Algebra, Vol.1 by Redei L.

By Redei L.

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Look at Figure 2-11. Figure 2-11 The boundary is dashed. y 4 3 2 1 −3 −2 −1 1 2 3 −1 x 4 3x + 4y = 12 −2 −3 Now, select a point not on the boundary, say (0, 0). Substitute this into the original inequality: 3x + 4y < 12 ? 3(0) + 4 (0) < 12 ? 0 + 0 < 12 0 < 12 ✓ This is a true statement. This means that the “(0, 0) side” of the boundary line is the desired region to be shaded. Now, shade that region as shown in Figure 2-12. Chapter 2: Segments, Lines, and Inequalities 37 Figure 2-12 The shading is below the line.

If you use the slope-intercept method, the equation, when written in y = mx + b form becomes y= - 34 x + 3 Because the original inequality is <, the boundary line will be a dashed line. Look at Figure 2-11. Figure 2-11 The boundary is dashed. y 4 3 2 1 −3 −2 −1 1 2 3 −1 x 4 3x + 4y = 12 −2 −3 Now, select a point not on the boundary, say (0, 0). Substitute this into the original inequality: 3x + 4y < 12 ? 3(0) + 4 (0) < 12 ? 0 + 0 < 12 0 < 12 ✓ This is a true statement. This means that the “(0, 0) side” of the boundary line is the desired region to be shaded.

See where the shading of the sentences overlaps. The overlapping region is the solution to the system of inequalities. The solution to the system is the region with both shadings (see Figure 3-2). 52 CliffsQuickReview Algebra II Figure 3-2 A system of inequalities. No points in this region will satisfy x + 2y ≤12 y 3x − y = 6 or 3x − y ≥ 6 (0,6) 6 Any point in this region will satisfy 5 Any point in this region will satisfy x + 2y ≤12 3x − y ≥ 6 but not 4 x + 2y ≤12 3 but not 2 3x − y ≥ 6 1 −6 −5 −4 −3 −2 −1 −1 (12,0) (2,0) 1 2 3 4 5 6 7 8 9 10 11 12 x + 2y =12 −2 Any point in this region will satisfy −3 −4 x + 2y ≤12 −5 3x − y ≥ 6 and (0,−6) −6 The solution to the following is, therefore, as shown in Figure 3-3.

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