Math 140 Quiz 5 - Summer 2006 Solution Review

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Math 140Quiz 5 - Summer 2006 Solution Review

• (Small white numbers next to problem number
  represent its difficulty as per cent getting it
  wrong.)

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Problem 1 (05)

• Solve the system. (7/3)x (5/4)y 4
  (a)
• (5/6)x -
  2y 21 (b)

Using the method of substitution, select an
equation, say (b), to solve for a selected
variable, say, y y
(5/12)x - (21/2). (c) Then,
substitute this for y in (a) solve for x
(7/3)x (5/4) (5/12)x - (21/2) 4
Hence, x (4105/8)/(7/3)(5/4) (5/12)
6. Put this in (c) to get y (5/12)(6) - (21/2)
-8. (6, -8)
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Problem 2 (33)

• Determine the number of solutions for the
  given system without solving the system.
• 4x - 3y 5
  (a)
• 16x - 12y 20 (b)

Replace (b) by (b) minus 4 times (a) 0
0. This is indicates a consistent system with an
infinite number of solutions. Check by noting
(b) is a line of same slope (4/3) and y-intercept
(5/3) as (a). Thus, it is the same line.
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Problem 3 (33)

• Determine the number of solutions for the
  given system without solving the system.
• 3x 3y -2
  (a)
• 12x 12y 3
  (b)

Replace (b) by (b) minus 4 times (a) gt 0
11 !!! This is indicates an inconsistent
system with no solution. Note (a) (b) are
lines of same slope (-4/3) but differing
y-intercepts (2/3) (-1/4). Thus, they are
parallel lines and do not intersect.
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Problem 4 (19)

• Use a calculator to solve the system of
  equations.
• y 2.12x - 31
  (a)
• y -0.7x 24
  (b)

Substitute for y in (a) the y given in (b)
solve for x.
Then, from (b) y -0.7(33.09153) 24
0.83593
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Problem 5 (43)

• Solve A twin-engine aircraft can fly 1190
  miles from city A to city B in 5 hours with the
  wind and make the return trip in 7 hours against
  the wind. What is the speed of the wind?

Let speeds be p for plane w for wind. Then,
5(p w) 1190 (a)
7(p - w) 1190
(b)
After dividing (a) by 5 (b) by 7, subtract
them. 2w 238 170 68
w 34 miles per hour
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Problem 6 (24)

• Solve the system of equations.

Use augmented matrix for system manipulate to
row echelon form by row operations.
R3 -r1 r3, R1 r1/2, R2 -5r1
r2
R2 -r2/14, R3 3r2 r3
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Problem 6 contd (24)

• Solve the system of equations.

Use augmented matrix for system manipulate to
row echelon form by row operations.
z -2 y -7/2 - z/4 -7/2 -
(-2)/4 -3
R3 r3/(14/19)
x -4 - z/2 - 2y -4 - (-2)/2 -2(-3) 3
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Problem 7 (0!)

• Write the augmented matrix for the system.

The augmented matrix for the system is obtained
by just copying the coefficients in the standard
equations.
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Problem 8 (0!)

• Write a system of equations associated with
  the augmented matrix. Do not try to solve.

The standard equations are obtained from the
augmented matrix for the system by just copying
the coefficients into their places.
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Problem 9 (29)

• Perform in order (a), (b), and (c) on the
  augmented matrix. (a) R2 -2r1 r2
• (b) R3 4r1 r3 (c)
  R3 3r2 r3

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Problem 10 (19)

• Find the value of the determinant.

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Problem 11 (38)

• Find the value of the determinant.

Use, e.g., (a) R1 r1 r3, (b) R3 -4r2
r3, expand down column 2.
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Problem 12 (81)

• Use the properties of determinants to find
  the value of the second determinant, given the
  value of the first.

Note the matrix in D2 differs from that in D1
only by (a) a row swap R1 r3, R3 r1 (b)
by the factor of 3 in row 1 of D2. In the R1 row
expansion of D2 these yield a (-1) overall an
overall factor of 3 compared to the R3 row
expansion of D1. Details are on next slide. The
result is D2 -3D1 12.
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Problem 12 contd (81)

• Use the properties of determinants to find
  the value of the second determinant, given the
  value of the first.

The R1 row expansion of D2 is
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Problem 13 (05)

• Use Cramer's rule to solve the linear system.

Construct evaluate the determinants
Then, x Dx /D -0.41024/(-.05128) 8
y Dy /D - 0.2564/(-.05128) 5
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Problem 14 (10)

• Use Cramer's rule to solve the linear system.

Construct evaluate the determinants
Then, x Dx /D 3/3 1, y Dy /D
15/3 5,
______________ z Dz /D 3/3 1.
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Problem 15 (10)

• Solve the system. x2 y2 100
  (a)
• x y
  2 (b)

Using the method of substitution, select an
equation, preferably (b), to solve for a selected
variable, say, y y -x
2. (c) Then,
substitute this for y in (a) solve for x
x2 (-x 2)2 100 gt 2x2 4x 96 0.
Divide by 2 factor (x-8)(x6)0 gt x 8 or
-6. Put this in (c) y -6 or 8. gt (8,
-6), (-6, 8)
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Problem 16 (10)

• Solve the system. xy 20
  (a)
• x y 9
  (b)

Using the method of substitution, select an
equation, preferably (b), to solve for a selected
variable, say, y y
- x 9. (c) Then, substitute
this for y in (a) solve for x x(- x 9)
20 gt x2 - 9x 20 0. Factor (x
4)(x 5) 0 gt x 4 or 5. Put this in (c)
y 5 or 4. gt (4, 5), (5, 4) See graph.
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Problem 16 contd (10)

• Solve the system. xy 20
  (a)
• x y 9
  (b)

Solution is (4, 5), (5, 4).
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Problem 17 (10)

• Solve the system. x2 y2 25
  (a)
• x2 y2 25
  (b)

Using the method of elimination, add subtract
the equations 2x2 50 and 2y2
0. Thus, solving this for x y x 5 and y
0. (-5, 0), (5, 0)
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Problem 18 (33)

• Solve the system. 2x2 y2 66
  (a)
• x2 y2
  41 (b)

Using the method of elimination, subtract the
equations back substitute result x2 25
and 25 y2 41. Thus, solving this for
x y x 5 and y 4. (-5, 4), (5, 4),
(-5, -4), (5, -4)
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Problem 19 (52)

• Solve the system. x2 - xy y2 3
  (a)
• 2x2 xy 2y2
  12 (b)

Using the method of elimination, add the
equations, simplify, back substitute result
x2 y2 5 and xy 2. Then, solving this
for x y y 2/x gt x2 (2/x)2 5, x4 -
5x2 4 0 gt x2 1 or 4. (-1, -2), (1, 2),
(-2, -1), (2, 1)
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Problem 20 (33)

• Solve A rectangular piece of tin has area
  736 in.2. A square of 3 in. is cut from each
  corner, and an open box is made by turning up the
  ends and sides. If the volume of the box is 1200
  in.3, what were the original dimensions of the
  piece of tin?

Let sides of tin be h w. Then,
Tin area hw 736
(a) Box volume 3(h - 6)(w - 6) 1200
(b)
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Problem 20 contd (29)
hw 736
(a) 3(h - 6)(w - 6) 1200
(b)

• After dividing (b) by 3 expanding (b)s product
• hw - 6w - 6h 36 400.
• Simplifying this with use of (a) we replace (b)
  with
• h w 62 (c)

Solve (c) for w 62 - h substitute in (a), h
(62 -h) 736 gt h2 - 62 h 736 (h -16) (h
-46) 0 So h 16 or 46 w 46 or 16. Tin is
16 in. by 46 in.
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