Solutions Manual for Finite Mathematics and Calculus with Applications 10th Edition by Lial IBSN 9780133981070

2 2
Solutions Manual for Finite Mathematics and Calculus with Applications 10th
Edition by Lial IBSN 9780133981070
Download: http://downloadlink.org/p/solutions-manual-for-finite-mathematics-and-
calculus-with-applications-10th-edition-by-lial-ibsn-9780133981070/
Test Bank for Finite Mathematics and Calculus with Applications 10th Edition by
Lial IBSN 9780133981070
Download: http://downloadlink.org/p/test-bank-for-finite-mathematics-and-
Chapter 2
SYSTEMS OF LINEAR EQUATIONS AND MATRICES
2.1 Solution of Linear Systems by the
Echelon Method
Your Turn 1
2x + 3y = 12 (1)
Your Turn 3
Since there are 60 utensils with a total weight of 189.6
pounds, the equations of Example 5 become
x + y + z = 60 (1)
3x - 4y = 1 (2) 3.9x + 3.6y + 3.0z = 189.6 (2)
Use row transformations to eliminate x in equation (2).
2x + 3y = 12
where x = the number of knives, y = the number of
forks, and z = the number of spoons.
3R1 + (-2)R2 R2 17y = 34
x + y + z = 60
Now make the coefficient of the first term in each 3.9 R + (−1) R → R 0.3y + 0.9z = 44.4
equation equal to 1.
1 R R x +
3
y = 6
1 2 2
x + y + z = 60
2 1 1
2
1
R → R y + 3z =148
1
R R17
y = 2
0.3
2 2
Back-substitute to solve for x.
x +
3
(2) = 6
Solve for y in terms of the parameter z: y = 148 − 3z
2
x + 3 = 6
x = 3
The solution of the system is (3, 2).
Substitute this expression into equation (1).
x + (148 − 3z) + z = 60
Then solve for x.
Your Turn 2
Let x = number of shares of Kohl’s stock
x = 2z − 88
Since both x and y must be nonnegative, we have:
148
y = number of shares of Best Buy stock 3z ≤148 or z ≤ , so
3
z ≤ 49
x + y = 188 (1) 2z ≥ 88 or z ≥ 44
52x + 27y = 8601 (2)
The largest number of spoons is thus 49 ; the

÷
x + y = 188 corresponding number of knives is (2)(49) −88 = 10 ,
R2 + (-52) R1 R2
æ 1 ö
ç- ÷ R
- 25y = -1175
y = 7
x + y = 188
y = 47
and the number of forks is 148 − (3)(49) = 1.
2.1 Exercises
In Exercises 1–16 and 19–28, check each solution by
25 ø 2
substituting it in the original equations of the system.
1. x + y = 5 (1)
Substitue y = 47 in equation (1).
2x - 2y = 2 (2)
To eliminate x in equation (2), multiply equation
x + 47 = 188 (1) by -2 and add the result to equation (2). The
x = 141
new system is
x + y = 5 (1)
Olinda owns 141 shares of Kohl’s and 47 shares of Best
Buy.
-2R1 + R2 R2 -4y = -8. (3)
.77

2
2 2
7
78 Chapter 2 SYSTEMS OF LINEAR EQUATIONS AND MATRICES
Now make the coefficient of the first term in
each row equal 1. To accomplish this, multiply
equation (3) by -
1
.
Back-substitution of 3 for y in equation (4) gives
2
4
x + y = 5 (1)
x - (3) = -1
3
- 1 R R y = 2 (4)
x - 2 = -1
4 2 2 x = 1.
Substitute 2 for y in equation (1).
x + 2 = 5
y = 3
The solution is (3, 2).
4. 2x + 7 y = -8 (1)
-2x + 3 y = -12 (2)
2x + 7y = - 8 (1)
2. 4x + y = 9 (1) R1 + R2 R2 10y = -20 (3)
3x - y = 5 (2)
First use transformation 3 to eliminate the x-term
Make each leading coefficient equal 1.
7
from equation (2). Multiply equation (1) by 3 and
add the result to -4 times equation (2).
1 R1 R1 x + y = -4 (4)
2
4x + y = 9 (1)
1 R10 2 R 2 y = -2 (5)
3R1 + (-4)R2 R2 7y = 7 (3) Substitute -2 for y in equation (4).
Now use transformation 2 to make the coefficient
of the first term in each equation equal to 1.
x +
7
(-2) = -4
2
1 R R x +
1
y =
9
(4)
x - 7 = -4
4 1 1
4 4 x = 3
1 R2 R 2 y = 1 (5) The solution is (3, -2 ).
Complete the solution by back-substitution.
Substitute 1 for y in equation (4) to get
x +
1
(1) =
9
4 4
5. 3x + 2 y = - 6 (1)
5x - 2 y = -10 (2)
Eliminate x in equation (2) to get the system
3x + 2y = -6 (1)
x =
8
= 2. 5R + (-3)R R 16y = 0. (3)
4
3. 3x - 2 y = -3 (1)
1 2 2
Make the coefficient of the first term in each
equation equal 1.
2
1 R R x + y = -2 (4)
5x - y = 2 (2) 3 1 1
3
To eliminate x in equation (2), multiply equation 1 R R16
y = 0 (5)
(1) by -5 and equation (2) by 3. Add the results.
Substitute 0 for y in equation (4) to get x = -2.
The new system is
3x - 2y = -3 (1)
The solution is (-2, 0).
-5R1 + 3R2 R2 7y = 21. (3) 6. -3x + y = 4 (1)

7
Now make the coefficient of the first term in
each row equal 1. To accomplish this, multiply
2x - 2 y = -4 (2)
Eliminate x in equation (2).
equation (1) by
1
3
and equation (3) by
1
.7 -3x + y = 4 (1)
1 R R x -
2
y = -1 (4) 2R1 + 3R2 R2 - 4y = -4 (3)
3 1 1
3 Make the coefficient of the first term in each row
1 R2 R 2 y = 3 (5) equal 1.
-1 R R x -
1
y = -
4
(4)
3 1 1
3 3
- 1 R4 2 R 2 y = 1 (5)
.

2 2
Section 2.1 79
Back-substitution of 1 for y in equation (4) gives Substitute -2 for q in equation (4) to get p = 3.
- ).
x -
1
(1) = -
4 The solution is (3, 2
3 3 10. 12s - 5t = 9
x -
1
= -
4
3 3
x = -1.
3s - 8t = -18
12s - 5t = 9
R1 + (-4)R2 R
2
27t = 81
1 R R s -
5
t =
3
7. 6x - 2 y = -4
(1)
12 1 1
12 4
3x + 4 y
=
8 (2) 1 R R27
t = 3
6x - 2 y = -4
(1)
Back substitution gives
5 3
-1R1 + 2 R 2 R
2
10y = 20 (3)
s - (3) =
12 4
Make the coefficient of the first term in each row s -
5
=
3
equal 1.
1 R R x -
1
y = -
2
(4)
4 4
s =
8
= 2.
6 1 1
3 3 4
1 R10 2 R 2 y = 2 (5) The solution is (2, 3).
Substitute 2 for y in equation (4) to get x = 0. The
solution is (0, 2).
8. 4m + 3n = -1
11. 6x + 7y = -2 (1)
7x - 6y = 26 (2)
2m + 5n = 3
4m + 3n = -1 7 R1 + (-6) R2 R2
6x + 7y = -2 (1)
85y = -170 (3)
R1 + (-2) R2 R2 - 7n = -7 Make the coefficient of the first term in each
Make each leading coefficient equal 1. equation equal 1.
1 R R x +
7
y = -
1
(4)
1 R R m +
3
n = -
1 6 1 1
6 3
4 1 1
4 4 1 R R y = -2 (5)
- 1 R7 2 R 2 n = 1 85 2 2
Substitute -2 for y in equation (4) to get x = 2.
Back-substitution gives
m +
3
(1) = -
1
The solution is (2, -2 ).
12. 3a - 8b = 14 (1)
4 4
a - 2b = 2 (2)

= -m
4
= -1.
4 Eliminate a in equation (2).
3a - 8b = 14 (1)
R1 + (-3) R2 R2 -2b = 8 (3)
9. 5p + 11q = -7 (1) Make the coefficient of the first term in each row
3p - 8q = 25 (2) equal 1.
R R a -
8
b =
14
(4)
Eliminate p in equation (2).
1
3 1 1
3 3
-3R1 + 5R2 R2
5p + 11q = -7 (1)
- 73q = 146 (3)
- 1 R2 2 R 2 b = -4 (5)
Make the coefficient of the first term in each row Back substitute -4 for b in equation (4) to get
equal 1. a = -6. The solution is (-6, -4).
1 R R p +
11
q = -
7
(4) 13. 3x + 2y = 5 (1)
5 1 1
5 5 6x + 4y = 8 (2)
- 1
R73 2 R 2 q = - 1 (5)
.

R
8
3
, y
R R
2
3x + 2y = 5 (1)
æ -5y - 2
çè 3
, y÷ö
,÷ø
-2R1 + R2 R2 0 = -2 (3) where y is any real number.
Equation (3) is a false statement.
17. x -
3y
=
5
(1)
The system is inconsistent and has no solution. 2 2
14. 9x - 5y = 1 4x
+
2y
= 6 (2)
-18x + 10y = 1
9x - 5y = 1
3 3
Rewrite the equations without fractions.
2R R 2x - 3 y = 5 (3)
2R1 + R2 R2 0 = 3 1 1
3R2 R2 4x + 2 y = 18 (4)
The equation 0 = 3 is a false statement, which
indicates that the system is inconsistent and has no
solution.
2x - 3y = 5 (3)
15. 3x - 2y = -4 (1) -2R1 + R2 R2 8y = 8 (5)
-6x + 4y = 8 (2) Make the coefficient of the first term in each
equation equal 1.
3x - 2y = -4 (1)
1
1 R1 x -
3
y =
5
2 2
(6)
2R1 + R2 R2 0 = 0 (3) 1 R2 R 2 y = 1 (7)
The true statement in equation (3) indicates that
there are an infinite number of solutions for the
system. Solve equation (1) for x.
3x - 2y = -4 (1)
3x = 2y - 4
18.
Substitute 1 for y in equation (6) to get x = 4.
x
+ 3y = 31
5
x =
2y - 4
(4)
3
2x -
y
= 8
5
For each value of y, equation (4) indicates that Multiply each equation by 5 to eliminate fractions.
2y-4 x + 15y = 155
x =
3
, and all ordered pairs of the form
10x - y = 40
(2y-4
) are solutions. x + 15y = 155
10R1 + (-1)R2 R2 151y = 1510
16. 3x + 5y + 2 = 0
x + 15y = 155
9x + 15y + 6 = 0
Begin by rewriting the equations in standard form,
1
151 2 2
Back-substitution gives
y = 10
3x + 5y = -2
x + 15(10) = 155
9x + 15y = -6
x = 5.

3x + 5y = -2
3R1 + (-1)R2 R2 0 = 0
The true statement, 0 = 0, shows that the two 19.
x
+ y =
3
(1)
equations have the same graph, which means that
there are an infinite number of solutions for the
system. All ordered pairs that satisfy the equation
2 2
x
+ y =
1
(2)
3 3
3x + 5y = -2 are solutions. Solve this equation Rewrite the equations without fractions.
for x. 2R R x + 2 y = 3 (3)
3x = -5y - 2
1 1
3R2 R 2 x + 3y = 1 (4)
x =
-5y - 2
3
The general solution is the set of all ordered pairs
x + 2y = 3 (3)
of the form -1R1 + R2 R2 y = -2 (5)
.

3
ç ÷
z -
Section 2.1 81
Substitute -2 for y in equation (3) to get x = 7.
x +
3
(3 - 5z) -
1
z =
1
The solution is (7, -2 ). 2 2 2
20.
x
+
y
=
1
(1) x +
9
-
15
z -
1
z =
1
9 6 3 2 2 2 2
2x +
8 y
=
2
(2)
x - 8z = -4
5 5
Rewrite the equations without fractions.
x = 8z - 4
The solution is (8z - 4, 3 - 5z, z).
18R1 R1 2x + 3y = 6 (3) 26. 3x + y - z = 0 (1)
5R2 R 2 10x + 8 y = 2 (4) 2x - y + 3z = -7 (2)
2x + 3y = 6 (3)
3x + y - z = 0 (1)
-5R1 + R2 R2 -7y = -28 (5)
2R + (-3) R R 5y - 11z = 21 (3)
Make the coefficient of the first term in each row
equal 1.
1 2 2
equation equal 1.
1 R R x +
3
y = 3 (6) 1 1
2 1 1
2 1 R1 R1 x + y - z =
3 3
0 (4)
- 1
R7 2 R 2 y = 4 (7)
1 R R y -
11
z =
21
(5)
Substitute 4 for y in equation (6) to get x = -3. 5 2 2
5 5
21. An inconsistent system has no solutions.
22. The solution of a system with two dependent
equations in two variables is an infinite set of
ordered pairs.
Solve equation (5) for y in terms of z.
y =
11
z +
21
5 5
Substitute this expression for y in equation (4), and
solve the equation for x.
25. 2x + 3y - z = 1 (1) x +
1 æ11
z +
21 ö÷
-
1
z = 0
3x + 5y + z = 3 (2)
3 5 5 ø 3
2x + 3y - z = 1 (1)
x +
11
z +
7
-
1
z = 0
5 5 3
2 7
-3R1 + 2R2 R2 y + 5z = 3 (3)
x + z = -
5 5
Since there are only two equations, it is not possible
to continue with the echelon method as in
the previous exercises involving systems with
three equations and three variables. To complete the
The solution is
2 7
x = -
5 5
solution, make the coefficient of the first term in æ 2
z
7
,
11
z
21
, z or

the each equation equal 1. çè
-
5
-
5 5
+
5 ÷ø
1 R R x +
3
y +
1
z =
1
(4)
æ -2z - 7 11z + 21 ö÷
ç , , z÷.
2 1 1
2 2 2 5 5
y + 5z = 3 (3)
Solve equation (3) for y in terms of the parameter z.
27. x + 2 y + 3z = 11 (1)
y + 5z = 3
y = 3 - 5z
2x - y + z = 2 (2)
x + 2y + 3z = 11 (1)
Substitute this expression for y in equation (4) to
solve for x in terms of the parameter z.
-2R1 + R2 R2 -5y - 5z = -20 (3)
x + 2 y + 3z = 11 (1)
- 1 R5 2 R2 y + z = 4 (4)
Since there are only two equations, it is not possible
to continue with the echelon method. To complete
.

ç
z +
÷
( )
5
the solution, solve equation (4) for y in terms of
the parameter z.
y = 4 - z
solve for x in terms of z.
x + 2
æ
12 -
4
z
ö÷÷ + 3z = 90
Now substitute 4 - z for y in equation (1) and 3
x + 2(4 - z) + 3z = 11
x + 8 - 2z + 3z = 11
x = 3 - z
The solution is ( 3 - z, 4 - z, z).
x + 24 -
8
z + 3z = 90
3
x +
1
z = 66
3
x = 66 -
1
z
3
28. -x + y - z = -7 (1)
Thus the solutions are (66 - 1 z,12 - 4 z,z ),
2x + 3 y + z
=
7 (2) 3 3
where z is any real number. Since the solutions have
-x + y - z = -7 (1)
to be nonnegative integers, set 66 -
Solving for z gives z £ 198.
1
z ³ 0.3
2R1 + R2 R2 5y - z = -7 (3)
equation equal 1.
Since y must be nonnegative, we have
12 -
4
z ³ 0. Solving for z gives z £ 9.3
-1R1 R1 x - y + z = 7 (4)
Since z must be a multiple of 3 for x and y to be
integers, the permissible values of z are 0, 3, 6, and
1 R2 R 2 y -
1
z = -
7
5 5
(5)
30.
9, which gives 4 solutions.
x - 7y + 4z = 75 (1)
Solve equation (5) for y in terms of z.
y =
1
z -
7
5 5
Substitute this expression for y in equation (4), and
solve the equation for x.
2y + 7z = 60 (2)
Let z be the parameter and solve equation (2) for
y in terms of z.
2y + 7z = 60
2y = 60 - 7z
æ 1 7 ö÷x - ç z - ÷ + z = 7 7
5 5 ø
x -
1
z +
7
+ z = 7
5 5
y = 30 - z
2
x +
4
z =
28
5 5 æ 7 ö÷x - 7ç30 - z ÷ + 4z = 75
4 28
x = -
5 5
çè
x - 210 +
2 ÷ø
49
z + 4z = 75
The solution of the system is
æ 4 28 1 7 ö÷ç- z + , z - , z
÷
or
2
x +
57
z = 285
2
çè 5 5 5 5 ø
æ -4z + 28 z - 7 ö
, , z .
x = 285 -
57
z
2
ç ÷5 5 ø
Thus the solutions are 285 - 57 z, 30 - 7 z, z ,2 2
29. x + 2 y + 3z = 90 (1) 3y + 4z = 36 (2)

Let z
be
the
para
mete
r
and
solv
e
equa
tion
(2)
for
y in
ter
ms
of z.
3
y
+
4
z
=
3
6
where z is any real number. Since the
solutions have to be nonnegative integers, set
285 -
57
7
3y = 36 - 4z
y = 12 -
4
z
3
30 - 2
z ³ 0. Solving for z gives z £ 8.57.
.

2
ê
ê ú
Section 2.1 83
Since z must be a multiple of 2 for x and y to be
integers, the permissible values of z are 0, 2, 4, 6,
and 8, which gives 5 solutions.
31. 3x + 2y + 4z = 80 (1)
y - 3z = 10 (2)
Let z be the parameter and solve equation (2) for y
in terms of z.
y - 3z = 10
have to be nonnegative integers, set
15 -
3
3
z + 6 ³ 0. Solving for z gives z £ -4.2
Since -4 £ z £ 10 and z must be a multiple of 2
for x and y to be integers, the permissible values of
z are 0, 2, 4, 6, 8, and 10, which gives 6 solutions.
y = 3z + 10
3x + 2(3z + 10) + 4z = 80
3x + 6z + 20 + 4z = 80
33. nb + (å x)m = å y
(å x)b + (å x2
)m = å xy
Multiply equation (1) by 1 .n
å x
(1)
(2)
å y
3x + 10z = 60
3x = 60 - 10z
x = 20 -
10
z
3
b + m =
n n
(å x)b + (å x2
)m = å xy
Eliminate b from equation (2).
(3)
(2)
Thus the solutions are (20 - 10 z, 3z + 10, z ), b +
å x
m =
å y
(3)
3
where z is any real number. Since the solutions
have to be nonnegative integers, set
10
n n
(-å x)R1 + R 2 R 2
20 - 3
z ³ 0. Solving for z gives z £ 6. é (å x)2
2
ù -(å x)(å y)
ê- + å x ú m = + å xy (4)
3z + 10 ³ 0. Solving for z gives z ³ -10 1 .3
ëê
ê n úû
ú n
1
Since z must be a multiple of 3 for x to be an integer,
the permissible values of z are 0, 3, and, 6, which
gives 3 solutions.
Multiply equation (4) by .
(å x)2
- + å x
n
32. 4x + 2 y + 3z = 72
(1)
b +
å x
m =
å y
(3)
2y - 3z = 12 (2) n n
Let z be the parameter and solve equation (2) for é ù
y in terms of z. é -(å x)(å y) ù ê 1 ú
m = ê + å xy ú ê ú (5)
2y - 3z = 12
2y = 3z + 12
y =
3
z + 6
2
n ê (å x)2
2
ú
- + å x ú
ë n úû
Simplify the right side of equation (5).
é - å å + å ù é ù
Substitute this expression for y in equation (1) to m = ê
( x)( y) n xy
ú ê n ú
solve for x in terms of z. ëê n ûú êë
ê
-(å x)2
+ n(å x2
) úû
ú
æ 3 ö÷4x + 2ç z + 6 ÷ + 3z = 72 m =
n å xy - (å x)(å y)
2 ø n(å x2
) - (å x)2
4x + 3z + 12 + 3z = 72

4x + 6z = 60
4x = 60 - 6z
x = 15 -
3
z
2
Thus the solutions are (15 - 3 z, 3 z + 6, z ),
From equation (3) we have
b =
å y
-
åx
m
n n
b =
å y - m(å x)
.
n
2 2
where z is any real number. Since the solutions
35. Let x = the cost per pound of rice, and
y = the cost per pound of potatoes.
.

8
6
6
The system to be solved is
20x + 10y = 16.20 (1)
Make the coefficient of the first term in equation (l)
equal 1.
30x + 12y = 23.04. (2) 1 R1 R1 x +
5
y =
8
525 (3)
Multiply equation (1) by 1
.20
1
2x + 2 y = 1200 (2)
20
R1 R1 x + 0.5y = 0.81 (3)
30x + 12 y = 23.04
(2) Eliminate x in equation (2).
x +
5
y = 525 (3)
8
6
x + 0.5y = 0.81 (3) -2R1 + R 2 R
2
y = 150 (4)
8
-30R1 + R2 R2 -3y = -1.26 (4) Make the coefficient of the first term in equation (4)
Multiply equation (4) by - 1 .3
x + 0.5y = 0.81 (3)
equal 1.
x +
5
y =
8
525 (3)
- 1 R3 2 R2 y = 0.42 (5) 8 R2 R 2 y = 200 (5)
Substitute 0.42 for y in equation (3).
x + 0.5(0.42) = 0.81
x + 0.21 = 0.81
x = 0.60
The cost of 10 pounds of rice and 50 pounds of
potatoes is
10(0.60) + 50(0.42) = 27,
that is, $27.
36. Let x = the number of new releases Blake
downloaded
y = the number of older songs Blake
downloaded.
x +
5
(200) = 525
8
x + 125 = 525
x = 400
There are 400 main floor seats and 200 balcony
seats.
38. Let x = the number of skirts originally in the
store, and
y = the number of blouses originally in the
store.
45x + 35 y = 51, 750
(1)
x + y = 33 (1) æ 1
45 x
æ 2
+ 35 y = 30, 600. (2)
1.29x +0.99y = 35.97 (2) èç 2 ÷ø èç 3 ÷ø
Eliminate x in eqatuon (2). Simplify each equation. Multiply equation (1) by
x + y = 33 (1)
1
and equation (2) by .5 5
R2 + (−1.29) R1 → R2
Solve equation (3) for y.
y =
−6.6
= 22
−0.3
− 0.3y = −6.67 (3) 9x + 7y = 10,350 (3)
27x + 28y = 36,720 (4)
Eliminate x from equation (4).
9x + 7y = 10,350 (3)
Substitute 22 for y in equation (1) and solve for x. -3R1 + R2 R2 7y = 5670 (5)
x + 22 = 33 Make each leading coefficient equal 1.

7
x = 11 1 R R x +
7
y = 1150 (6)
Blake bought 11 new releses and 22 older songs.
9 1 1
9
37. Let x = the number of seats on the main floor, and
y = the number of seats in the balcony.
8x + 5y = 4200 (1)
0.25(8x) + 0.40(5y) = 1200. (2)
1 R2 R 2 y = 810 (7)
.

Section 2.1 85
x +
7
(810) = 1150
9
x + 630 = 1150
41. (a) Let x = the number of fives
y = the number of tens
z = the number of twenties
We want to solve the following system.
x = 520 x + y + z = 70 (1)
Half of the skirts are sold, leaving half in the
store, so
5x + 10y + 20z = 960 (2)
1
x =
1
(520) = 260.
2 2
x + y + z = 70
Two-thirds of the blouses are sold, leaving one-
-5R1 + R2 R2 5y + 15z = 610
third in the store, so
1
y =
1
(810) = 270.
3 3
There are 260 skirts and 270 blouses left in the
store.
39. Let x = the number of model 201 to make each
day, and
y = the number of model 301 to make each
day.
Let z be the parameter. Solve for y and for x
in terms of z.
5y + 15z = 610
5y = 610 - 15z
y = 122 - 3z
x + (122 - 3z) + z = 70
x - 2z = -52
x = 2z - 52
2x + 3y = 34 (1) The solutions are (2z - 52, 122 - 3z, z),
18x + 27y = 335. (2)
Make the coefficient of the first term in equation (1)
where z is any real number. All variables must
represent nonnegative integers.
equal 1.
1 R R x +
3
y = 17 (3)
122 - 3z ³ 0
-3z ³ -122
2z - 52 ³ 0
2z ³ 52
2 1 1
2 z £ 40.67 z ³ 26
18x + 27y = 335 (2)
x +
3
y = 17 (3)
2
Therefore, z must be an integer such that
26 £ z £ 40. Thus, there are a total of
15 solutions.
(b) The smallest value of z is 26.
-18R1 + R 2 R
2
0 = 29 (4) x = 2z - 52
= 2(26) - 52 = 0
y = 122 - 3z
= 122 - 3(26) = 44
Since equation (4) is false, the system is inconsistent.
Therefore, this situation is impossible.
40. Let x = the number of shares of Disney stock, and
y = the number of shares of Intel stock.
The solutions with the smallest number of
five-dollar bills is no fives, 44 tens, and
26 twenties.
(c) The largest value of z is 40.
30x + 70y = 16,000 (1)
45x + 105y = 25,500 (2)
x = 2z - 52
= 2(40) - 52 = 28
y = 122 - 3z
= 122 - 3(40) = 2
Simplify each equation. Multiply equation (1) by The solutions with the largest number of five-
1 1
10
and equation (2) by
15
.
3x + 7y = 1600 (3)
3x + 7y = 1700 (4)

EZ Compact Commercial Totals
Weight 10 20 60 440
Space 10 8 28 248
dollar bills is 28 fives, 2 tens,
and 40 twenties.
42. Let x = the number of EZ models,
y = the number of compact models, and
Since 3x + 7y cannot equal both 1600 and 1700
z = the number of commercial models.
for one point (x, y), we have an inconsistent
system. Therefore, this situation is not possible.
Make a table.
.

Buffet Chair Table Totals
Construction 30 10 10 350
Finishing 10 10 30 150
R
R
ê ú
1 (
2 (
10
12
)
)
10x + 20y + 60z = 440 (1)
10x + 8y + 28z = 248 (2)
10x + 20y + 60z = 440 (1)
Ignore solutions containing values that are not
integers. There are three possible solutions:
1. 12 EZ models, 16 compact models, and
0 commercial models;
R1 + (-1) R2 R2 12y + 32z = 192 (3) 3 commercial models; or
Make the leading coefficients equal 1.
6 commercial models.
1
1 R1
1
2 R 2
x + 2 y + 6z = 44 (4)
y +
8
z = 16 (5)
3
43. Let x = the number of buffets produced each
week,
y = the number of chairs produced each
week.
Solve equation (5) for y.
y = 16 -
8
z
3
z = the number of tables produced each
week.
Make a table.
y =
48 - 8z
3
y =
8(6 - z)
3
Substitute this expression for y into equation (4)
and solve for x.
x + 2
é 8(6 - z) ù
+ 6z = 44
30x + 10y + 10z = 350 (1)
10x + 10y + 30z = 150. (2)
Make the coefficient of the first term in equation
3 (1) equal 1.
x = 44 - 6z -
16(6 - z) 1 R R x +
1
y +
1
z =
35
(3)
3 30 1 1
3 3 3
x =
132 - 18z - 96 + 16z
3
x =
36 - 2z
10x + 10 y + 30z = 150 (2)
1 1 35
3 x + y + z = (3)
x =
2(18 - z)
3 3 3
20 80 100
3 -10R1 + R 2 R
2
y + z =
3 3 3
(4)
The solution of the system is
æ 2(18 - z) 8(6 - z) ö÷ç
3
,
3
, z
÷ø
.
Solve equation (4) for y. Multiply by 3.
20y + 80z = 100
y + 4z = 5
The solutions must be nonnegative integers.
Therefore, 0 £ z £ 6. (Any larger values of z
would cause y to be negative, which would make
no sense in the problem.)
Substitute 5 -
4z
for x.
y = 5 - 4z
for y in equation (1) and solve
Values of z Solutions
0 (12, 16, 0)
34 , 40 ,13 3
30x + 10(5 - 4z) + 10z = 350
30x + 50 - 40z + 10z = 350
30x = 300 + 30z
x = 10 + z
32 , 32 ,23 3 The solution is (10 + z, 5 - 4z, z). All variables

4 (
5 (
)
)
3 (10, 8, 3)
28 , 16 ,43 3
26 , 8 ,53 3
6 (8, 0, 6)
must be nonnegative integers. Therefore,
5 - 4z ³ 0
5 ³ 4z
z £
5
,
4
.

for y in equation (1).
÷
will be 10,550,000
Section 2.1 87
so z = 0 or z = 1. (Any larger value of z would (b) y = 759 - 0.338x (1)
cause y to be negative, which would make no y = 1637 - 0.779x (2)
sense in the problem.) If z = 0, then the solution
is (10, 5, 0). If z = 1, then the solution is (11, 1, 1). Rewrite equations so that variables are on the
left side and constant term is on the right side.
Therefore, the company should make either
10 buffets, 5 chairs, and no tables or 11 buffets,
1 chair, and 1 table each week.
44. (a) The system to be solved is
43,500x - y = 1,295,000 (1)
0.338x + y = 759 (3)
0.779x + y = 1637 (4)
Eliminate y from equation (4).
0.338x + y = 759 (3)
27,000x - y = 440,000 (2)
-1R1 + R2 R2 0.441x = 878 (5)
43,500x - y = 1,295,000 (1)
Make leading coefficient for equation (5)
equal 1.
The two estimates agree in the year closest to
-
27,000
R + R R -
11
y = -
10,550,000
(3) x = 878 » 1990.93, so they agree in 1991.
43,500 1 2 2
29 29 0.441
The estimated number of days into the year
Make the coefficient of the first term in equation
(3) equal 1.
when a robin can be expected is
æ 878
0.338
ö÷ + y = 759
43,500x - y = 1,295,000 (1) çè 0.441÷ø
- 29
R11 2 R 2 y =
10, 550,
000
11
(4) y » 86.
Substitute 10,550,000
11
46. (a) We are given the equation
2
y = ax + bx + c.
43,500x -
10, 550, 000
=
1, 295, 000
11
Since a car traveling at 0 mph has a stopping
distance of 0 feet, then y = 0 when x = 0.
43,500x =
24,795,000
11 Substituting these values into y = ax2
+ bx + c
yields
x =
570
11
æ 570 10,550,000 ö
The solution is , .
0 = a(0)2
c = 0.
+ b(0) + c, so
ç ÷ Therefore, we have11 11 ø
y = ax2 + bx.The profit/loss will be equal after 570
11 weeks
or about 51.8 weeks. At that point, the profit
11
or about $959,091.
(b) If the show lasts longer than 51.8 weeks,
Broadway is a more profitable venue. If it lasts
After substituting the given values for the
stopping distances (y) and speeds (x) in mph,
the system to be solved is
61.7 = a(25)2
+ b(25) (1)
2
less than 51.8 weeks, off Broadway is a more 106 = a(35) + b(35). (2)
profitable venue.

45. (a) For the first equation, the first sighting in 2000 These equations can be written as
625a + 25b = 61.7 (1)
was on day y = 759 - 0.338(2000) = 83, 1225a + 35b = 106. (2)
or during the eighty-third day of the year.
Since 2000 was a leap year, the eighty-third
day fell on March 23.
For the second equation, the first sighting in
Multiply equation (1) by 1 ; also eliminate625
the decimal in 61.7 by multiplying the numerator
and denominator of the fraction by 10.
2000 was on day y = 1637 - 0.779(2000) = 1 R R a +
1
b =
617
(3)
79, or during the seventy-ninth day of the year. 625 1 1
25 6250
Since 2000 was a leap year, the seventy-ninth
day fell on March 19.
1225a + 35b =
Eliminate a in equation (2).
106 (2)
.

a +
1
b =
617
(3)
x + y = 13
-1225R + R R
25 6250
-14b = -
3733
(4)
(x +1) − y = 2
or
1 2 2
Multiply equation (4) by - 1 .14
250 x − y = 1 (1)
x + y = 13 (2)
x - y = 1
a +
1
b =
617
(3)
25 6250
(-1)R1 + R 2 R
2
2y = 12
- 1 R14 2 R 2 b =
3733
3500
(5)
1
R R
x - y = 1
y = 6
Substitute 3733
3500
for b in equation (3). 2
2 2
1 æ 3733 ö÷a + ç ÷ =
617 Substitute y = 6 into (1).
25 3500 ø 6250 x − 6 = 1
Therefore,
a =
4905
87,500
a =
4905
87,500
» 0.056057,
x = 7
The flight time is 7 hours, and the difference in
time zones is 6 hours.
49. (a) Since 8 and 9 must be two of the four numbers
and b =
3733
» 1.06657.
3500
combined using addition, subtraction,
multiplication, and/or division to get 24, begin
by finding two numbers to use with 8 and 9.
(b) Substitute the values from part (a) for a and b One possibility is 8 and 3 since
and 55 for x in the equation
Solve for y.
y = ax2
+ bx. (9 - 8) ⋅ 8 ⋅ 3 = 24.
If we can find values of x and y such that
y = 0.056057(55)2
+ 1.06657(55)
y » 228
The stopping distance of a car traveling
either x + y = 8 and 3x + 2y = 3, or
x + y = 3 and 3x + 2y = 8, we will have
found a solution. Solving the first system gives
55 mph is approximately 228 ft. x = -13 and y = 21. This, however, does
47. Let x = number of field goals, and
not satisfy the condition that x and y be single-
digit positive integers. Solving the second
y = number of foul shots. system gives x = 2 and y = 1. Since both
Then x + y = 64 (1)
2x + y = 100 (2).
of these values are single-digit positive
integers, we have one possible system. Thus,
one system is
Eliminate x in equation (2). ìïï x + y = 3
x + y = 64 (1)
ï
í
î3x + 2 y = 8
-2R1 + R2 R2 - y = -28 (3)
Its solution is (2, 1). These values of x and y give
Make the coefficients of the first term of each
equation equal 1.
x + y = 64 (1)
the numbers 8, 9, 8, and 3 on the game card. These
numbers can be combined as (9 - 8) ⋅ 8 ⋅ 3 to
make 24.

-1R2 R2 y = 28 (4)
Substitute 28 for y in equation (1) to get x = 36.
Wilt Chamberlain made 36 field goals and 28 foul
shots.
48. Let x = flight time eastward
y = difference in time zones
The flight time westward is one hour longer than
the time eastward, so flight time westward is
x +1. The system to be solved is
.

10 5
5 -4 6 -4
ë û
ê ú
ê ú
ê ú
1
ê ú
2
ê ú
-
Section 2.2 89
2.2 Solution of Linear Systems by the 6R3 + R1 R1 é 1 0 0 7ù
ê ú
Gauss-Jordan Method - 9 R3 + R2 R2
ê0 1 0 2ú
2 ê ú
Your Turn 1
ê0 0 1 -3ú
4x + 5y = 10
7x + 8y = 19
Write the augmented matrix and transform the matrix.
é 4 5 10 ù
ê ú
ê 7 8 19 ú
ë û
The solution is x = 7, y = 2, and z = -3, or
(7, 2, -3).
Your Turn 3
2x - 2 y + 3z - 4w =
6
3x + 2 y + 5z - 3w = 7
1 é 5 5 ù 4x + y + 2z - 2w = 8
4
R1 R1 ê 1
4 2 ú
ê ú
ê
ë 7 8 19 ú
û
Write the augmented matrix and transform the matrix.
é 2 -2 3 -4 6 ù
é
1
5 5 ù ê ú
ê 4 2 ú ê 3 2 5 -3 7 ú
ê ú ê ú
-7R1 + R 2 R 2
ê
0
- 3 3 ú ê 4 1 2 -2 8 ú
4 2 ë û
é 5 5 ù 1 é 3 ù
ê 1
4 2 ú 2
R1 R1 ê 1 -1
2
-2 3 ú
4 ê ú ê ú
- 3
R2 R 2 ëê 0 1 -2 ûú ê 3 2 5 -3 7 ú
ê ú
ê 4 1 2
5
-2 8 ú
- 4
R2 + R1 R1
é 1 0 5 ù ë û
ê 0 1 -2 ú é ù
ë û
ê 1 -1
ê
3
-2 3 ú2
ú
The solution is x = 5 and y = -2, or (5, -2).
-3R1 +R2 R2
ê
0 5
1
3 -2
ú
Your Turn 2 -4R +R R
ê ú
0 5 -4 6 -4
x + 2y + 3z = 2 1 3 3
ê ú
ë û
2x + 2y - 3z = 27
3x + 2y + 5z = 10
é
1 -1
3
-2 3
ù
2
ê ú
Write the augmented matrix and transform the matrix. 5
R2 R2
ê
ê 0 1 1 3 2 ú
5 ú
ê ú
é 1 2 3 2ù ê 0 ú
ê ú ê ú
ê2 2 -3 27ú ë û
ê ú
ê 3 2 5 10ú R + R R
é
1 0 8
-
7 13 ù
ë û 2 1 1 ê 5 5 5 ú
ê ú
é 1 2 3 2ù ê 1 3 2 ú
ê ú ê 0 1 10 5 - 5 ú
-2R1 + R2 R2
ê0 -2 -9 23ú ê ú
ê ú -5R + R R ê
0 0 -
9
3 -2
ú
-3R1 + R3 R3
ê0 -4 -4 4ú 2 3 3 2
ë û

ê ú
-
é 1 2 3 2ù
é
ê 1 0 8 7 13 ù
ú
-1 R R ê0 1 9 - 23 ú ê 5 5 5
ú
2 2 2 ê 2 2 ú ê 1 3 2 ú
ê ú ê 0 1 10 5 - 5 ú
êë0 -4 -4 4 ê ú
2
- R R ê
0 0 1 -
2 4 ú
-2R2 + R1 R1
é 1 0 -6 25ù 9 3 3 ê 3 9 ú
ê ú ë û
ê0 1 9 - 23 ú
ê 2 2 ú
-
8
R +R R
é
1 0 0 -
1 17 ù
4R2 + R3 R3
ê
0 0 14
ú
-42
5 3 1 1 ê 3 9 ú
ê ú
1 ê 2 4 ú
é 1 0 -6 25ù -10
R3 +R2 R2 ê 0 1 0 3
- 9 ú
ê ú ê ú
ê0 1 9 - 23 ú ê
0 0 1 -
2 4 ú
ê 2 2 ú ê 3 9 ú
1 R R
ê ú ë û
14 3 3 ëê0 0 1 -3
.

ë û
ë û
ë û
We cannot change the values in column 4 without
changing the form of the other three columns. So, let w be
the parameter. The last matrix gives these equations.
é3 1 6 ù
ê ú.
ê2 5 15ú
x -
1
w =
17
, or x =
17
+
1
w 2. 4x - 2y = 8
3 9 9 3
-7y = -12
y +
2
w = -
4
, or y = -
4
-
2
w
3 9 9 3 The equations are already in proper form. The
z -
2
w =
4
, or z =
4
+
2
w
augmented matrix obtained from the coefficients
3 9 9 3 and the constants is
é4 -2 8ù
The solution is (17 + 1 w, -4 - 2 w, 4 + 2 w, w ), ê ú.
9 3 9 3 9 3 ê0 -7 -12ú
where w is a real number.
3. 2x +
ë û
y + z = 3
3x - 4y + 2z = -7
2.2 Warmup Exercises
W1. Using z as the parameter, the general solution is
(24 + 3z,10 − 2z, z).
x + y + z = 2
leads to the augmented matrix
é2 1 1 3ù
ê ú
For nonnegative solutions we require
ê 3 -4 2 -7ú.
24 +3z ≥ 0, 10 − 2z ≥ 0, and z ≥ 0.
ê ú
ê 1 1 1 2ú
Solving each inequality for z, we have
z ≥ −8, z ≤ 5, and z ≥ 0.
Thus, 0 ≤ z ≤ 5 and there are 6 solutions in
nonegative intergers.
W2. Using z as the parameter, the general solution is
(25 − 2z, −14 + 3z, z).
4. 2x - 5y + 3z = 4
-4x + 2y - 7z = -5
3x - y = 8
The equations are already in proper form. The
augmented matrix obtained from the coefficients
and the constants is
é 2 -5 3 4ù
ê ú
ê-4 2 -7 -5ú.
ê úFor nonnegative solutions we require
25 − 2z ≥ 0, −14 + 3z ≥ 0, and z ≥ 0.
Solving each inequality for z, we have
z ≤
25
, z ≥
14
, and z ≥ 0.
2 3
Thus,
14
≤ z ≤
25
, and for interger
3 2
z, 5 ≤ z ≤ 12, so there are 8 solutions in
nonnegative intergers.
ê 3 -1 0 8ú
ë û
5. We are given the augmented matrix
é 1 0 2ù
ê ú.
ê0 1 3ú
This is equivalent to the system of equations
x = 2
y = 3,
or x = 2, y = 3.
é 1 0 5ù
6. ê ú
ê0 1 -3ú
2.2 Exercises
ë û
is equivalent to the system

1. 3x +
y = 6
x = 5
2x + 5y = 15
The equations are already in proper form. The
augmented matrix obtained from the coefficients and
the constants is
y = -3.
.

0 0
1 0
0 1
2 6
-2 5
-2 9
6 4
3 2
5 3
0 3 2 5
0 5 3 7
é 1 0
ê
0 -3ù
ú
= ê0 3ê 2 5ú
ú
ê0 5 3 7ú
7 4
2 3
4 5
0 0
5 0
0 4
ê ú
ë û
ê ú
ë û
ë û
Section 2.2 91
é 1 4ù The resulting matrix is
ê ú
7. ê0 -5ú é 3 18ù
ê ú ê ú
ê0 1ú ê2 7ú.
ë û ê ú
The system associated with this matrix is
x = 4
ê0 42ú
ë û
é 1 7ù
y = -5 ê ú
13. 0 5
ê ú
ê úz = 1, ê0 7ú
or x = 4, y = -5, z = 1.
ë û
Find (-2)R2 + R1 R1
é 1 0 0 4ù
ê ú
8. ê0 1 0 2ú
ê0 0 1 3ú
is equivalent to the system
x = 4
y = 2
z = 3.
é(-2)0 + 1 (-2)3 + 6 (-2)2 + 4 (-2)5 + 7 ù
ê úê úê
ú ê ú ë
û
ë û
or x = 4, y = 2, z = 3.
14. Replace R1 by R3 + (-3)R1.
9. Row operations on a matrix correspond to
transformations of a system of equations. The original matrix is
é 1 0 4 21ù
é 3 10ù ê ú
ê ú ê0 6 5 30ú.
11. ê 1 6ú ê ú
ê ú ê0 0 12 15ú
ê0 11ú ë û
ë û
Find R1 + (-3)R2 .
The resulting matrix is
é-3 0 0 -48ù
In row 2, column 1, ê ú
0 6 5 30 .
ê ú
ê ú
3 + (-3)1 = 0. ê 0 0 12 15ú
In row 2, column 2,
7 + (-3)2 = 1.
ë û
é 3 18ù
ê ú
In row 2, column 3,
15. ê0 9ú
ê ú
ê0 8ú
4 + (-3)3 = -5. ë û
In row 2, column 4, 1
R R
é 1
(3)
1
(0) 1
(0) 1
(18)ù
é 1 0 0 6ù
3 1 1 ê 3 3 3 3 ú
ê ú
10 + (-3)6 = -8.
ê
0 5 0 9
ú
ê0 5 0 9ú
ê ú =
ê ú
Replace R2 with these values. The new matrix is
é 3 7 4 10ù
ê ú
ê 0 0 4 8 ú
ëê ûú
ê0 0 4 8ú
ê0 1 -5 -8ú. 16. Replace R3 by 1 R3.
ê ú
ê0 4 5 11ú
12. Replace R3 by -1R1 + 3R3.

2 6
-2 5
0 5
0 0
1 0
0 1
ê ú
ë û
The original matrix is 6
é 1 0 0 30ù
ê ú
ê0 1 0 17ú.
ê0 0 6 162ú
é 3 18ù The resulting matrix is
ê ú
ê2 7ú. é 1 30ù
ê ú ê úê 1 20ú ê0 17ú.
ë û ê ú
ê0 27ú
ë û
.

Solutions Manual for Finite Mathematics and Calculus with Applications 10th
Edition by Lial IBSN 9780133981070
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Test Bank for Finite Mathematics and Calculus with Applications 10th Edition by
Lial IBSN 9780133981070
Download: http://downloadlink.org/p/test-bank-for-finite-mathematics-and-calculus-
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