\begin{problembox}
    \subsection{Problems:}
    \begin{questions}
        \question[1] David weighs \(5 \mathrm{~kg}\) less than Christos. If Christos weighs \(w \mathrm{~kg}\), express David's weight in terms of \(w\).
        \begin{solutionordottedlines}[1in]
        David's weight = \(w - 5\) kg.
        \end{solutionordottedlines}
        \question[1] When John's age is doubled, the number is 5 more than Kayla's age. If John is \(x\) years old, what is Kayla's age in terms of \(x\) ?
        \begin{solutionordottedlines}[1in]
        Kayla's age = \(2x - 5\) years.
        \end{solutionordottedlines}
        \question[2] The length of a rectangle is \(10 \mathrm{~m}\) greater than the width of the rectangle.
        \begin{parts}
            \part If \(w \mathrm{~m}\) is the width of the rectangle, express the length of the rectangle in terms of \(w\).
        \begin{solutionordottedlines}[1in]
        Length of the rectangle = \(w + 10\) m.
        \end{solutionordottedlines}
            \part If the length of the rectangle is \(\ell \mathrm{m}\), express the width of the rectangle in terms of \(\ell\).
        \begin{solutionordottedlines}[1in]
        Width of the rectangle = \(\ell - 10\) m.
        \end{solutionordottedlines}
        \end{parts}
        \question[6] Solve these equations:
        \begin{parts}\begin{multicols}{2}
            \part \(a+7=5\)
        \begin{solutionordottedlines}[1in]
        \(a = 5 - 7\)
        \(a = -2\)
        \end{solutionordottedlines}
            \part \(-3 m=18\)
        \begin{solutionordottedlines}[1in]
        \(m = \frac{18}{-3}\)
        \(m = -6\)
        \end{solutionordottedlines}
            \part \(6 a+21=-1\)
        \begin{solutionordottedlines}[1in]
        \(6a = -1 - 21\)
        \(6a = -22\)
        \(a = \frac{-22}{6}\)
        \(a = -\frac{11}{3}\)
        \end{solutionordottedlines}
            \part \(7 b-27=-16\)
        \begin{solutionordottedlines}[1in]
        \(7b = -16 + 27\)
        \(7b = 11\)
        \(b = \frac{11}{7}\)
        \end{solutionordottedlines}
            \part \(18-7 x=3 x-15\)
        \begin{solutionordottedlines}[1in]
        \(18 + 15 = 3x + 7x\)
        \(33 = 10x\)
        \(x = \frac{33}{10}\)
        \(x = 3.3\)
        \end{solutionordottedlines}
            \part \(-5 p+8=6-7 p\)
        \begin{solutionordottedlines}[1in]
        \(-5p + 7p = 6 - 8\)
        \(2p = -2\)
        \(p = \frac{-2}{2}\)
        \(p = -1\)
        \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[8] Now solve these:
        \begin{parts}\begin{multicols}{2}
            \part \(5(x+3)=18\)
        \begin{solutionordottedlines}[1in]
        \(5x + 15 = 18\)
        \(5x = 18 - 15\)
        \(5x = 3\)
        \(x = \frac{3}{5}\)
        \end{solutionordottedlines}
            \part \(5(x-2)=15\)
        \begin{solutionordottedlines}[1in]
        \(5x - 10 = 15\)
        \(5x = 15 + 10\)
        \(5x = 25\)
        \(x = \frac{25}{5}\)
        \(x = 5\)
        \end{solutionordottedlines}
            \part \(12(x-1)=96\)
        \begin{solutionordottedlines}[1in]
        \(12x - 12 = 96\)
        \(12x = 96 + 12\)
        \(12x = 108\)
        \(x = \frac{108}{12}\)
        \(x = 9\)
        \end{solutionordottedlines}
            \part \(7(3-x)=20\)
        \begin{solutionordottedlines}[1in]
        \(21 - 7x = 20\)
        \(-7x = 20 - 21\)
        \(-7x = -1\)
        \(x = \frac{-1}{-7}\)
        \(x = \frac{1}{7}\)
        \end{solutionordottedlines}
            \part \(6(11-x)=17\)
        \begin{solutionordottedlines}[1in]
        \(66 - 6x = 17\)
        \(-6x = 17 - 66\)
        \(-6x = -49\)
        \(x = \frac{-49}{-6}\)
        \(x = \frac{49}{6}\)
        \end{solutionordottedlines}
            \part \(6(a+1)-4(a+2)=24\)
        \begin{solutionordottedlines}[1in]
        \(6a + 6 - 4a - 8 = 24\)
        \(2a - 2 = 24\)
        \(2a = 24 + 2\)
        \(2a = 26\)
        \(a = \frac{26}{2}\)
        \(a = 13\)
        \end{solutionordottedlines}
            \part \(15(3 a-1)=2(6 a+7)\)
        \begin{solutionordottedlines}[1in]
        \(45a - 15 = 12a + 14\)
        \(45a - 12a = 14 + 15\)
        \(33a = 29\)
        \(a = \frac{29}{33}\)
        \(a = \frac{29}{33}\)
        \end{solutionordottedlines}
            \part \(-4(5 m+6)=-3(4-5 m)\)
        \begin{solutionordottedlines}[1in]
        \(-20m - 24 = -12 + 15m\)
        \(-20m - 15m = -12 + 24\)
        \(-35m = 12\)
        \(m = \frac{12}{-35}\)
        \(m = -\frac{12}{35}\)
        \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] The length of a rectangular lawn is \(15 \mathrm{~m}\) more than four times its width.
        \begin{parts}
            \part If \(x \mathrm{~m}\) is the width of the lawn, express the length of the lawn in terms of \(x\).
        \begin{solutionordottedlines}[1in]
        Length of the lawn = \(4x + 15\) m.
        \end{solutionordottedlines}
            \part If the perimeter of the lawn is \(265 \mathrm{~m}\), find the length and width of the lawn.
        \begin{solutionordottedlines}[1in]
        Perimeter = \(2(\text{length}) + 2(\text{width})\)
        \(265 = 2(4x + 15) + 2x\)
        \(265 = 8x + 30 + 2x\)
        \(265 = 10x + 30\)
        \(10x = 265 - 30\)
        \(10x = 235\)
        \(x = \frac{235}{10}\)
        \(x = 23.5\) m (width)
        Length = \(4(23.5) + 15\)
        Length = \(94 + 15\)
        Length = \(109\) m.
        \end{solutionordottedlines}
        \end{parts}
        \question[4] Solve for the pronumeral:
        \begin{parts}
            \part \[\frac{x}{3}-2=-7\]
        \begin{solutionordottedlines}[1in]
        \(\frac{x}{3} = -7 + 2\)
        \(\frac{x}{3} = -5\)
        \(x = -5 \times 3\)
        \(x = -15\)
        \end{solutionordottedlines}
            \part \[-2(z-1)=5(z+6)\]
        \begin{solutionordottedlines}[1in]
        \(-2z + 2 = 5z + 30\)
        \(-2z - 5z = 30 - 2\)
        \(-7z = 28\)
        \(z = \frac{28}{-7}\)
        \(z = -4\)
        \end{solutionordottedlines}
            \part \[\frac{3 s-2}{5}+1=6 s\]
        \begin{solutionordottedlines}[1in]
        \(\frac{3s - 2}{5} = 6s - 1\)
        \(3s - 2 = 30s - 5\)
        \(-2 + 5 = 30s - 3s\)
        \(3 = 27s\)
        \(s = \frac{3}{27}\)
        \(s = \frac{1}{9}\)
        \end{solutionordottedlines}
            \part \[\frac{x+2}{5}+\frac{x-1}{2}=\frac{x+1}{3}\]
        \begin{solutionordottedlines}[1in]
        Multiply through by the common denominator, \(30\):
        \(6(x + 2) + 15(x - 1) = 10(x + 1)\)
        \(6x + 12 + 15x - 15 = 10x + 10\)
        \(6x + 15x - 10x = 10 - 12 + 15\)
        \(11x = 13\)
        \(x = \frac{13}{11}\)
        \end{solutionordottedlines}
        \end{parts}
        \question[3] A person has a number of 10 -cent and 20 -cent coins. If their total value is \(\$ 18\), how many are there of each coin if there are:
        \begin{parts}
            \part equal numbers of each coin?
        \begin{solutionordottedlines}[1in]
        Let the number of each coin be \(n\).
        \(0.10n + 0.20n = 18\)
        \(0.30n = 18\)
        \(n = \frac{18}{0.30}\)
        \(n = 60\)
        There are 60 of each coin.
        \end{solutionordottedlines}
            \part twice as many 10 -cent coins as there are 20 -cent coins?
        \begin{solutionordottedlines}[1in]
        Let the number of 20-cent coins be \(n\), then there are \(2n\) 10-cent coins.
        \(0.20n + 0.10(2n) = 18\)
        \(0.20n + 0.20n = 18\)
        \(0.40n = 18\)
        \(n = \frac{18}{0.40}\)
        \(n = 45\)
        There are 45 20-cent coins and 90 10-cent coins.
        \end{solutionordottedlines}
            \part twice as many 20 -cent coins as there are 10 -cent coins?
        \begin{solutionordottedlines}[1in]
        Let the number of 10-cent coins be \(n\), then there are \(2n\) 20-cent coins.
        \(0.10n + 0.20(2n) = 18\)
        \(0.10n + 0.40n = 18\)
        \(0.50n = 18\)
        \(n = \frac{18}{0.50}\)
        \(n = 36\)
        There are 36 10-cent coins and 72 20-cent coins.
        \end{solutionordottedlines}
        \end{parts}
        \question[3] The distance between two towns \(A\) and \(B\) is \(450 \mathrm{~km}\). Find \(x\) if:
        \begin{parts}
            \part the trip takes \(x\) hours at an average speed of \(90 \mathrm{~km} / \mathrm{h}\)
        \begin{solutionordottedlines}[1in]
        \(450 = 90x\)
        \(x = \frac{450}{90}\)
        \(x = 5\) hours.
        \end{solutionordottedlines}
            \part the trip takes \(5 \frac{1}{4}\) hours at an average speed of \(x \mathrm{~km} / \mathrm{h}\)
        \begin{solutionordottedlines}[1in]
        \(450 = x \times 5.25\)
        \(x = \frac{450}{5.25}\)
        \(x = 85.71\) km/h (approximately).
        \end{solutionordottedlines}
            \part the trip takes 5 hours, travelling \(x\) hours at \(110 \mathrm{~km} / \mathrm{h}\) and the remaining time at \(60 \mathrm{~km} / \mathrm{h}\)
        \begin{solutionordottedlines}[1in]
        \(450 = 110x + 60(5 - x)\)
        \(450 = 110x + 300 - 60x\)
        \(450 = 50x + 300\)
        \(50x = 450 - 300\)
        \(50x = 150\)
        \(x = \frac{150}{50}\)
        \(x = 3\) hours at \(110\) km/h and \(2\) hours at \(60\) km/h.
        \end{solutionordottedlines}
        \end{parts}
        \question[3] A room has a length \(8 \mathrm{~m}\) shorter than its width and a perimeter of \(80 \mathrm{~m}\). Find the length and width of the room.
        \begin{solutionordottedlines}[1in]
        Let the width be \(w\) and the length be \(w - 8\).
        Perimeter = \(2(\text{length}) + 2(\text{width})\)
        \(80 = 2(w - 8) + 2w\)
        \(80 = 2w - 16 + 2w\)
        \(80 = 4w - 16\)
        \(4w = 80 + 16\)
        \(4w = 96\)
        \(w = \frac{96}{4}\)
        \(w = 24\) m (width)
        Length = \(24 - 8\)
        Length = \(16\) m.
        \end{solutionordottedlines}
        \question[4] Solve these equations for \(x\).
        \begin{parts}\begin{multicols}{2}
            \part \[\frac{m x+n}{a}=b\]
        \begin{solutionordottedlines}[1in]
        \(mx + n = ab\)
        \(mx = ab - n\)
        \(x = \frac{ab - n}{m}\)
        \end{solutionordottedlines}
            \part \[m(x-n)=p\]
        \begin{solutionordottedlines}[1in]
        \(mx - mn = p\)
        \(mx = p + mn\)
        \(x = \frac{p + mn}{m}\)
        \end{solutionordottedlines}
            \part \[a(b x-c)=d\]
        \begin{solutionordottedlines}[1in]
        \(abx - ac = d\)
        \(abx = d + ac\)
        \(x = \frac{d + ac}{ab}\)
        \end{solutionordottedlines}
            \part \[\frac{m(x+n)}{a}=b\]
        \begin{solutionordottedlines}[1in]
        \(m(x + n) = ab\)
        \(mx + mn = ab\)
        \(mx = ab - mn\)
        \(x = \frac{ab - mn}{m}\)
        \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[6] Copy each statement and insert the correct symbol, \(\geq\) or \(\leq\).
        \begin{parts}\begin{multicols}{2}
            \part \(11 \geq 5\)
            \part \(-8 \geq -12\)
            \part \(-14 \leq -16\)
            \part \(-8 \leq -7\)
            \part \(-11 \leq 5\)
            \part \(-10 \geq -100\)
        \end{multicols}\end{parts}
        \question[12] Graph each set on the number line.
        \begin{parts}
            \part \[\{x: x>2\}\]
                \begin{solutionorbox}[1in]
                A number line with an open circle at 2 and an arrow pointing to the right.
                \end{solutionorbox}
            \part \[\{x: x<3\}\]
                \begin{solutionorbox}[1in]
                A number line with an open circle at 3 and an arrow pointing to the left.
                \end{solutionorbox}
            \part \[\left\{x: x<-1 \frac{1}{2}\right\}\]
               
