Exercise 42 Page 151

Let's begin by graphing each inequality. Then, we will identify the overlapping region and find the coordinates of the vertices.

Inequality I

Let's graph the first inequality. By exchanging the inequality symbol to an equals sign, we can find the boundary line. Then, we will rewrite the equation in slope-intercept form. Inequality: 2y-x≥- 20 Boundary Line: 2y-x=- 20 Slope-Intercept Form: y=1/2x-10 The y-intercept of this boundary line is - 10 and the slope is 12. The line will be solid as the inequality is not strict. Let's test the point ( 0, 0). Since the point satisfies the inequality, we will shade the region that contains the point.

Inequality II

Let's graph the second inequality. This time the boundary line is y=- 3x-6. Again, the line will be solid and we will use ( 0, 0) as our test point.

y≥- 3x-6

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Substitute coordinates and evaluate

SubstituteII

x= 0, y= 0

0? ≥- 3( 0)-6

ZeroPropMult

Zero Property of Multiplication

0? ≥0-6

SubTerm

Subtract term

0≥- 6

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

Inequality III

We will follow the same process for the third inequality. The boundary line is y=- 2x+2. Next, to determine which region to shade, we will test the point ( 0, 0).

y≤- 2x+2

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Substitute coordinates and evaluate

SubstituteII

x= 0, y= 0

0? ≤- 2( 0)+2

ZeroPropMult

Zero Property of Multiplication

0? ≤0+2

IdPropAdd

Identity Property of Addition

0≤ 2

Yet again, the point satisfied the inequality so we will shade the region containing the point. Remember that the line will be solid.

Inequality IV

Finally, we will graph the fourth inequality. Its boundary line is y=2x+14 and it is solid. Let's test the point ( 0, 0).

y≤ 2x+14

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Substitute coordinates and evaluate

SubstituteII

x= 0, y= 0

0? ≤2( 0)+14

ZeroPropMult

Zero Property of Multiplication

0? ≤0+14

IdPropAdd

Identity Property of Addition

0≤ 14

Thus, we will shade the region that includes the point.

Combining the Inequality Graphs

Let's draw the graphs of the inequalities on the same coordinate plane and highlight the vertices.

To determine the first vertex we have to find the points of intersection of the boundary lines. Let's calculate the vertex formed by y=- 3x-6 and y=2x+14. To do this, we have to solve the system of equations related to these lines.

y=- 3x-6 & (I) y=2x+14 & (II)

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Solve by substitution

Substitute

(I): y= 2x+14

2x+14=- 3x-6 y=2x+14

AddEqn

(I): LHS+3x=RHS+3x

5x+14=- 6 y=2x+14

SubEqn

(I): LHS-14=RHS-14

5x=- 20 y=2x+14

DivEqn

(I): .LHS /5.=.RHS /5.

x=- 4 y=2x+14

Substitute

(II): x= - 4

x=- 4 y=2( - 4)+14

MultPosNeg

(II): a(- b)=- a * b

x=- 4 y=- 8+14

AddTerms

(II): Add terms

x=- 4 y=6

The first vertex is (- 4,6). We can determine the other vertices in the same way.

Line I	Line II	Intersection
y=- 2x+2	y=2x+14	(- 3,8)
2y-x=- 20	y=- 2x+2	(4.8,- 7.6)
2y-x=- 20	y=- 3x-6	(8/7,-66/7)

Thus, the remaining vertices are (- 3,8), (4.8,- 7.6), and ( 87,- 667).