McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
8. Graphing Linear and Absolute Value Inequalities
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Exercise 28 Page 120

Practice makes perfect
a We have been told that x is the number of bracelets and y is the number of necklaces. Mei has enough beads to make 50 pieces. This means that the sum of the number of bracelets and necklaces is less than or equal to 50.
x+ y ≤ 50
b We can graph the inequality by finding its boundary line. It can be found by replacing the inequality symbol with the equals sign.

&Inequality &&Boundary Line &x+y ≤ 50 &&x+y = 50 Now we will rewrite the equation in slope-intercept form. Thus, we can immediately identify its slope and y-intercept to make the sketch bit easier. &Boundary Line &&Slope-Intercept Form &x+y = 50 &&y=-1x+50 We will plot the y-intercept and find a second point using the slope to draw a line.

Next, we will connect the points with line segment. The line will be solid because the inequality is non-strict.

In the final step, we will test an arbitrary point to decide which region we should shade. We will test the point (0,0) to make the math bit easier.

x+y≤ 50
0+ 0? ≤50
0≤ 50

The point satisfied the inequality, so we will shade the region that contains the point.

c For the possible solutions, we will choose three points in the shaded region.

Thus, the possible solutions for the number of necklaces and bracelets that can be made can be listed as shown below. 10 necklaces, 20 bracelets 20 necklaces, 20 bracelets 20 necklaces, 10 bracelets