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Section 1.4 #19: Find an equation of the line that passes through the points (1,5) and (2,4).
Solution: We will first find its slope and the use the point-slope form of the line. The slope is m=452(1)=93=3. Using the point-slope form of the equation of the line with slope m=3 and the point (x1,y1)=(1,5) we get y5=3(x(1)) or y5=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=27x+1.
Solution: Since the slope of the given line is 27 the slope of the line parallel is also 27. Therefore using the point-slope form we get the following equation for the parallel line: y5=27(x3), or in slope-intercept form y=27x+297. Since the slope of the given line is 27 the slope of the line parallel is 72. Therefore using the point-slope form we get the following equation for the perpendicular line: y5=72(x3), or in slope-intercept form y=72x+312.

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 1560P and substitute our other values to get 1560P=P(0.04)(1). Add P to both sides to get 1560=1.04P. Hence the amount originally invested is P=15601.04=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 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+(x2)=180. Simplify the left-hand isde to get 7x=182, or and solve for the value of angle A: x=1827=26. Therefore angle B is 265=130 and angle C is 262=24.

Section 1.6 #35: Solve and graph the inequality 3x6orx1>0.
Solution: Divide the left piece 3x6 by 3 to get x2. Add 1 to the right piece to get x>1. Now we graph both inequalities simultaneously: