5 - solving systems of inequalities by graphing

8/16/2019 5 - Solving Systems of Inequalities by Graphing

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3-2 Study Guide and InterventionSolving Systems of Inequalities by Graphing

Systems of Inequalities To solve a system of inequalities, graph the inequalities in the same coordinate plane.The solution of the system is the region shaded for all of the inequalities.

Example: Solve the system of inequalities.

y ≤ 2 x – 1 and y > x

3 + 2

The solution of y ≤ 2 x – 1 is Regions 1 and 2.

The solution of y > x

3 + 2 is Regions 1 and .

The intersection of these regions is Region 1, !hich is

the solution set of the system of inequalities.

Exerises

Solve eah system of inequalities !y "raphin".

1. x – y ≤ 2 2. x – 2 y ≤ –1 #. y ≤ 1 x + 2 y " 1 x + # y " –12 x > 2

$. y " x

2 – %. y $

x

3 + 2 &. y " –

x

4

1

y $ 2 x y $ –2 x + 1 y $ x – 1

'. x + y " # (. x + y $ ). x – 2 y > %

2 x – y > 2 x – 2 y " # x + # y $ –#

Chapter 3 12 Glencoe Algebra


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3-2 Study Guide and Intervention (continued)Solving Systems of Inequalities by Graphing

*ind erties of an Enlosed ,e"ion &ometimes the graph of a system of inequalities produces an enclosed region ithe form of a polygon. 'ou can find the vertices of the region (y a com(ination of the methods used earlier in this chapte

graphing, su(stitution, and*or elimination.

Example: *ind the oordinates of the verties of the trian"le formed !y % x + $ y - 2/ y - 2 x + #/ and x – # y - $.raph each inequality. The intersections of the (oundary lines are the

vertices of a triangle. The verte -#, / can (e determined from the graph.

To find the coordinates of the second and third vertices, solve the t!o

systems of equations

y=2 x+3

5 x+4 y=20 and

y=2 x+3

x−3 y=4

0or the first system of equations, re!rite the first equation

in standard form as 2 x – y –. Then multiply that

equation (y # and add to the second equation.

2 x – y – Multiply by 4. x – # y –12

3 x + # y 2 - +/ 3 x + # y 2

1 x

x

8

13

Then su(stitute x 8

13 in one of the original equations

and solve for y.

2 ( 813 ) – y –16

13 – y –

y 55

13

The coordinates of the second verte are ( 813 ,4 313 ) .

0or the second system of equations, use su(stitution.

&u(stitute 2 x + for y in the second equation to get

x – -2 x + / #

x – % x – 4 #

–3 x 1

x –13

5

Then su(stitute x –13

5 in the first equation to solve

for y.

y 2 (−135 ) + y –

26

5 +

y –11

5

The coordinates of the third verte are (−2 35 ,−2 15 )

Chapter 3 13 Glencoe Algebra


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Thus, the coordinates of the three vertices are -#, /, ( 813 ,4 313 ) and (−2 35 ,−21

5 ) .

Exerises

*ind the oordinates of the verties of the trian"le formed !y eah system of inequalities.

1. y ≤ – x + 5 2. x > – #. y $ –1

2 x +

y $1

2 x y $ –

1

3 x + y >

1

2 x + 1

y > –2 y > x – 1 y $ x + 1

5 - solving systems of inequalities by graphing

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