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Section 9.1: Exercises 6, 12, 20, 28, 32, 50
6.) Explain why each graph is that of a function.
Each graph is that of a function because at each point in a graph we get a value for ycorresponding to a value for x. This is just the pictorial reflection of what we get in case of a
function.
Solve each system by substitution
12.) 1154 yx
52 yx
Solution
2x + y = 5 y = 5- 2x
Substituting in first equation
4x5y = -11
4x5(5-2x) = -11
4x25 + 10x = -11
14 x = -11 + 25
14x = 14
x = 1
Therefore, we get :
So, y = 5-2x = 5 -2*1 = 3
x = 1, y =3
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Solve each system by elimination.
20.) 234 yx
172 yx
Solution
4x +y = -23 eq 1
x - 2y = -17.eq 2
Performing 2*eq 1 + eq 2 we get :
8x + 2y = -46
x2y = -17
This Gives us :
8x + x = -46 -17
9x = -63
x = -7
So,
x2y = -17
-7 -2y = -17
2y = 17-7 =10
y = 5
So, X = -7, Y =5
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28.) 222
3
yx
022
yx
Solution
3x/2 + y/2 = -2
3x + y = -4 .eq 1
x/2 + y/2 = 0
x + y = 0 ..eq2
Performing eq 1eq 2
3x + y = -4
(-)x + (-)y = (-)0
3xx = -4
2x = -4 X = -2
So,
X + y = 0
-2 + y = 0 Y = 2
So, x = - 2, y = 2
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Solve each system. State whether it is inconsistent or has infinitely many solutions. If the system has
infinitely many solutions, write the solution set and y arbitrary.
32) 523 yx
846 yx
Solution
3x + 2y = 5 .eq 1
6x + 4y = 8 .eq2
3x + 2y = 4 ..eq 2
This system is inconsistent as the coefficients of one equation does not match the other equations
coefficients
Solve each system.
50.) 234 zyx
1553 xyx
1442 zyx
Solution
X -1
Y = 4
Z 2
Section 9.2: Exercises 4, 8, 10, 20, 32
Use the given row transformation to change each matrix as indicated.
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4.)
3
4
2
7
1
5
1
2
6
; -6 times row 3 added to row 1
Solution
3
4
6*32
7
1
7*65
1
2
6*16
3
4
16
7
1
37
1
2
0
Write each augmented matrix for each system and give its size. Do not solve the system.
8.) 04324 zyx
0745
0753
zyx
zyx
Solution
5
3
4
1
5
2
4
1
3
7
7
4
Write the system of equations associated with each augmented matrix. Do not solve.
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10.)
5
4
2
0
3
1
4
0
3
11
10
12
Solution
01232 zyx
01145
01034
zx
yx
Use the Gauss-Jordan method to solve each of equations. For systems in two variables with infinitely
many solutions, give the solution with y arbitrary; for systems in three variables with infinitely many
solutions, give the solution with z arbitrary.
20.)0153
01052
yx
yx
Solution
Writing in augmented form
3
2
1
5
15
10
R1= R1
3
1
1
2/5
15
5
R23R1 = R2
0
1
2/17
2/5
0
5
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2/17* R2= R2
0
1
1
2/5
0
5
R1 + 5/2 R2 = R1
0
1
1
0
0
5
So, X = 5, Y = 0
32.) 1yx
22
02
zy
zx
Solution
Writing in augmented form
0
2
1
1
0
1
2
1
0
2
0
1
R22R1 = R2
0
0
1
1
2
1
2
1
0
2
2
1
-1/2 *R2 = R2
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0
0
1
1
1
1
2
2/1
0
2
1
1
R1R2 = R1
R3R2 = R3
0
0
1
0
1
0
2/3
2/1
2/1
3
1
0
2/3*R3 = R3
0
0
1
0
1
0
1
2/1
2/1
2
1
0
R1 + R3 = R1
R2 R3 = R2
0
0
1
0
1
0
1
0
0
2
2
1
So,
X = -1
Y = 2
Z = -2
Section 9.3: Exercises 6, 14, 18, 40, 52, 62, 74, 82
Find the value of each detriment.
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6.)1
0
5
2
Determinant = 5*02*1 = -2
Find the cofactor of each element in the 2nd
row for each detriment.
14.) 2 -1 4
3 0 1
-2 1 4
Co factors for each of the elements in the 2 ndrow :-
For 3 = - (-1*41*4) = +8
For 0 = 2*4(-2)*(4) = +16
For 1 = - (2*1(-2)(-1) ) = 0
Find the value of each detriment.
18.) 2 1 -1
4 7 -2
2 4 0
Determinant = 2(7 * 0 - 4 *(-2)) - 4(1*0 (4) *(-1))+2(1 *(-2) - 7*(-1)) = 10
Solve each equation for x.
40.) 4 3 0
2 0 1 = 5
-3 x -1
Determinant = 4*(0*-1x*1)3(2*-1(-3)*1) +0 = -4x -3
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-4x3 = 5
-4x = 8
X = -2
Use the determinant theorems to find the value of each detriment.
52.) 4 8 0
1 -2 1
2 4 3
Determinant = 4(-2*3-4*1)-1(8*3-4*0)+2(8*12*0) = -48
Use Cramers rule to solve each equation. If D = 0, use another method to determine the solution set.
62.) 423 yx
52 yx
SOLN :
D = -7
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D1 = 14
D2 = -7
X = D1/D 14/-7 = -2
Y = D2/D = -7/-7 = 1
74.) x + y + z4 = 0
2xy + 3z4 = 0
4x + 2yz + 15 = 0
SOLUTION:
D=17
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D1 = -68
D2 = 51
D3=85
X= D1/D = -68/17 = -4
Y=D2/D = 51/17 = 3
Z = D3/D = 85/17 = 5
82.) 3x + 5y = -7
2x + 7z = 2
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4y + 3z = -8
Here the determinant is 0, so we use Gauss Jordan Rule :
Section 9.5: Exercises 12, 22, 48
Give all solutions of each nonlinear system of equations, including those with nonreal complex
components.
12.) 962 xxy
22 yx
Using the value of y in eqn 2 we get :
X+ 2x2+12x+18 = -2
Or, 2x2+13x+20 = 0
Or, 2x2+8x+5x+20 =0
Or, 2x(x+4) + 5(x+4) = 0
Or, (2x+5)(x+4) = 0
Or, x = -5/2, -4
Therefore y = , 1
22.) 92 22 yx
2743 22 yx
Multiply eqn 1 by 2 to get :1842
22 yx
Adding eqn (ii) with eqn(i) we get: 5x2= 45
Solving this we get, X = (+
- 3)
And putting the value of x in eqn (i) we get : Y = 0
Therefore, X=+-3 and Y=0
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Solve each problem using a system of equations in two variables.
48.) Unknown numbers Find two numbers whose sum is 10 and who squares differ by 20.
Let the numbers be x and y
X + y = 10 .eq 1
X2y2= 20. Eq2
(x+y)(x-y) = 20Since x+y =10, we replace it in the eqn : (x+y)(x-y) =20 , to get
10*(x-y) = 20 X-y = 2 . Eq 3
Now,
We add eqn1 and Eqn 3 to get :
Eq1 + eq 3
X+ y + Xy = 10 - 2
2x = 12
X = 6
X + y = 10
Y = 10x = 10-6 = 4
Section 9.6: Exercises 4, 32, 38
Graph each inequality.
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4.) 4y3x < 5
Graph the solution set of each system of inequalities.
32.) yx 34 < 12
y + 4x > -4
SOLUTION :
X< 3-3y/4 and x> -1/4y -1
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38.) x + y 9
x 2
y
SOLUTION :
X 9-y
X -y2
Section 9.7: Exercises 6, 14, 22, 36, 52, 58, 66, 88
Find the values of the variables for which each statement is true, if possible.
6.)
6
2
5
0
3
4
x
5
8
9
=
6
4
3
z
y
0
3
2
w
8
9
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X-4 = 2 . Therefore x = 6
Y+3 = 5 . Therefore Y = 2
Z+4 = 2 . Therefore Z = -2
W = 5 . Given
Find the size of each matrix. Identify any square, column, or row matrices.
14.)
4
9
1
6
8
2
Size = 2*3. It is neither a row or a column or a square matrix.
Perform each operation when possible.
22.)
8
9
2
4
4
3
7
2 = 6 6
-12 9
36.)
nm
ak
xz
yk
nm
ak
xz
yk
24
64
52
65
24
52
36
84
=
Find each matrix product when possible.
-k -14y
4z -8x
-2k -a
-8m 4n
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52.)
7
1
2
6
0
5 =
58.)
2
7
28
0
2
32
6
7 =
Given A =
3
4
1
2, B =
3
0
5
7
2
1
, and C =
0
5
3
4
6
1, find each product when possible.
66.) AC
Yes it is possible and the solution is :
4*-5 -2*0 4*4-2*3 1*1-2*6 = -20 10 -8
3*-5+1*0 3*4+1*3 3*1+1*6 -15 15 9
A =
21
11
a
a
22
12
a
a, B =
21
11
b
b
22
12
b
b, and C =
21
11
c
c
22
12
c
c, where all the elements are real numbers.
Use these materials to show each statement is true for 2 x 2matrices.
88.) A + (B + C) = (A + B) + C (associative property)
B+C = b11 + c11 b12+c12
b21+c21 b22+c22
-1*6 + 5*2 4
7*6 0*2 = 42
7*23+ 0*2 7*-7 + 0*6 221 -7
2*23 + 28*2 2*-7 +28*6 = 4(7+3) 107
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A+(B+C)= a11+b11+c11 a12+b12+c12
a21+b21+c21 a22+b22+c22
A+B = a11+b11 a12+b12
A21+b21 a22+b22
a11+b11+c11 a12+b12+c12
(A+B)+C = a21+b21+c21 a22+b22+c22
Therefore , (A+B)+C = A+(B+C)