Exercise 22 Page 149

We want to determine the vertices of the triangle formed by the given system. 8y-19x<74 & (I) 38y+26x≤ 119 & (II) 54y-12x≥- 198 & (III) Let's write the equation of the boundary line for each inequality. We can do it by changing the inequality symbol to an equals sign. 8y-19x=74 & (I) 38y+26x=119 & (II) 54y-12x=- 198 & (III) Each vertex is created by the intersection of two boundary lines from this system. Therefore, we have to find the point of intersection of each pair of lines. To find a point, without graphing, we have to solve the system of equations related to these lines. Let's do it separately for each pair.

Lines (I) and (II)

We will begin by finding the vertex created by the lines 8y-19x=74 and 38y+26x=199. Below we write the system related to these lines. 8y-19x=74 & (I) 38y+26x=119 & (II) Since there is no isolated variable, we will use the Elimination Method.

We have that the first vertex, or point of intersection, is (- 2,4.5).

Lines (II) and (III)

Let's write the system corresponding to the second and third boundary line. 38y+26x=119 & (I) 54y-12x=- 198 & (II) Again, we will use the Elimination Method.

The second vertex is (7.5,- 2).

Lines (I) and (III)

Finally, we will find the last vertex. 8y-19x=74 & (I) 54y-12x=- 198 & (II) Let's use the Elimination Method one more time.

The third vertex is (- 6,- 5).

Graph

You can use a graphing calculator to help you visualize the triangle formed by these vertices. Below, we have included the graph of our system with the vertices highlighted.