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course mth 164
3:45pm 11/28
SOLUTIONS/COMMENTARY ON QUERY 14**** Query problem 10.1.22 (5th ed 10.1.24) solve the equation 3x - y
= 7, 9x - 3y = 21.
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14:24:23
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Your solution:
After you multiply the 1st one by 3 and subtract them they cancel each other out. I don’t know what the answer would be then.
confidence rating #$&*: 1
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Given Solution:
** The two equations coincide. If we subtract 3 times the first from the second we get 0 = 0.
This tells us that one equation is a multiple of the other, and that they are therefore equivalent. Their graphs coincide.
A solution to one equation is a solution to the other. So the solution set consists of all (x, y) satisfying the first equation 3 x - y = 7. These solutions lie on the line y = 3x - 7. **
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Self-critique (if necessary):So does that mean that the answer could be 3(3) - (2) = 7.
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Self-critique Rating:2
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(3, 2) would be a point on the graph of the equation.
The equation, though, is a linear equation with infinitely many possible solutions (for example, (4, 5) is another solution).
3x - y = 7 is the equation of a striaght line. All solutions lie on that straight line, and evey point on that straight line corresponds to a solution.
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Query problem 10.2.12 (5th edition 10.1.42) solve the equation
3x-2y+2z=6, 7x-3y+2x=-1, 2x-3y+4z=0.
** That's 2z in the second equation, not 2x. **
** If we row reduce the coefficient matrix [[3,-2,2,6],[7,-3,2,-1],[2,-3,4,0]] we get [[1, 0, -0.4] [0; 0, 1, -1.6] [0; 0, 0, 0, 1]], indicating inconsistent system.
Using the equations:
From the equations
3 x - 2 y + 2 z = 6 and
2 x - 3 y + 4 z = 0
we can eliminate z (add -2 times the first to the second) to get
-4 x + y = -12 or
y = 4x - 12.
From the equations
3 x - 2 y + 2 z = 6 and
7x-3y+2z=-1
we eliminate z (just subtract the equations) to get
y = 4 x + 7 .
y cannot be equal to 4x - 12 at the same time it's equal to 4x + 7. The two expressions could only be equal if -12 = 7.
We conclude that the system is inconsistent. No solution exists.
The first thing I did was number the equations as follows:
x - y + z = -4 (Equ #1)
2x -3y + 4z = -15 (Equ #2)
5x +y - 2z = 12 (Equ #3)
then, I replaced #1 with product of #1 & -2 making it -2x + 2y -2z = 8, then I replaced #2 with the sum of #1 & #2 making #2 : -y + 2z = -7, then I multiplied #1 by 5 and multiplied #3 by 2 making #1 -10x + 10y - 10z = 40 and #3: 10x + 2y - 4z = 24, next I replaced #3 with the sum of #1 & #3 making #3: 12y - 14z = 64, then I multiplied #2 by 6 and multiplied #3 by 1/2 making #2: -6y = 12z = -42 and #3: 6y -7z = 32, I then replaced #3 with the sum of #2 & #3 making #3 5z = -10 which then solves to z = -2, Next I back substituted -2 for z in #1 & #2 making #1: -10x + 10y = 20 and #2: y = 3, I then substituted 3 for y in #1 and solved it for x: x = 1.
** Your system differs from the one in my notes. Your solution x = 1, y = 3, z = -2 is correct for your equations, which I suspect are the correct equations from the book. **
** Solution to the system as given in my note as x - y - z = -4, 2•x - 3•y + 4•z = -15, 5•x + y - 2•z = 12:
We start with
x - y - z = -4
2x -3y + 4z = -15
5x +y - 2z = 12.
We multiply the first equ by -1 and add it to the second:
-2x + 2y + 2z = 8
2x - 3y + 4z = -15.
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-y + 6z = -7
Add -5 times first to third to get
-5x + 5y + 5z = 20
5x + y - 2z = 12
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6y + 3 z = 32.
Solve the system -y + 2z = -7, 6y - 7z = 32. First add 6 times first to second:
-6y + 36 z = -42
6y + 3 z = 32
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39 z = -10
z = -10/39.
Substituting into the equation 6y + 3z = 32 we get
6y + 3 * -10/39 = 32 which gives solution
y = 71/31.
Substituting z = -10/39 and y = 71/13 into the first original equation x - y - z = -4 we get x = 47/39. **
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Self-critique (if necessary):
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Self-critique Rating:
**** Query problem 10.1.44 (5th ed 10.1.60) ticket $9 adult $7
senior, 325 people paid $2495.
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14:38:27
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Your solution:
Set up the equations 9x + 7y = 2495 9x + 7y = 2495
(x + y = 325)-7, -7x - 7y = -2275 Add these together to get 2x = 220, then divide both sides by 2 to get x = 110, then we have 7y + 990 = 2495 so we subtract 990 from both sides to get 7y = 1505 then we divide both sides by 7 to arrive at y = 215
9(110) + 7(215) = 2495
There were 110 adults and 215 seniors.
confidence rating #$&*: 3
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Given Solution:
Good student solution:
We can set up a formula for the price of the tickets by saying
x= adult ticket price and
y=senior citizen ticket price
so we have 9x+7y=2495
We can then write an equation for the amount of tickets purchased by saying
x+y=325
We can then multiply the second equation by -9 and we get
-9x-9y=-2925
so when we add the sum of these two equations we get
-2y=-430 so
y=215
so there were 215 senior citizens and 110 adults.
INSTRUCTOR COMMENT:
Good. x adults and y seniors at $9 per adult and $7 per senior yields 9 x + 7 y dollars. This is equal to the $2495, giving us the equation
9x + 7y = $2495.
The total number of tickets is
x + y = 325,
giving us the second equation.
This equation is solved as indicated by the student.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#