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.

8y-19x=74 38y+26x=119

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Solve by elimination

MultEqn

(I): LHS * 19=RHS* 19

152y-361x=1406 38y+26x=119

MultEqn

(II): LHS * 4=RHS* 4

152y-361x=1406 152y+104x=476

SysEqnSub

(I): Subtract (II)

152y-361x-( 152y+104x)=1406- 476 152y+104x=476

Distr

(I): Distribute - 1

152y-361x-152y-104x=1406-476 152y+104x=476

SubTerms

(I): Subtract terms

- 465x=930 152y+104x=476

DivEqn

(I): .LHS /(- 465).=.RHS /(- 465).

x=- 2 152y+104x=476

Substitute

(II): x= - 2

x=- 2 152y+104( - 2)=476

MultPosNeg

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

x=- 2 152y-208=476

AddEqn

(II): LHS+208=RHS+208

x=- 2 152y=585

DivEqn

(II): .LHS /152.=.RHS /152.

x=- 2 y=4.5

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.

38y+26x=119 54y-12x=- 198

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Solve by elimination

MultEqn

(I): LHS * 6=RHS* 6

228y+156x=714 54y-12x=- 198

MultEqn

(II): LHS * 13=RHS* 13

228y+156x=714 702y-156x=- 2574

SysEqnAdd

(II): Add (I)

228y+156x=714 702y-156x+ 228y+156x=- 2574+ 714

AddTerms

(II): Add terms

228y+156x=714 930y=- 1860

DivEqn

(II): .LHS /930.=.RHS /930.

228y+156x=714 y=- 2

Substitute

(I): y= - 2

228( - 2)+156x=714 y=- 2

MultPosNeg

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

- 456+156x=714 y=- 2

AddEqn

(I): LHS+456=RHS+456

156x=1170 y=- 2

DivEqn

(I): .LHS /156.=.RHS /156.

x=7.5 y=- 2

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.

8y-19x=74 54y-12x=- 198

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Solve by elimination

MultEqn

(I): LHS * 27=RHS* 27

216y-513x=1998 54y-12x=- 198

MultEqn

(II): LHS * (- 4)=RHS* (- 4)

216y-513x=1998 - 216y+48x=792

SysEqnAdd

(II): Add (I)

216y-513x=1998 - 216y+48x+ 216y-513x=792+ 1998

AddSubTerms

(II): Add and subtract terms

216y-513x=1998 - 465x=2790

DivEqn

(II): .LHS /(- 465).=.RHS /(- 465).

216y-513x=1998 x=- 6

Substitute

(I): x= - 6

216y-513( - 6)=1998 x=- 6

MultNegNegOnePar

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

216y+3078=1998 x=- 6

SubEqn

(I): LHS-3078=RHS-3078

216y=- 1080 x=- 6

DivEqn

(I): .LHS /216.=.RHS /216.

y=- 5 x=- 6

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.