New Discovering Mathematics 3A Textbook solutions
Chapter 2 Try it yourself 1-9
Please email me if you spot any mistakes or have any questions.
Solutions
Click to display or to hide
(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}
(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} \\ {6(3x + 1) \over 24} & \ge {4(5x - 2) \over 24} \\ 6(3x + 1) & \ge 4(5x - 2) \\ 18x + 6 & \ge 20x - 8 \\ 18x - 20x & \ge -8 - 6 \\ -2x & \ge -14 \\ x & \le {-14 \over -2} \\ x & \le 7 \end{align}
(b)
$$ x = 7 $$
(a) A rational number is a number that can be expressed as a fraction
\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)
\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)
\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}
\begin{align} 5x - 8 & < 12 &&& 7x + 3 & < 17 \\ 5x & < 12 + 8 &&& 7x & < 17 - 3 \\ 5x & < 20 &&& 7x & < 14 \\ x & < {20 \over 5} &&& x & < {14 \over 7} \\ x & < 4 &&& x & < 2 \end{align}
$$ x < 2 $$
(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 \\
-y & \le 126 - 180
&&&
y & < 240 - 126 \\
-y & \le -54
&&&
y & < 114 \\
y & \ge 54
\end{align}
\begin{align}
54 \le y & < 114 \\
\\
\text{Since max. score is 100, } & 54 \le y \le 100
\end{align}