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Solutions to various mathematical problems involving graphing linear equations, finding intercepts, sketching solution sets of inequalities, computing values of quadratic functions, and applying math to real-life situations. Students will learn how to label axes, identify intercepts, and graph lines. They will also practice solving inequalities and understanding the concept of half-planes.
Typology: Assignments
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3/ x
y
2y+4x=
3/
Figure 1. Illustration for Problem 1
Problem #1 (5 points): Sketch the graph of 2y + 4x = 3. For full credit, your axes and graph will be appropriately labeled. Also, find both intercepts.
Solution Notice that when y = 0, then the equation becomes 4x = 3 and
so x = 43. Thus, the x-intercept is
3 4 ,^0
. Similarly, when x = 0,
the equation becomes 2( y = 3 and so y = 32. Hence, the y-intercept is
0 , (^32)
. Since the equation is linear, we can now sketch its graph. We
do so in figure 1.
x
y
−
2/
Figure 2. Illustration for Problem 2
Problem #2 (5 points): Sketch the solution set of the inequality y < 3 x − 2, labeling your axes and the intercepts of the boundary line.
Solution We first consider the equality y = 3x − 2. This is the equation of a line in slope-intercept form. The y-intercept is (0, −2). The x-intercept can be found by setting y equal to zero and solving for x. Doing so, we obtain the equation 0 = 3x − 2. Adding two to both sides, we get
2 = 3x and so x = 23. Hence, the x-intercept is
2 3 ,^0
. Now the point
(0, 0) is not on the line. Plugging x = y = 0 into the original inequality leads us to the inequality 0 < 0 −2. This is clearly an invalid inequality, and hence (0, 0) lies in the half-plane given by the inequality y > 3 x−2. Thus, the region we must shade is the half-plane with boundary line y = 3x − 2 that does not include the origin. The boundary line will
Problem #4 (5 points): Khamˆul gives fell beast rides at the Ephel D´uath Ranch. A 3 hour ride costs 50 gold pieces. Assuming that the time of the ride varies directly as the cost, find the cost of a 5 hour ride.
Solution Let t be the time in hours. Let c be the cost in gold pieces. We are given that t is proportional to c; that is, there is a k such that c = kt. When t = 3, c = 50. Thus, 50 = k3 and so k = 503. Hence, c = 503 t. If t = 5, then c = 503 · 5 = 2503. In conclusion, a five hour ride costs exactly 2503 gold pieces.
