3-1 Study GuideSolving Systems of Equations

Solve Systems Graphically A system of equations is two or more equations with the same variables. You can solve a system of linear equations by using a table or by graphing the equations on the same coordinate plane. If the lines intersect, the solution is that intersection point. The following chart summarizes the possibilities for graphs of two linearequations in two variables.Example: Graph the system of equations x – 3y = 6

2x – y = –3

Write each equation in slope-intercept form.

x – 3 y = 6 → y = 13 x – 2

2 x – y = –3 → y = 2x + 3The graphs intersect at (–3, –3).

ExercisesGraph each system of equations

1. 3x + y = –2 2. x + 2y = 5 3. 2x – 3y = 06x + 2y = 10 3x – 15 = –6y 4x – 6y = 3

4. 2x – y = 3 5. 4x + y = –2 6. 3x – y = 2

x+ 2y = 4 2x + y2 = –1 x + y = 6

ExercisesSolve each system of equations using substitution or elimination.

1. 3x + y = 7 2. 2x + y = 5 3. 2x + 3y = –34x + 2y = 16 3x – 3y = 3 x + 2y = 2

Chapter 3 Test Review Name________________________________ Block_______

4. 2x – y = 7 5. 7x + 2y = –1 6. 5x + 2y = 126x – 3y = 14 4x – 3y = –13 –6x – 2y = –14

3-2 Study GuideSolving Systems of Inequalities by GraphingSystems 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 ≤ 2x – 1 and y > x3 + 2

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

The solution of y > x3 + 2 is Regions 1 and 3.

The intersection of these regions is Region 1, which isthe solution set of the system of inequalities.

Exercises

Solve each system of inequalities by graphing.

1. x – y ≤ 2 2. 3x – 2y ≤ –1 3. y ≤ 1x + 2y ≥ 1 x + 4y ≥ –12 x > 2

4. y ≥ x2 – 3 5. y <

x3 + 2 6. y ≥ –

x4 + 1

y < 2x y < –2x + 1 y < 3x – 1

7. x + y ≥ 4 8. x + 3y < 3 9. x – 2y > 62x – y > 2 x – 2y ≥ 4 x + 4y < –4

Find Vertices of an Enclosed Region Sometimes the graph of a system of inequalities produces an enclosed region in the form of a polygon. You can find the vertices of the region by a combination of the methods used earlier in this chapter: graphing, substitution, and/or elimination.

Example: Find the coordinates of the vertices of the triangle formed by 5x + 4y < 20, y < 2x + 3, and x – 3y < 4.Graph each inequality. The intersections of the boundary lines are thevertices of a triangle. The vertex (4, 0) can be determined from the graph.To find the coordinates of the second and third vertices, solve the twosystems of equations

y=2 x+35 x+4 y=20 and

y=2 x+3x−3 y=4

For the first system of equations, rewrite the first equation in standard form as 2x – y = –3. Then multiply that equation by 4 and add to the second equation.2x – y = –3 Multiply by 4. 8x – 4y = –125x + 4y = 20 ( + ) 5 x + 4 y = 20

13x = 8

x =8

13

Then substitute x = 8

13 in one of the original equations and

solve for y.

2 ( 813 ) – y = –3

1613

– y = –3

y = 5513

The coordinates of the second vertex are ( 813

, 4 313 ).

For the second system of equations, use substitution.Substitute 2x + 3 for y in the second equation to getx – 3(2x + 3) = 4

x – 6x – 9 = 4–5x = 13

x = – 135

Then substitute x = – 135

in the first equation to solve for y.

y = 2 (−135 ) + 3

y = – 265

+ 3

y = – 115

The coordinates of the third vertex are (−2 35

,−2 15 ).

Thus, the coordinates of the three vertices are (4, 0), ( 813

, 4 313 ) and (−2 3

5,−2 1

5 ).

ExercisesFind the coordinates of the vertices of the triangle formed by each system of inequalities.

1. y ≤ –3x + 7 2. x > –3

y < 12x y < –

13x + 3

y > –2 y > x – 1

3-3 Study GuideOptimization with Linear ProgrammingMaximum and Minimum Values When a system of linear inequalities produces a bounded polygonal region, the maximum or minimum value of a related function will occur at a vertex of the region.

Example: Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function f(x, y) = 3x + 2y for this polygonal region.

y ≤ 4

y ≤ –x + 6

y ≥ 12x –

32

y ≤ 6x + 4

First find the vertices of the bounded region. Graph the inequalities.

The polygon formed is a quadrilateral with vertices at(0, 4), (2, 4), (5, 1), and (–1, –2). Use the table to find themaximum and minimum values of f(x, y) = 3x + 2y.

(x, y) 3x + 2y f (x, y)(0, 4) 3(0) + 2(4) 8

(2, 4) 3(2) + 2(4) 14

(5, 1) 3(5) + 2(1) 17

(–1, –2) 3(–1) + 2(–2) –7

The maximum value is 17 at (5, 1). The minimum value is –7 at (–1, –2).

ExercisesGraph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the given function for this region.

1. y ≥ 2 2. y ≥ –2 3. x + y ≥ 21 ≤ x ≤ 5 y ≥ 2x – 4 4y ≤ x + 8y ≤ x + 3 x – 2y ≥ –1 y ≥ 2x – 5f(x, y) = 3x – 2y f(x, y) = 4x – y f(x, y) = 4x + 3y

3-4 Study GuideSystems of Equations in Three Variables

Systems in Three Variables Use the methods used for solving systems of linear equations in two variables to solve systems of equations in three variables. A system of three equations in three variables can have a unique solution, infinitely many solutions, or no solution. A solution is an ordered triple.

ExercisesSolve each system of equations.

1. 2x + 3y – z = 0 2. 2x – y + 4z = 11 x – 2y – 4z = 14 x + 2y – 6z = –11 3x + y – 8z = 17 3x – 2y –10z = 11

4. 2x – 4y – z = 10 3. x – 2y + z = 8 4x – 8y – 2z = 16 2x + y – z = 0 3x + y + z = 12

3x – 6y + 3z = 24

3-5 Study Guide Operations with MatricesAlgebraic Operations Matrices with the same dimensions can be added together or one can be subtracted from the other.

Addition of Matrices [a b cd e fg h i ] + [ j k l

m n op q r ] = [ a+ j b+k c+l

d+m e+n f +og+p h+q i+r ]

Subtraction of Matrices [a b cd e fg h i ] – [ j k l

m n op q r ] = [ a− j b−k c−l

d−m e−n f −og−p h−q i−r ]

You can multiply an m × n matrix by a scalar k.

Scalar Multiplication k [a b cd e f ] = [ka kb kc

kd ke kf ]Exercises

Perform the indicated operations. If the matrix does not exist, write impossible.

1. [ 8 7−10 −6] – [−4 3

2 −12] 2. [ 6 −5 9−3 −4 5] + [−4 3 2

6 9 −4]

3. 3[−4 52 3] 4. –2 [8 −1

0 56 −6] + 4 [−1 0 −1

5 10 15 ]

5. If A=[3 01 −5] , B=[ 8 3

−2 3] ,∧C=[6 40 1] , find the following .

4A – C -B+C 3A+5C

3-6 Study GuideMultiplying MatricesMultiply Matrices You can multiply two matrices if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix.

Multiplication of Matrices A ⋅ B = AB

[a bc d ] ⋅ [ e f

g h] = [ae+bg af +bhce+dg cf +dh ]

Example: Find AB if A = [−4 32 −21 7 ] and B = [ 5 −2

−1 3 ]AB = [−4 3

2 −21 7 ] ⋅ [ 5 −2

−1 3 ] Substitution

= [ −4 (5 )+3 (−1 ) −4 (−2 )+3 (3 )2(5)+(−2)(−1) 2(−2)+(−2)(3)

1 (5 )+7 (−1 ) 1 (−2 )+7 (3 ) ] Multiply columns by rows.

= [−23 1712 −10−2 19 ] Simplify.

ExercisesFind each product, if possible.

1. [ 4 1−2 3] ⋅ [3 0

0 3 ] 2. [−1 03 7 ] ⋅ [ 3 2

−1 4] 3. [3 −12 4 ] ⋅ [3 −1

2 4 ]

4. [−3 15 −2] ⋅ [ 4 0 −2

−3 1 1 ] 5. [ 3 −20 4

−5 1 ] ⋅ [1 22 1] 6. [5 −2

2 −3 ] ⋅ [ 4 −1−2 5 ]

7. [ 6 10−4 3−2 7 ] ⋅ [0 4 –3] 8. [7 −2

5 −4 ] ⋅ [ 1 −3−2 0 ] 9. [ 2 0 −3

1 4 −2−1 3 1 ] ⋅ [ 2 −2

3 1−2 4 ]

3-7 Study GuideSolving Systems of Equations Using Cramer’s RuleDeterminants A 2×2 matrix has a second-order determinant; a 3×3 matrix has a third-order determinant.

Second-OrderDeterminant For the matrix [a b

c d ], the determinant is |a bc d| = ad – bc.

Third-OrderDeterminant

For the matrix [a b cd e fg h i ] the determinant is found using the diagonal rule.

[a b cd e fg h i ]

a bd eg h [a b c

d e fg h i ]

a bd eg h

Area of a Triangle

The area of a triangle having vertices (a, b), (c, d), and (e, f) is ⎪A ⎥ ,

where A = 12 |a b 1

c d 1e f 1|

Example: Evaluate each determinant.

a. | 6 3−8 5|| 6 3−8 5| = 6 (5) – 3 (–8)

= 54

b. |4 5 −21 3 02 −3 6 |

|4 5 −21 3 02 −3 6 |4 5

1 32 −3

|4 5 −21 3 02 −3 6 |4 5

1 32 −3

= [4(3)6 + 5(0)2 + (–2)1(–3)] – [(–2)3(2) + 4(0)(–3) + 5(1)6]

= [72 + 0 + 6] – [–12 + 0 + 30]

= 78 – 16 or 60

ExercisesEvaluate each determinant.

1. |6 −25 7 | 2. |3 2

9 6|

3. | 3 −2 −20 4 1

−1 4 −3| 4. |1 2 45 −4 13 0 6|

5. Find the area of a triangle with vertices (2, –3), (7, 4), and (–5, 5).

Cramer’s Rule Determinants provide a way for solving systems of equations.

Cramer’s Rule forTwo-Variable Systems

Let C be the coefficient matrix of the system ax+by=mfx+gy=n → |a b

f g|The solution of this system is x = |m b

n g||c|

, y = |a mf n||c|

, if C ≠ 0.

Example: Use Cramer’s Rule to solve the system of equations. 5x – 10y = 810x + 25y = –2

x = |m bn g||c|

Cramer’s Rule y = |a mf n||c|

= | 8 −10−2 25 |5 −10

10 25

a = 5, b = –10, f = 10, g = 25, m = 8, n = –2 = | 5 810 −2|

| 5 −1010 25 |

= 8(25)−(−2)(−10)5(25)−(−10)(10)

Evaluate each determinant. = 5(−2)−8(10)

5(25)−(−10)(10)

= 180225 or

45 Simplify. = –

90225 or –

25

The solution is ( 45

, 25 ).

ExercisesUse Cramer’s Rule to solve each system of equations.

1. 3x – 2y = 7 2. x – 4y = 17 3. 2x – y = –22x + 7y = 38 3x – y = 29 4x – y = 4

4. 2x – y = 1 5. 4x + 2y = 1 6. 6x – 3y = –35x + 2y = –29 5x – 4y = 24 2x + y = 21

3-8 Study GuideSolving Systems of Equations Using Inverse MatricesIdentity and Inverse Matrices The identity matrix for matrix multiplication is a square matrix with 1s for every element of the main diagonal and zeros elsewhere.

Identity Matrix forMultiplication If A is an n × n matrix and I is the identity matrix, then A ⋅ I = A and I ⋅ A = A.

If an n × n matrix A has an inverse A−1, then A ⋅ A−1 = A−1 ⋅ A = I.

Example: Determine whether X = [ 7 410 6] and Y = [ 3 −2

−5 72 ] are inverse matrices.

Find X ⋅ Y.

X ⋅ Y = [ 7 410 6] ⋅ [ 3 −2

−5 72 ]

= [21−20 −14+1430−30 −20+21] or [1 0

0 1]Find Y ⋅ X.

Y ⋅ X = [ 3 −2

−5 72 ] ⋅ [ 7 4

10 6]

= [ 21−20 12−12−35+35 −20+21] or [1 0

0 1]Since X ⋅ Y = Y ⋅ X = I, X and Y are inverse matrices.

ExercisesDetermine whether the matrices in each pair are inverses of each other.

1. [4 53 4] and [ 4 −5

−3 4 ] 2. [3 25 4] and [ 2 −1

−52

32 ] 3. [2 3

5 −1] and [ 2 3−1 −2]

4. [8 113 14 ] and [−4 11

3 −8 ] 5. [4 −15 3 ] and [1 2

3 8 ] 6. [ 5 211 4] and [−2 1

112

−52 ]

Matrix Equations A matrix equation for a system of equations consists of the product of the coefficient and variable matrices on the left and the constant matrix on the right of the equals sign.

Example: Use a matrix equation to solve a system of equations.3x – 7y = 12x + 5y = –8Determine the coefficient, variable, and constant matrices.

[3 −71 5 ] ⋅ [ x

y ] = [ 12−8]

Find the inverse of the coefficient matrix.

13(5)−1(−7)

[ 5 7−1 3] = [ 5

22722

−122

322 ]

Rewrite the equation in the form of X = A−1B

[ xy ]=[ 5

227

22−122

322 ] [ 12

−8]Solve.

[ xy ] = [ 2

11−18

11 ]ExercisesUse a matrix equation to solve each system of equations.

1. 7x + 5y = 3 2. 4x – 2y = -63x – 2y = 22 -7x + 2y = 12

3. -x + y = 4 2x - 2y = -26