 54 & \le y &&& y & < 114 \end{align}
\begin{align} 54 \le y & < 114 \\ \\ \text{Since max. score is 100, } & 54 \le y \le 100 \end{align}

(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) Note: The product of two negative numbers is positive! Using the two most negative values gives (-7)(-2) = 14, which is larger than using the two largest positive values, (3)(1) = 3.

\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 & = \underbrace{ (-4)^3}_\text{Smallest} - \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) From (a), x ≥ 20½. For x ≥ 20½, both (x + 2) and (x - 5) are positive and increase as x increases, so 4(x + 2)(x - 5) also increases. Hence the smallest value occurs at x = 20½.

\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 \\ \\ \therefore \text{Smallest value} & = 1395 \end{align}

Note: x is the number of people, so it must be a whole number (an integer).

\begin{align} \text{Largest possible (before rounding)} & = 174 \\ \\ \text{Smallest possible (before rounding)} & = 165 \\ \\ \therefore 165 \le x \le 174 & \\ \\ \text{(equivalently, } 165 \le x < 175 \text{)} & \end{align}

(a)

\begin{align} x + 1 & > 3 &&& 3x & > 15 \\ x & > 3 - 1 &&& x & > {15 \over 3} \\ x & > 2 &&& x & > 5 \end{align}

$$ x > 5 $$

(b)

\begin{align} x - 1 & \le - 4 &&& 3x - 5 & \le 4 \\ x & \le -4 + 1 &&& 3x & \le 4 + 5 \\ x & \le -3 &&& 3x & \le 9 \\ & &&& x & \le {9 \over 3} \\ & &&& x & \le 3 \end{align}

$$ x \le -3 $$

(c)

\begin{align} 2x + 5 & \le 0 &&& 7x - 1 & \ge 13 \\ 2x & \le -5 &&& 7x & \ge 13 + 1 \\ x & \le {-5 \over 2} &&& 7x & \ge 14 \\ x & \le -2.5 &&& x & \ge {14 \over 7} \\ & &&& x & \ge 2 \end{align}

$$ \text{No solutions} $$

(d)

\begin{align} 5x - 3 & < 2 &&& 4x + 7 & \ge x + 3 \\ 5x & < 2 + 3 &&& 4x - x & \ge 3 - 7 \\ 5x & < 5 &&& 3x & \ge - 4 \\ x & < {5 \over 5} &&& x & \ge -{4 \over 3} \\ x & < 1 \end{align}

$$ -1{1 \over 3} \le x < 1 $$

(e)

\begin{align} 6x + 2 & \ge 3x + 11 &&& 9x - 4 & > 5x \\ 6x - 3x & \ge 11 - 2 &&& 9x - 5x & > 4 \\ 3x & \ge 9 &&& 4x & > 4 \\ x & \ge {9 \over 3} &&& x & > {4 \over 4} \\ x & \ge 3 &&& x & > 1 \end{align}

$$ x \ge 3 $$

(f)

\begin{align} 9 - 2x & > 7 - x &&& 11 - 6x & < 8 - 10x \\ -2x + x & > 7 - 9 &&& -6x + 10x & < 8 - 11 \\ -x & > -2 &&& 4x & < -3 \\ x & < 2 &&& x & < -{3 \over 4} \end{align}

$$ x < -{3 \over 4} $$

(g)

\begin{align} {x \over 2} - 1 & < 3 &&& {2x \over 5} -3 & \le x \\ {x \over 2} & < 3 + 1 &&& {2x \over 5} & \le x + 3 \\ {x \over 2} & < 4 &&& {2x \over 5} & \le {x + 3 \over 1} \\ {x \over 2} & < {8 \over 2} &&& {2x \over 5} & \le {5(x + 3) \over 5} \\ x & < 8 &&& 2x & \le 5(x + 3) \\ & &&& 2x & \le 5x + 15 \\ & &&& 2x - 5x & \le 15 \\ & &&& -3x & \le 15 \\ & &&& x & \ge {15 \over -3} \\ & &&& x & \ge -5 \end{align}

$$ -5 \le x < 8 $$

(h)

\begin{align} {x - 1 \over 4} & \le {x + 1 \over 3} &&& {5 - 2x \over 7} & > 1 \\ {3(x - 1) \over 12} & \le {4(x + 1) \over 12} &&& {5 - 2x \over 7} & > {7 \over 7} \\ 3(x - 1) & \le 4(x + 1) &&& 5 - 2x & > 7 \\ 3x - 3 & \le 4x + 4 &&& -2x & > 7 - 5 \\ 3x - 4x & \le 4 + 3 &&& -2x & > 2 \\ -x & \le 7 &&& x & < {2 \over -2} \\ x & \ge - 7 &&& x & < -1 \end{align}

$$ -7 \le x < -1 $$

(a)

\begin{align} -5 & < 2x + 1 &&& 2x + 1 & \le 11 \\ -5 - 1 & < 2x &&& 2x & \le 11 - 1 \\ -6 & < 2x &&& 2x & \le 10 \\ {-6 \over 2} & < x &&& x & \le {10 \over 2} \\ -3 & < x &&& x & \le 5 \end{align}

$$ -3 < x \le 5 $$

(b)

\begin{align} 3x - 5 & < 31 &&& 31 & < 7 - 6x \\ 3x & < 31 + 5 &&& 6x & < 7 - 31 \\ 3x & < 36 &&& 6x & < -24 \\ x & < {36 \over 3} &&& x & < {-24 \over 6} \\ x & < 12 &&& x & < -4 \end{align}

$$ x < -4 $$

(c)

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

$$ x \ge {1 \over 4} $$

(d)

\begin{align} 3x & \le 5x + 13 &&& 5x + 13 & < 4x + 9 \\ 3x - 5x & \le 13 &&& 5x - 4x & < 9 - 13 \\ -2x & \le 13 &&& x & < -4 \\ x & \ge {13 \over -2} \\ x & \ge -6{1 \over 2} \end{align}

$$ -6{1 \over 2} \le x < -4 $$

(a)

\begin{align} 2(4x + 1) & \ge 3(5x - 4) &&& 3(2x - 6) & < 5(x + 1) \\ 8x + 2 & \ge 15x - 12 &&& 6x - 18 & < 5x + 5 \\ 8x - 15x & \ge -12 - 2 &&& 6x - 5x & < 5 + 18 \\ -7x & \ge -14 &&& x & < 23 \\ x & \le {-14 \over -7} \\ x & \le 2 \end{align}

$$ x \le 2 $$

(b)

\begin{align} {3 \over 4}(x + 2) & \le x + 5 &&& x - 4 & > {2 \over 3}(3x + 1) \\ {3 \over 4}x + {3 \over 2} & \le x + 5 &&& x - 4 & > 2x + {2 \over 3} \\ {3 \over 4}x - x & \le 5 - {3 \over 2} &&& x - 2x & > {2 \over 3} + 4 \\ -{1 \over 4}x & \le {7 \over 2} &&& -x & > {14 \over 3} \\ x & \ge { {7 \over 2} \over -{1 \over 4}} &&& x & < - {14 \over 3} \\ x & \ge -14 &&& x & < - 4{2 \over 3} \end{align}
$$ -14 \le x < -4{2 \over 3} $$

(c)

\begin{align} 3(x - 2) & < 5(x + 1) &&& 5(x + 1) & \le 3 - x \\ 3x - 6 & < 5x + 5 &&& 5x + 5 & \le 3 - x \\ 3x - 5x & < 5 + 6 &&& 5x + x & \le 3 - 5 \\ -2x & < 11 &&& 6x & \le -2 \\ x & > {11 \over -2} &&& x & \le {-2 \over 6} \\ x & > -5{1 \over 2} &&& x & \le -{1 \over 3} \end{align}
$$ -5{1 \over 2} < x \le -{1 \over 3} $$

(d)

\begin{align} {x \over 3} & \le {x - 2 \over 2} &&& {x - 2 \over 2} & < {3x + 7 \over 4} \\ {2x \over 6} & \le {3(x - 2) \over 6} &&& {2(x - 2) \over 4} & < {3x + 7 \over 4} \\ 2x & \le 3(x - 2) &&& 2(x - 2) & < 3x + 7 \\ 2x & \le 3x - 6 &&& 2x - 4 & < 3x + 7 \\ 2x - 3x & \le -6 &&& 2x - 3x & < 7 + 4 \\ -x & \le -6 &&& -x & < 11 \\ x & \ge 6 &&& x & > -11 \end{align}

$$ x \ge 6 $$

\begin{align} -2 & < {5x - 1 \over 8} &&& {5x - 1 \over 8} & \le 2 \\ {-16 \over 8} & < {5x - 1 \over 8} &&& {5x - 1 \over 8} & \le {16 \over 8} \\ -16 & < 5x - 1 &&& 5x - 1 & \le 16 \\ -16 + 1 & < 5x &&& 5x & \le 16 + 1 \\ -15 & < 5x &&& 5x & \le 17 \\ {-15 \over 5} & < x &&& x & \le {17 \over 5} \\ -3 & < x &&& x & \le 3{2 \over 5} \end{align}
\begin{align} -3 < & \phantom{.} x \le 3{2 \over 5} \\ \\ \text{Integers: } & -2, -1, 0, 1, 2, 3 \end{align}

(a)

\begin{align} -6 & < 5x - 6 &&& 5x - 6 & \le 36 \\ -6 + 6 & < 5x &&& 5x & \le 36 + 6 \\ 0 & < 5x &&& 5x & \le 42 \\ {0 \over 5} & < x &&& x & \le {42 \over 5} \\ 0 & < x &&& x & \le 8{2 \over 5} \end{align}
$$ 0 < x \le 8{2 \over 5} $$

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

$$ \text{Prime numbers: } 2, 3, 5, 7 $$

(a)

\begin{align} {x - 6 \over 2} & \le {x + 5 \over 4} &&& {x + 5 \over 4} & < {2x - 11 \over 3} \\ {2(x - 6) \over 4} & \le {x + 5 \over 4} &&& {3(x + 5) \over 12} & < {4(2x - 11) \over 12} \\ 2(x - 6) & \le x + 5 &&& 3(x + 5) & < 4(2x - 11) \\ 2x - 12 & \le x + 5 &&& 3x + 15 & < 8x - 44 \\ 2x - x & \le 5 + 12 &&& 3x - 8x & < -44 - 15 \\ x & \le 17 &&& -5x & < -59 \\ & &&& x & > {-59 \over 5} \\ & &&& x & > 11{4 \over 5} \end{align}
\begin{align} & 11{4 \over 5} < x \le 17 \\ \\ \text{Integers: } & 12, 13, 14, 15, 16, 17 \end{align}

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

\begin{align} 11{4 \over 5} < x \le 17 & \\ \\ \text{Prime numbers: } & 13, 17 \end{align}

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

\begin{align} 11{4 \over 5} < x \le 17 & \\ \\ \text{Perfect square: } & 4^2 = 16 \end{align}

Referring to the 4th quadrant in the diagram, the x-coordinate is positive while the y-coordinate is negative

\begin{align} {2k - 1 \over 3} & > 0 &&& {7 - k \over 5} & < 0 \\ {2k - 1 \over 3} & > {0 \over 3} &&& {7 - k \over 5} & < {0 \over 5} \\ 2k - 1 & > 0 &&& 7 - k & < 0 \\ 2k & > 1 &&& -k & < -7 \\ k & > {1 \over 2} &&& k & > 7 \end{align}

$$ k > 7 $$

\begin{align} \text{For } x \text{ to be } 1, 2, 3, 4 \text{ or } 5, & \phantom{0} 5 < 1 - 2k \le 6 \end{align}
\begin{align} 5 & < 1 - 2k &&& 1 - 2k & \le 6 \\ 2k & < 1 - 5 &&& -2k & \le 6 - 1 \\ 2k & < -4 &&& -2k & \le 5 \\ k & < {-4 \over 2} &&& k & \ge {5 \over -2} \\ k & < -2 &&& k & \ge -2.5 \end{align}
$$ -2.5 \le k < -2 $$

(a) The combined solution must be x < -2, so one inequality must give x < -2 and the other must include every x < -2

\begin{align} ax + 3 & > 2x + b &&& cx - 1 & > 5x - 4 \\ ax - 2x & > b - 3 &&& cx - 5x & > - 4 + 1 \\ (a - 2)x & > b - 3 &&& (c - 5)x & > - 3 \\ \\ \text{If } a = 1,& \phantom{.} b =5, &&& \text{If } & c = 1, \\ (1 - 2)x & > 5 - 3 &&& (1 - 5)x & > - 3 \\ - x & > 2 &&& -4x & > - 3 \\ x & < - 2 &&& x & < {3 \over 4} \phantom{000000} [\text{Combined: } x < -2] \end{align}
$$ \therefore a = 1, b = 5, c = 1 $$

(b) One inequality must give x > -3 and the other inequality must give x < 4

\begin{align} (a - 2)x & > b - 3 &&& (c - 5)x & > - 3 \\ \\ \text{If } a = 1 &, b = -1, &&& \text{If } c & = 6, \\ (1 - 2)x & > -1 - 3 &&& (6 - 5)x & > -3 \\ -x & > - 4 &&& x & > - 3 \\ x & < 4 \end{align}
$$ \therefore a = 1, b = -1, c = 6 $$

\begin{align} \text{Length of } n \text{ bricks} & = 5 \times n \\ & = 5n \text{ cm} \end{align} \begin{align} 5n & > 28 &&& 5n & < 40 \\ n & > {28 \over 5} &&& n & < {40 \over 5} \\ n & > 5.6 &&& n & < 8 \end{align}
\begin{align} 5.6 < n < 8 & \\ \\ \text{Possible values of } n & = 6, 7 \end{align}

\begin{align} \text{Volume of cylinder} & = \text{Base area} \times \text{Height} \\ & = 50 \times h \\ & = 50h \text{ cm}^3 \end{align} \begin{align} 50h & \ge 200 &&& 50h & \le 320 \\ h & \ge {200 \over 50} &&& h & \le {320 \over 50} \\ h & \ge 4 &&& h & \le 6.4 \end{align}
$$ 4 \le h \le 6.4 $$

(a)

\begin{align} \text{Temperature of water} & = 24 + 7 \times t \\ & = (24 + 7t)^\circ \text{C} \end{align}

(b)

\begin{align} 24 + 7t & \ge 90 &&& 24 + 7t & \le 98 \\ 7t & \ge 90 -24 &&& 7t & \le 98 - 24 \\ 7t & \ge 66 &&& 7t & \le 74 \\ t & \ge {66 \over 7} &&& t & \le {74 \over 7} \\ t & \ge 9{3 \over 7} &&& t & \le 10{4 \over 7} \end{align}
$$ 9{3 \over 7} \le t \le 10{4 \over 7} $$

(a)

\begin{align} \text{Total amount} & = 3 \times 50 + n \times 10 \\ & = \$ (150 + 10n) \end{align}

(b)(i)

\begin{align} & 260 \le 150 + 10n < 310 \end{align}

(b)(ii)

\begin{align} 260 & \le 150 + 10n &&& 150 + 10n & < 310 \\ 260 - 150 & \le 10n &&& 10n & < 310 - 150 \\ 110 & \le 10n &&& 10n & < 160 \\ {110 \over 10} & \le n &&& n & < {160 \over 10} \\ 11 & \le n &&& n & < 16 \end{align}
\begin{align} 11 \le n < 16 & \\ \\ \text{Min. no of \$ 10 notes } (n) & = 11 \end{align}

(a)(i)

$$ 19.5 \le x < 20.5 $$

(a)(ii)

$$ 12.5 \le y < 13.5 $$

(b)(i)

\begin{align} \text{Perimeter (smallest possible)} & = 2(19.5) + 2(12.5) \\ & = 64 \text{ m} \\ \\ \text{Perimeter (largest possible)} & = 2(20.5) + 2(13.5) \\ & = 68 \text{ m} \\ \\ \\ 64 \le & \text{ Perimeter} < 68 \end{align}

(b)(ii)

\begin{align} \text{Area (smallest possible)} & = 19.5 \times 12.5 \\ & = 243.75 \text{ m}^2 \\ \\ \text{Area (largest possible)} & = 20.5 \times 13.5 \\ & = 276.75 \text{ m}^2 \\ \\ \\ 243.75 \le & \text{ Area} < 276.75 \end{align}

\begin{align} \text{Perimeter} & = 2(2x + 1) + 2(x) \\ & = 4x + 2 + 2x \\ & = (6x + 2) \text{ cm} \end{align} \begin{align} 6x + 2 & \ge 38 &&& 6x + 2 & \le 80 \\ 6x & \ge 38 - 2 &&& 6x & \le 80 - 2 \\ 6x & \ge 36 &&& 6x & \le 78 \\ x & \ge {36 \over 6} &&& x & \le {78 \over 6} \\ x & \ge 6 &&& x & \le 13 \end{align}
$$ 6 \le x \le 13 $$

\begin{align} \text{Mean PSI} & = {40 + 54 + 44 + 38 + x \over 5} \\ & = {176 + x \over 5} \end{align} \begin{align} {176 + x \over 5} & > 45 &&& {176 + x \over 5} & < 48 \\ {176 + x \over 5} & > {45 \over 1} &&& {176 + x \over 5} & < {48 \over 1} \\ {176 + x \over 5} & > {225 \over 5} &&& {176 + x \over 5} & < {240 \over 5} \\ 176 + x & > 225 &&& 176 + x & < 240 \\ x & > 225 - 176 &&& x & < 240 - 176 \\ x & > 49 &&& x & < 64 \end{align}
\begin{align} 49 < x < 64 & \\ \\ \text{Smallest integer: } & 50 \end{align}

(a)

\begin{align} \text{No. of matches won} & = 38 - x - 2x \\ & = 38 - 3x \\ \\ \text{Total points} & = 3(38 - 3x) + 2x \\ & = 114 - 9x + 2x \\ & = 114 - 7x \end{align}

(b)(i)

\begin{align} 114 - 7x & \ge 85 &&& 114 - 7x & < 94 \\ -7x & \ge 85 - 114 &&& -7x & < 94 - 114 \\ -7x & \ge -29 &&& -7x & < -20 \\ x & \le {-29 \over -7} &&& x & > {-20 \over -7} \\ x & \le 4{1 \over 7} &&& x & > 2{6 \over 7} \end{align}
\begin{align} 2 {6 \over 7} < x \le 4{1 \over 7} & \\ \\ \text{Largest value of } x & = 4 \\ \\ \text{Matches drawn} & = 2x \\ & = 2(4) \\ & = 8 \end{align}

(b)(ii) To find the greatest number of matches won (38 - 3x), the value of x must be the smallest

\begin{align} 2 {6 \over 7} < & \phantom{.} x \le 4{1 \over 7} \\ \\ \text{Smallest value of } x & = 3 \\ \\ \text{Matches won} & = 38 - 3(3) \\ & = 29 \end{align}

\begin{align} \text{Let } x \text{ represent the num} & \text{ber of hours} \\ \\ \text{Charges from company } A & = 25 \times x \\ & = \$ 25x \\ \\ \text{Charges from company } B & = 35 + 18 \times x \\ & = \$ (35 + 18x) \\ \\ 35 + 18x & < 25x \\ 18x - 25x & < - 35 \\ -7x & < -35 \\ x & > {-35 \over -7} \\ x & > 5 \\ \\ \\ \text{Minimum hours} & = \text{Smallest integer value of } x \\ & = 6 \text{ hours} \end{align}

\begin{align} \text{Let } x \text{ represent the num} & \text{ber of senior citizens} \\ \\ \text{Total number of bottles} & = 5 \times x + 12 \\ & = 5x + 12 \\ \\ \text{No. of senior citizens with 6 bottles} & = x - 1 \\ \\ 5x + 12 - 6(x - 1) & < 4 \phantom{000000} [\text{Last senior citizen receives less than 4}] \\ 5x + 12 - 6x + 6 & < 4 \\ 5x - 6x & < 4 - 12 - 6 \\ -x & < -14 \\ x & > 14 \\ \\ \text{Smallest no. of senior citizens} & = 15 \end{align}

\begin{align} \text{Time} & = {\text{Distance} \over \text{Speed}} \\ \\ \text{Let } x \text{ represent } & \text{distance of route } X \\ \\ \text{Time taken (route } X) & = {x \over 60} \\ \\ \text{Time taken (route } Y) & = {x + 1 \over 50} \\ \\ 8 \text{ mins} & = {8 \over 60} \text{ h} \\ & = {2 \over 15} \text{ h} \\ \\ {x + 1 \over 50} - {x \over 60} & \ge {2 \over 15} \\ {6(x + 1) \over 300} - {5x \over 300} & \ge {40 \over 300} \\ {6(x + 1) - 5x \over 300} & \ge {40 \over 300} \\ 6(x + 1) - 5x & \ge 40 \\ 6x + 6 - 5x & \ge 40 \\ 6x - 5x & \ge 40 - 6 \\ x & \ge 34 \\ \\ \text{Shortest possible distance (route } X) & = 34 \text{ km} \end{align}

(a)(i)

\begin{align} 5.7 + 0.11 & = 5.81 \text{ million} \end{align}

(a)(ii)

\begin{align} 5.81 + 0.11 & = 5.92 \text{ million} \end{align}

(a)(iii)

\begin{align} (5.7 + 0.11x) \text{ million} \end{align}

(b)

\begin{align} 5.7 + 0.11x & \ge 6.5 \\ 0.11x & \ge 6.5 - 5.7 \\ 0.11x & \ge 0.8 \\ x & \ge {0.8 \over 0.11} \\ x & \ge 7{3 \over 11} \\ \\ \text{Smallest integer value of } & x = 8 \\ \\ \text{Population first reaches } & 6.5 \text{ million 8 years after 2020} \\ \\ \therefore \text{End of } & 2028 \end{align}

(c)

No. Births, deaths, migration, limited resources or events such as pandemics can change how fast the population grows, so it is unlikely to grow by the same amount every year.

\begin{align} \text{Cost for tables} & = 20 \times x \\ & = \$ 20 x \\ \\ \text{Cost for chairs} & = 50 \times 3 \\ & = \$ 150 \\ \\ \text{Cost for fans} & = 1000 - 20x - 150 - \underbrace{40}_\text{Transportation} \\ & = \$ (810 - 20x) \\ \\ \text{Max. no. of fans} & = {810 - 20x \over 70} \end{align}
\begin{align} \text{Max. no. of fans} & \ge 3 &&& \text{Cost of tables and chairs} & \ge \$ 600 \\ {810 - 20x \over 70} & \ge 3 &&& 20x + 150 & \ge 600 \\ {810 - 20x \over 70} & \ge {210 \over 70} &&& 20x & \ge 600 - 150 \\ 810 - 20x & \ge 210 &&& 20x & \ge 450 \\ -20x & \ge 210 - 810 &&& x & \ge {450 \over 20} \\ -20x & \ge -600 &&& x & \ge 22.5 \\ x & \le {-600 \over -20} \\ x & \le 30 \end{align}
\begin{align} 22.5 \le & \phantom{.} x \le 30 \\ \\ \text{Max. no. of fans} & = {810 - 20x \over 70} \\ & = {810 - 20(23) \over 70} \phantom{00000000} [\text{Choose smallest integer value of } x ] \\ & = 5 \end{align}

(a)

\begin{align} {7 - x \over 4} & \le {1 - 2x \over 6} \\ {3(7 - x) \over 12} & \le {2(1 - 2x) \over 12} \\ 3(7 - x) & \le 2(1 - 2x) \\ 21 - 3x & \le 2 - 4x \\ -3x + 4x & \le 2 - 21 \\ x & \le -19 \end{align}

(b)

\begin{align} {x + 3 \over 2} & < 1 - {2x - 1 \over 5} \\ {5(x + 3) \over 10} & < {10 \over 10} - {2(2x - 1) \over 10} \\ {5(x + 3) \over 10} & < {10 - 2(2x - 1) \over 10} \\ 5(x + 3) & < 10 - 2(2x - 1) \\ 5x + 15 & < 10 - 4x + 2 \\ 5x + 4x & < 10 + 2 - 15 \\ 9x & < -3 \\ x & < {-3 \over 9} \\ x & < -{1 \over 3} \end{align}

(a)

\begin{align} -12 & < 3x - 8 &&& 3x - 8 & < 13 \\ -12 + 8 & < 3x &&& 3x & < 13 + 8 \\ -4 & < 3x &&& 3x & < 21 \\ {-4 \over 3} & < x &&& x & < {21 \over 3} \\ -1{1 \over 3} & < x &&& x & < 7 \end{align}
$$ -1{1 \over 3} < x < 7 $$

(b)

\begin{align} -1{1 \over 3} < x < 7 & \\ \\ \text{Smallest integer value of } x & = -1 \\ \\ \text{Greatest integer value of } x & = 6 \end{align}

\begin{align} 7 - 3(3 - 2x) & < 8x + 11 &&& 8x + 11 & < 4x + 3 \\ 7 - 9 + 6x & < 8x + 11 &&& 8x - 4x & < 3 - 11 \\ 6x - 8x & < 11 - 7 + 9 &&& 4x & < -8 \\ -2x & < 13 &&& x & < {-8 \over 4} \\ x & > {13 \over -2} &&& x & < -2 \\ x & > -6{1 \over 2} \end{align}
\begin{align} -6{1 \over 2} & \phantom{.} < x < -2 \\ \\ \text{Integers: } & -6, -5, -4, -3 \end{align}

(a)

\begin{align} -x & \le {3x - 2 \over 4} &&& {3x - 2 \over 4} & \le 13 \\ {-4x \over 4} & \le {3x - 2 \over 4} &&& {3x - 2 \over 4} & \le {52 \over 4} \\ -4x & \le 3x - 2 &&& 3x - 2 & \le 52 \\ -4x - 3x & \le -2 &&& 3x & \le 52 + 2 \\ -7x & \le -2 &&& 3x & \le 54 \\ x & \ge {-2 \over -7} &&& x & \le {54 \over 3} \\ x & \ge {2 \over 7} &&& x & \le 18 \end{align}
$$ {2 \over 7} \le x \le 18 $$

(b)(i) A rational number can be expressed as a fraction a/b, where a and b are integers and b ≠ 0

$$ {2 \over 7} $$

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

$$ 17 $$

(b)(iii) Square number (or perfect square): 1², 2², 3², 4², ...

$$ 4^2 = 16 $$

(a)

\begin{align} 3(x - 1) & < 2(9 + 5x) &&& {2x \over 5} - {3x \over 4} & < 1 - {x \over 2} \\ 3x - 3 & < 18 + 10x &&& {8x \over 20} - {15x \over 20} & < {20 \over 20} - {10x \over 20} \\ 3x - 10x & < 18 + 3 &&& {8x - 15x \over 20} & < {20 - 10x \over 20} \\ -7x & < 21 &&& 8x - 15x & < 20 - 10x \\ x & > {21 \over -7} &&& -7x & < 20 - 10x \\ x & > -3 &&& -7x + 10x & < 20 \\ & &&& 3x & < 20 \\ & &&& x & < {20 \over 3} \\ & &&& x & < 6{2 \over 3} \end{align}
$$ -3 < x < 6{2 \over 3} $$

(b)

\begin{align} -8 & \le 5x + 2 &&& 5x + 2 & < 37 \\ -8 - 2 & \le 5x &&& 5x & < 37 - 2 \\ -10 & \le 5x &&& 5x & < 35 \\ {-10 \over 5} & \le x &&& x & < {35 \over 5} \\ -2 & \le x &&& x & < 7 \end{align}
$$ -2 \le x < 7 $$

(c)

\begin{align} 13 - (1 - x) & < 2x &&& 2x & \le 6(7 - x) \\ 13 - 1 + x & < 2x &&& 2x & \le 42 - 6x \\ 12 + x & < 2x &&& 2x + 6x & \le 42 \\ x - 2x & < -12 &&& 8x & \le 42 \\ -x & < -12 &&& x & \le {42 \over 8} \\ x & > 12 &&& x & \le 5.25 \end{align}

$$ \text{No solutions} $$

(a)

\begin{align} x + y & = \underbrace{2}_\text{Largest} + \underbrace{3}_\text{Largest} \\ & = 5 \end{align}

(b) The difference can be x - y or y - x. The largest value of y - x is 3 - (-3) = 6, so we use x - y

\begin{align} x - y & = \underbrace{2}_\text{Largest} - \underbrace{ (-7)}_\text{Smallest} \\ & = 9 \end{align}

(c) Note: The product of a positive and a negative number is negative! Compare (2)(-7) = -14 and (-3)(3) = -9, so the smallest is -14.

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

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

\begin{align} x^2 - y^2 & = \underbrace{ (0)^2 }_\text{Smallest} - \underbrace{ (-7)^2 }_\text{Largest} \\ & = -49 \end{align}

(a)

\begin{align} {9 + x \over 5} & < {5 + x \over 1} &&& 5 + x & \le 6 \\ {9 + x \over 5} & < {5(5 + x) \over 5} &&& x & \le 6 - 5 \\ 9 + x & < 5(5 + x) &&& x & \le 1 \\ 9 + x & < 25 + 5x \\ x - 5x & < 25 - 9 \\ -4x & < 16 \\ x & > {16 \over -4} \\ x & > -4 \end{align}
$$ -4 < x \le 1 $$

(b) Taking x as an integer, x = -3, -2, -1, 0, 1

\begin{align} \text{Smallest integer value of } x^2 & = (0)^2 \\ & = 0 \\ \\ \text{Largest integer value of } x^2 & = (-3)^2 \\ & = 9 \end{align}

(a)

\begin{align} v & = u + at \\ 30 & = 18 + a(4) \\ 30 & = 18 + 4a \\ 30 - 18 & = 4a \\ 12 & = 4a \\ {12 \over 4} & = a \\ 3 & = a \end{align}

(b)

\begin{align} v & = u + at \\ v & = 10 + 3t \end{align} \begin{align} 10 + 3t & > 16 &&& 10 + 3t & < 49 \\ 3t & > 16 - 10 &&& 3t & < 49 - 10 \\ 3t & > 6 &&& 3t & < 39 \\ t & > {6 \over 3} &&& t & < {39 \over 3} \\ t & > 2 &&& t & < 13 \end{align}
$$ 2 < t < 13 $$

\begin{align} \text{Let } x \text{ represent } & \text{the number of rental days} \\ \\ \text{Cost (Company } A) & = 45 \times x \\ & = \$ 45x \\ \\ \text{Cost (Company } B) & = 75 + 38 \times x \\ & = \$ (75 + 38x) \\ \\ 75 + 38x & < 45x \\ 38x - 45x & < -75 \\ -7x & < - 75 \\ x & > {-75 \over -7} \\ x & > 10{5 \over 7} \\ \\ \text{Minimum no. of days} & = 11 \end{align}

\begin{align} \text{Let } x \text{ represent time taken for} & \text{ a man's haircut (in mins)} \\ \\ \text{Time taken for woman's haircut} & = (x + 20) \text{ mins} \\ \\ \text{Total time} & = 8x + 5(x + 20) \\ & = 8x + 5x + 100 \\ & = (13x + 100) \text{ mins} \end{align}
\begin{align} 13x + 100 & > 295 &&& 13x + 100 & \le 503 \\ 13x & > 295 - 100 &&& 13x & \le 503 - 100 \\ 13x & > 195 &&& 13x & \le 403 \\ x & > {195 \over 13} &&& x & \le {403 \over 13} \\ x & > 15 &&& x & \le {31} \end{align}
\begin{align} & 15 < x \le 31 \\ \\ 15 \text{ mins} & < \text{Time taken (man)} \le 31 \text{ mins} \end{align}

\begin{align} \text{Marks scored from correct answers} & = (n + 8) \times 3 \\ & = 3(n + 8) \\ & = 3n + 24 \\ \\ \text{No. of wrong answers} & = 2n - (n + 8) \\ & = 2n - n - 8 \\ & = n - 8 \\ \\ \text{Marks deducted from wrong answers} & = -1 \times (n - 8) \\ & = -n + 8 \\ \\ \text{Total marks} & = 3n + 24 - n + 8 \\ & = 2n + 32 \end{align}
\begin{align} 2n + 32 & > 40 &&& 2n + 32 & < 50 \\ 2n & > 40 - 32 &&& 2n & < 50 - 32 \\ 2n & > 8 &&& 2n & < 18 \\ n & > {8 \over 2} &&& n & < {18 \over 2} \\ n & > 4 &&& n & < 9 \end{align}
$$ 4 < n < 9 $$

Note: The number of wrong answers, n - 8, cannot be negative, so n ≥ 8. Combined with 4 < n < 9, this gives n = 8.