\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]
                    \ineqLine{0}{5}{2}{5}{0}
                \end{solutionorbox}
            \part \[\{x: x<3\}\]
                \begin{solutionorbox}[1in]
                    \ineqLine{5}{-5}{3}{-5}{0}
                \end{solutionorbox}
            \part \[\left\{x: x<-1 \frac{1}{2}\right\}\]
                \begin{solutionorbox}[1in]
                    \ineqLine{0}{-5}{-1}{-5}{0}
                \end{solutionorbox}
            \part \[\{x: x \geq 4\}\]
                \begin{solutionorbox}[1in]
                    \ineqLine{0}{8}{4}{8}{1}
                \end{solutionorbox}
            \part \[\left\{x: x \geq \frac{1}{3}\right\}\]
                \begin{solutionorbox}[1in]
                    \ineqLine{0}{5}{0.33}{5}{1}
                \end{solutionorbox}
            \part \[\left\{x: x \leq 2 \frac{1}{4}\right\}\]
                \begin{solutionorbox}[1in]
                    \ineqLine{3}{-5}{2}{-4}{1}
                \end{solutionorbox}
        \end{parts}
        \question[4] Use \textit{correct} set notation to describe each interval.
        \begin{parts}
            \part \ineqLine{-3}{4}{-1}{3}{0}
            \begin{solutionordottedlines}[1cm]
                $x>-1$
            \end{solutionordottedlines}
            \part \ineqLine[1.5]{-3}{4}{1.5}{-2.5}{1}
            \begin{solutionordottedlines}[1cm]
                $x\leq 1.5$
            \end{solutionordottedlines}
        \end{parts}
        \question[12] Solve these inequalities.
        \begin{parts}\begin{multicols}{2}
            \part \[x+7 \geq 17\]
        \begin{solutionordottedlines}[1in]
        Subtract 7 from both sides: \\
        \(x \geq 17 - 7\) \\
        \(x \geq 10\)
        \end{solutionordottedlines}
            \part \[x-7 \leq-4\]
        \begin{solutionordottedlines}[1in]
        Add 7 to both sides: \\
        \(x \leq -4 + 7\) \\
        \(x \leq 3\)
        \end{solutionordottedlines}
            \part \[-2 x \geq 16\]
        \begin{solutionordottedlines}[1in]
        Divide both sides by -2 (and reverse the inequality): \\
        \(x \leq \frac{16}{-2}\) \\
        \(x \leq -8\)
        \end{solutionordottedlines}
            \part \[\frac{x}{5}-\frac{1}{2} \geq 2\]
        \begin{solutionordottedlines}[1in]
        Add \(\frac{1}{2}\) to both sides: \\
        \(\frac{x}{5} \geq 2 + \frac{1}{2}\) \\
        \(\frac{x}{5} \geq \frac{5}{2}\) \\
        Multiply both sides by 5: \\
        \(x \geq \frac{5}{2} \times 5\) \\
        \(x \geq \frac{25}{2}\) \\
        \(x \geq 12.5\)
        \end{solutionordottedlines}
            \part \[6(x-3) \leq 15\]
        \begin{solutionordottedlines}[1in]
        Divide both sides by 6: \\
        \(x - 3 \leq \frac{15}{6}\) \\
        \(x - 3 \leq 2.5\) \\
        Add 3 to both sides: \\
        \(x \leq 2.5 + 3\) \\
        \(x \leq 5.5\)
        \end{solutionordottedlines}
            \part \[7-4 x \leq-12\]
        \begin{solutionordottedlines}[1in]
        Subtract 7 from both sides: \\
        \(-4x \leq -12 - 7\) \\
        \(-4x \leq -19\) \\
        Divide both sides by -4 (and reverse the inequality): \\
        \(x \geq \frac{-19}{-4}\) \\
        \(x \geq 4.75\)
        \end{solutionordottedlines}
            \part \[11-3 x>18\]
        \begin{solutionordottedlines}[1in]
        Subtract 11 from both sides: \\
        \(-3x > 18 - 11\) \\
        \(-3x > 7\) \\
        Divide both sides by -3 (and reverse the inequality): \\
        \(x < \frac{7}{-3}\) \\
        \(x < -\frac{7}{3}\) \\
        \(x < -2.\overline{3}\)
        \end{solutionordottedlines}
            \part \[6(1-x) \leq 2.4(15-12 x)\]
        \begin{solutionordottedlines}[1in]
        Expand both sides: \\
        \(6 - 6x \leq 36 - 28.8x\) \\
        Add \(6x\) to both sides and subtract 36: \\
        \(6 \leq -22.8x + 36\) \\
        \(-30 \leq -22.8x\) \\
        Divide both sides by -22.8 (and reverse the inequality): \\
        \(x \geq \frac{30}{22.8}\) \\
        \(x \geq \frac{5}{3.8}\) \\
        \(x \geq 1.\overline{315789473684210526}\)
        \end{solutionordottedlines}
            \part \[\frac{x-3}{3} \leq \frac{3+x}{2}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(2(x - 3) \leq 3(3 + x)\) \\
        \(2x - 6 \leq 9 + 3x\) \\
        Subtract \(2x\) and add 6 to both sides: \\
        \(-6 \leq 9 + x\) \\
        \(-15 \leq x\) \\
        \(x \geq -15\)
        \end{solutionordottedlines}
            \part \[\frac{2 x+3}{2}-\frac{x-4}{3}>2\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{3(2x + 3) - 2(x - 4)}{6} > 2\) \\
        \(\frac{6x + 9 - 2x + 8}{6} > 2\) \\
        \(\frac{4x + 17}{6} > 2\) \\
        Multiply both sides by 6: \\
        \(4x + 17 > 12\) \\
        Subtract 17 from both sides: \\
        \(4x > -5\) \\
        Divide both sides by 4: \\
        \(x > -\frac{5}{4}\) \\
        \(x > -1.25\)
        \end{solutionordottedlines}
            \part \[\frac{4-x}{2}+\frac{3-x}{4}<1\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{2(4 - x) + (3 - x)}{4} < 1\) \\
        \(\frac{8 - 2x + 3 - x}{4} < 1\) \\
        \(\frac{11 - 3x}{4} < 1\) \\
        Multiply both sides by 4: \\
        \(11 - 3x < 4\) \\
        Subtract 11 from both sides: \\
        \(-3x < -7\) \\
        Divide both sides by -3 (and reverse the inequality): \\
        \(x > \frac{7}{3}\) \\
        \(x > 2.\overline{3}\)
        \end{solutionordottedlines}
            \part \[\frac{3-2 x}{2} \geq \frac{7-10 x}{5}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(5(3 - 2x) \geq 2(7 - 10x)\) \\
        \(15 - 10x \geq 14 - 20x\) \\
        Add \(10x\) to both sides: \\
        \(15 \geq 14 - 10x + 10x\) \\
        \(15 \geq 14\) \\
        The inequality is always true, so \(x\) can be any real number.
        \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] At 2 p.m., two aeroplanes leave airports \(2880 \mathrm{~km}\) apart and fly towards one another. The average speed of one plane is twice that of the other. If they pass each other at 5 p.m., what is the average speed of each plane?
        \begin{solutionordottedlines}[1in]
        Let the speed of the slower plane be \(v\) km/h. \\
        Then the speed of the faster plane is \(2v\) km/h. \\
        They meet after 3 hours, so the total distance covered is: \\
        \(3v + 3(2v) = 2880\) \\
        \(3v + 6v = 2880\) \\
        \(9v = 2880\) \\
        \(v = 320\) km/h \\
        The speed of the slower plane is 320 km/h, \\
        and the speed of the faster plane is \(2 \times 320 = 640\) km/h.
        \end{solutionordottedlines}
        \question[3] When a mathematics teacher was asked her age she replied, 'One-fifth of my age three years ago, when added to half my age last year, gives my age 11 years ago.' How old is she?
        \begin{solutionordottedlines}[1in]
        Let her current age be \(x\) years. \\
        One-fifth of her age three years ago: \(\frac{x - 3}{5}\) \\
        Half her age last year: \(\frac{x - 1}{2}\) \\
        Her age 11 years ago: \(x - 11\) \\
        The equation is: \\
        \(\frac{x - 3}{5} + \frac{x - 1}{2} = x - 11\) \\
        Multiply through by 10 (LCM of 5 and 2): \\
        \(2(x - 3) + 5(x - 1) = 10(x - 11)\) \\
        \(2x - 6 + 5x - 5 = 10x - 110\) \\
        \(7x - 11 = 10x - 110\) \\
        Subtract \(7x\) from both sides: \\
        \(-11 = 3x - 110\) \\
        Add 110 to both sides: \\
        \(99 = 3x\) \\
        Divide by 3: \\
        \(x = 33\) \\
        The teacher is 33 years old.
        \end{solutionordottedlines}
        \question[3] \(3 \$ 420\) is divided between \(A, B\) and \(C . B\) receives \(\$ 20\) less than \(A\), and \(C\) receives half as much as \(A\) and \(B\) together. How much does each receive?
        \begin{solutionordottedlines}[1in]
        Let \(A\) receive \(x\) dollars. \\
        Then \(B\) receives \(x - 20\) dollars. \\
        \(C\) receives \(\frac{1}{2}(x + (x - 20))\) dollars. \\
        The total amount is: \\
        \(x + (x - 20) + \frac{1}{2}(x + (x - 20)) = 3420\) \\
        \(2x - 20 + x + (x - 20) = 3420\) \\
        \(4x - 40 = 3420\) \\
        Add 40 to both sides: \\
        \(4x = 3460\) \\
        Divide by 4: \\
        \(x = 865\) \\
        So, \(A\) receives \$865, \(B\) receives \(865 - 20 = \$845\), \\
        and \(C\) receives \(\frac{1}{2}(865 + 845) = \$855\).
        \end{solutionordottedlines}
        \question[3] In a printing works, 75000 leaflets are run off by two printing presses in \(18 \frac{3}{4}\) hours. One press delivers 200 more leaflets per hour than the other. Find the number of leaflets produced per hour by each of the machines.
        \begin{solutionordottedlines}[1in]
        Let the number of leaflets the slower press prints per hour be \(x\). \\
        Then the faster press prints \(x + 200\) leaflets per hour. \\
        Total time is \(18 \frac{3}{4}\) hours or \(18.75\) hours. \\
        The equation is: \\
        \(18.75x + 18.75(x + 200) = 75000\) \\
        \(18.75x + 18.75x + 3750 = 75000\) \\
        \(37.5x + 3750 = 75000\) \\
        Subtract 3750 from both sides: \\
        \(37.5x = 71250\) \\
        Divide by 37.5: \\
        \(x = 1900\) \\
        The slower press prints 1900 leaflets per hour, \\
        and the faster press prints \(1900 + 200 = 2100\) leaflets per hour.
        \end{solutionordottedlines}
        \question[4] A salesman makes a trip to visit a client. The traffic he encounters keeps his average speed to \(40 \mathrm{~km} / \mathrm{h}\). On the return trip, he takes a route \(6 \mathrm{~km}\) longer, but he averages \(50 \mathrm{~km} / \mathrm{h}\). If he takes the same time each way, how long was the total journey?
        \begin{solutionordottedlines}[1in]
        Let the distance of the initial trip be \(d\) km. \\
        Time for the initial trip: \(\frac{d}{40}\) hours. \\
        Time for the return trip: \(\frac{d + 6}{50}\) hours. \\
        Since the times are equal: \\
        \(\frac{d}{40} = \frac{d + 6}{50}\) \\
        Cross multiply: \\
        \(50d = 40(d + 6)\) \\
        \(50d = 40d + 240\) \\
        Subtract \(40d\) from both sides: \\
        \(10d = 240\) \\
        Divide by 10: \\
        \(d = 24\) km \\
        Total distance for both trips: \(24 + (24 + 6) = 54\) km \\
        Total time for both trips: \(\frac{24}{40} + \frac{30}{50} = 1.2\) hours
        \end{solutionordottedlines}
        \question[4] Solve these inequalities:
        \begin{parts}
            \part \[\frac{5-2 x}{3}-\frac{5+2 x}{4} \geq-1\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{4(5 - 2x) - 3(5 + 2x)}{12} \geq -1\) \\
        \(\frac{20 - 8x - 15 - 6x}{12} \geq -1\) \\
        \(\frac{5 - 14x}{12} \geq -1\) \\
        Multiply both sides by 12: \\
        \(5 - 14x \geq -12\) \\
        Add 14x to both sides: \\
        \(5 + 14x \geq -12 + 14x\) \\
        Subtract 5 from both sides: \\
        \(14x \geq -17\) \\
        Divide by 14: \\
        \(x \geq -\frac{17}{14}\) \\
        \(x \geq -1.\overline{214285714285714285}\)
        \end{solutionordottedlines}
            \part \[\frac{2 x-1}{7} \geq \frac{2 x+3}{4}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(4(2x - 1) \geq 7(2x + 3)\) \\
        \(8x - 4 \geq 14x + 21\) \\
        Subtract \(8x\) from both sides: \\
        \(-4 \geq 6x + 21\) \\
        Subtract 21 from both sides: \\
        \(-25 \geq 6x\) \\
        Divide by 6 (and reverse the inequality): \\
        \(-\frac{25}{6} \leq x\) \\
        \(x \geq -\frac{25}{6}\) \\
        \(x \geq -4.\overline{1}\)
        \end{solutionordottedlines}
            \part \[\frac{6 x+1}{3}>\frac{3 x-1}{2}+3\]
        \begin{solutionordottedlines}[1in]
        Find a common denominator and combine: \\
        \(\frac{2(6x + 1) - 3(3x - 1)}{6} > 3\) \\
        \(\frac{12x + 2 - 9x + 3}{6} > 3\) \\
        \(\frac{3x + 5}{6} > 3\) \\
        Multiply both sides by 6: \\
        \(3x + 5 > 18\) \\
        Subtract 5 from both sides: \\
        \(3x > 13\) \\
        Divide by 3: \\
        \(x > \frac{13}{3}\) \\
        \(x > 4.\overline{3}\)
        \end{solutionordottedlines}
            \part \[\frac{6-3 x}{2} \geq \frac{3 x-6}{5}\]
        \begin{solutionordottedlines}[1in]
        Cross multiply to compare: \\
        \(5(6 - 3x) \geq 2(3x - 6)\) \\
        \(30 - 15x \geq 6x - 12\) \\
        Add \(15x\) to both sides: \\
        \(30 \geq 21x - 12\) \\
        Add 12 to both sides: \\
        \(42 \geq 21x\) \\
        Divide by 21: \\
        \(2 \geq x\) \\
        \(x \leq 2\)
        \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{problembox}