Exercise 147 Page 301

Looking at the shaded figure, imagine the boundary lines that make up this polygon. They pass through the given points and create a system of four linear inequalities.

We will be able to use the given points to write equations for the boundary lines. From there, we will use these equations to write the inequalities for the system.

Inequality I

The first boundary line passes through the points (-3,-1) and (-3,5). This line is a vertical line, so each point that lies on it will have a x-coordinate of -3. Boundary Line I: x=-3 Because the boundary line is solid, and the shading is on the right side of it, we can conclude that all values greater than or equal to -3 will be included in the solution set of this inequality. Changing the equals sign in our equation to an inequality symbol will give us our first inequality. Boundary Line I: x=-3 Inequality I: x≥ -3

Inequality II

To write an equation for the second boundary line, we will first determine its slope m. We know that the line passes through the points ( -3, 5) and ( 3, 1), so we can use these in the Slope Formula.

Since we know the slope of the boundary line and at least one point through which it passes, we can write its equation in point-slope form. Let's use ( 3, 1). Point-slope form:& y- y_2=m(x- x_2) Boundary Line II:& y- 1=-2/3(x- 3) Now, to write the equation in slope-intercept form, let's isolate y.

Next, to determine the inequality sign, we first observe the boundary line to see if it is strict. In this case, the boundary line is solid, so the points that lie on it are included in the solution set.

We have identified a point on the graph that we know satisfies the inequality. By substituting this point into our inequality and simplifying, we can identify the correct inequality symbol.

We already know that the inequality is not strict and substituting a point from the solution set created a statement that requires a less than or equal to symbol to be true. We can now complete this inequality. Boundary Line II: y=-2/3x+3 Inequality II: y≤-2/3x+3

Inequality III

The third boundary line passes through the points (3,1) and (3,-5). This line is a vertical line, so each point that lies on it will have a x-coordinate of 3. Boundary Line III: x=3 Because the boundary line is solid, and the shading is on the left side of it, we can conclude that all values less than or equal to 3 will be included in the solution set of this inequality. Changing the equals sign in our equation to an inequality symbol will give us our third inequality. Boundary Line III: x=3 Inequality III: x≤ 3

Inequality IV

Looking at the given graph, we can see that the Boundary Line II is parallel to the Boundary Line IV. Therefore their slopes are the same and equal - 23. Once again, by substituting the slope and one of the points through which the line passes into point-slope form, we can write an equation for the boundary line. Let's use ( -3, -1). Point-slope form:& y- y_1=m(x- x_1) Boundary Line IV:& y-( -1)=-2/3(x-( -3)) Let's rewrite this equation in slope-intercept form as well.

To determine the inequality sign, we observe the boundary line to see if it is strict. In this case, the boundary line is solid, so the points that lie on it are included in the solution set.

We've identified a point on the graph that we know satisfies the inequality. By substituting this point into our inequality and simplifying, we can identify the correct inequality symbol.

We already know that the inequality is not strict and substituting a point from the solution set created a statement that requires a greater than or equal to symbol to be true. We can now complete this inequality. Boundary Line IV:& y=-2/3x-3 Inequality IV:& y≥ -2/3x-3

Writing the System

We can combine all of the inequalities to have a completed system of inequalities for the given shaded figure. x≥ -3 y≤ -2/3x+3 x≤ 3 y≥ -2/3x-3