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Section 1.4 #19: Find an equation of the line that passes through the points $\left( -1, 5 \right)$ and $\left(2 , -4 \right)$.
Solution: We will first find its slope and the use the point-slope form of the line. The slope is $$m=\dfrac{-4-5}{2-(-1)}=\dfrac{-9}{3} = -3.$$ Using the point-slope form of the equation of the line with slope $m=-3$ and the point $(x_1,y_1)=(-1,5)$ we get $$y-5 = -3(x-(-1))$$ or $$y-5 = -3(x+1),$$ or the slope-intercept form $$y=-3x+2.$$

Section 1.4 #43: Find an equation of the line passing through $(3,5)$ parallel to and perpendicular to the line $y=\dfrac{2}{7}x+1$.
Solution: Since the slope of the given line is $\dfrac{2}{7}$ the slope of the line parallel is also $\dfrac{2}{7}$. Therefore using the point-slope form we get the following equation for the parallel line: $$y-5 = \dfrac{2}{7}(x-3),$$ or in slope-intercept form $$y=\dfrac{2}{7}x+\dfrac{29}{7}.$$ Since the slope of the given line is $\dfrac{2}{7}$ the slope of the line parallel is $-\dfrac{7}{2}$. Therefore using the point-slope form we get the following equation for the perpendicular line: $$y-5 = -\dfrac{7}{2}(x-3),$$ or in slope-intercept form $$y=-\dfrac{7}{2}x + \dfrac{31}{2}.$$

Section 1.5 #36: Khalid makes an investment at $4\%$ simple interest. At the end of $1$ year, the total value of the investment is $\$1560$. How much was originally invested? Solution: Recall the formula$I=Prt$. We are told in the problem that$t=1$and$r=0.04$. We seek$P$but we are not told$I$. We are told the value of$P+I=1560$(note: interest is just the extra that accrues but we are told the total value after$1$year). Therefore in the equation$I=Prt$we may replace$I$with$1560-P$and substitute our other values to get $$1560-P=P(0.04)(1).$$ Add$P$to both sides to get $$1560 = 1.04 P.$$ Hence the amount originally invested is $$P = \dfrac{1560}{1.04} = 1500.$$ Section 1.5 #37: In triangle$ABC$, angle$B$is five times as large as angle$A$. The measure of angle$C$is$2^{\circ}$less than that of angle$A$. Find the measures of the angles. Solution: We add the three angles and use the fact that sum of the angles in a triangle is$180$to get $$5x+x+(x-2)=180.$$ Simplify the left-hand isde to get $$7x = 182,$$ or and solve for the value of angle$A$: $$x=\dfrac{182}{7}=26.$$ Therefore angle$B$is$26 \cdot 5 = 130$and angle$C$is$26-2=24$. Section 1.6 #35: Solve and graph the inequality $$3x \leq -6 \hspace{2pt} \mathrm{or} \hspace{2pt} x-1>0.$$ Solution: Divide the left piece$3x \leq -6$by$3$to get $$x \leq -2.$$ Add$1$to the right piece to get$x>1\$. Now we graph both inequalities simultaneously: 