New Discovering Mathematics 3A Textbook solutions

Ex 2.2

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Solutions

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Question 1

(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 & \le -2.5 &&& 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}

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

Question 2

(a)

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

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

Question 3

(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}

$$ x \ge 6 $$

Question 4

\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 \\ -5x & < -1 + 16 &&& 5x & \le 16 + 1 \\ -5x & < 15 &&& 5x & \le 17 \\ x & > {15 \over -5} &&& x & \le {17 \over 5} \\ x & > -3 &&& 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}

Question 5

(a)

\begin{align} -6 & < 5x - 6 &&& 5x - 6 & \le 36 \\ -5x & < -6 + 6 &&& 5x & \le 36 + 6 \\ -5x & < 0 &&& 5x & \le 42 \\ x & > {0 \over -5} &&& x & \le {42 \over 5} \\ x & > 0 &&& 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

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

Question 6

\begin{align} 2x - 3 & \le 4x + 6 &&& 8x - 1 & > 5x - 7 \\ 2x - 4x & \le 6 + 3 &&& 8x - 5x & > -7 + 1 \\ -2x & \le 9 &&& 3x & > -6 \\ x & \ge {9 \over -2} &&& x & > {-6 \over 3} \\ x & \ge -4{1 \over 2} &&& x & > -2 \end{align}

\begin{align} \therefore & \phantom{.} x > -2 \\ \\ \text{Smallest } & \text{integer: } -1 \end{align}

Question 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}

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

Question 8

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}

$$ k > 7 $$

Question 9

\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}
$$ -2.5 \le k < -2 $$

Question 10

(a) Both inequalities must satisfy 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{Both satisfies } x < -2] \end{align}
$$ \therefore a = 1, b = 5, c = 1 $$

(b) One equality must satisfy x > -3 and the other equality must satisfy 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}
$$ \therefore a = 1, b = -1, c = 6 $$

Question 11

\begin{align} x - 2a & > 1 &&& 6 - 3x & \ge -1 \\ x & > 1 + 2a &&& -3x & \ge -1 -6 \\ & &&& -3x & \ge - 7 \\ & &&& x & \le {-7 \over -3} \\ & &&& x & \le 2{1 \over 3} \end{align}

\begin{align} \text{For inequalities } & \text{to have no solution,} \\ \\ 1 + 2a & \ge 2{1 \over 3} \\ 2a & \ge 2{1 \over 3} - 1 \\ 2a & \ge 1{1 \over 3} \\ a & \ge { 1{1 \over 3} \over 2} \\ a & \ge {2 \over 3} \end{align}