#$&* course Mth163 Question `q001: What is the vertex of each of the other two parabolas depicted above?
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Given Solution: The vertex of the 'blue' parabola is at the point (2, -1), the 'lowest' point on the parabola. The vertex of the 'purple' parabola is at the point (-2, 3), the 'highest' point on the parabola. Self-critique: ------------------------------------------------ Self-critique rating:ok Question `q002: What are the coordinates of the other two 'basic points' of each parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For the purple parabola the other points are (-1,1) and (-3,1). For the blue parabola the other points are (1,0) and (3,0). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the 'blue' parabola, the points (1, 0) and (3, 0) are two units apart, and lie on the same horizontal line. The horizontal line is the x axis. For the 'purple' parabola, the points (-3, 1) and (-1, 1) are two units apart. These points lie on the horizontal line where y = 2. Self-critique: ------------------------------------------------ Self-critique rating:ok-3 Question `q003: For the first parabola, the one whose vertex is (2, 1), how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'? How far would we then have to move in the vertical direction to reach that point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If we moved parabola with the vertex of (2,1) to the right 1 space we will be below the point (3,3). If we move to the left one space we will be under the point (1,3). Once we move either left or right, we move upward 2 points and we will be directly at points (3,3) or (1,3). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the parabola with vertex (2, 1), if we move 1 unit to the right or left we will be at the point (3, 1) or (1, 1), putting us directly below one of the other two basic points. If we then move 2 units upward, we will be at the point (3, 3) or (1, 3). So if we move 1 unit to the right or left, we need to move 2 units upward to get to another basic point. Self-critique: ------------------------------------------------ Self-critique rating:ok-3 Question `q004: For each of the other two parabolas, how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'? How far would we then have to move in the vertical direction to reach that point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For both of the other parabolas we would only have to move 1 point left or right. But for the purple parabola we would have to move the vertex downward 2 points. For the blue parabola we would have to move upward 1 point. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For either of the other two parabolas, if we move 1 unit to the right or left we will be directly above or below one of its basic points (above in the case of the 'third' parabola, whose vertex is (-2, 3), below in the case of the 'second' parabola, whose vertex is (2, -1).) To get to the basic points of the 'third' parabola we will need to move 2 units downward. To get to the basic points of the 'second' parabola we will need to move 1 unit upward. Self-critique: ------------------------------------------------ Self-critique rating:ok-3 Question `q005. Solve the following system of simultaneous linear equations: 3a + 3b = 9 6a + 5b = 16. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To eliminate one of the variables we have to find the coefficients. The coefficient in the second equation, (a=6) is half of the first (a=3). So to make the first equation equal the second we have to multiply everything in the first equation by -2. -2 * (3a + 3b) = -2 * 9 -6a - 6b = -18 Then we take the two equations and add them together. -6a -6b = -18 6a + 5b = 16 Then we get -b = -2 or b = 2. Now we substitute b=2 into the first equation and solve. 3a + 3(2) = 9 3a + 6 = 9 ; subtract 6 from both sides 3a = 3 ; divide by 3 to get a by itself a = 1 so therefore our solution is a=1 and b=2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. We used the first equation in our last step, so we verigy this solution is by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16. STUDENT QUESTION I got my answer in a very different way than the solution given. I have been trying to remember things from the classes I took a long time ago and came up with this answer. Is it alright to use this method? INSTRUCTOR RESPONSE Here is a synopsis of your solution: I'll first solve the first equation for a: 3a+3b=9 so a+b=3 so a=3-b. Now I'll substitute this expression for a into the second equation 6 a + 5 b = 16 Replacing a with 3 - b: 6(3-b)+5b=1618-6b+5b=16-b=-2b=2 a = 3 - b so a=3 - 2 = 1 Substituting a = 1 and b = 2 into the two equations we get 3(1)+3(2)=9 so 9 = 96(1)+5(2)=16 so 16 = 16. The solution checks with the two equations. You have an excellent solution. The method you have used is performed correctly and is equally valid with the method used in the solutions. It is called the 'substitution method'. For these first few problems in this course the substitution method and the elimination method are equally efficient. However the elimination method is also important, and since elimination works better on most of the problems we'll be encountering in the near future, it is the method I use in the given solutions. You can use either method, as long as you know both. However you might find the given solutions easier to understand if you use the elimination method. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get the equations the same you have to multiply the first equation by -3 and the second equation by 2. (-3)*4a + (-3)*5b = 18*(-3) -12a -15b = -54 (2)*6a + (2)*9b = 30*(2) 12a + 18b = 60 Now we have to add the two equations together -12a -15b = -54 12a + 18b = 60 3b = 6 ; divide by 3 on each side b = 2 Now we have to substitute 2 into the first equation and solve for a. 4a + 5(2) = 18 4a + 10 = 18 ; subtract 10 from both sides 4a = 8 ; divide by 4 on both sides a = 2 Therefore our solutions are a = 2 and b = 2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12. • We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2. • Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well. Choosing to 'match' the coefficient of a, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, so the system becomes 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the equations we get -3 b = -6, so b = 2. Substituting this value of b into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. If y = 5x + 8, then for what value of x will we have y = 13? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get this answer we have to substitute y= 13 into our equation and solve for x. y = 5x + 8 13 = 5x + 8 ; subtract 8 from both sides 5 = 5x ; divide 5 from both sides x=1 When y = 13, x = 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We first substitute y = 13 into the equation y = 5 x + 8 to obtain 13 = 5 x + 8. Subtracting 8 from both equations and reversing the equality we obtain 5 x = 5, which we easily solve to obtain x = 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!