#$&* course Mth 163 1/14 Around 2:00pm 002.
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Given Solution: The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 which can be obtained from the system in the preceding problem, by eliminating the easiest variable. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For these two equations, b will obviously be the easiest variable to eliminate. To do that, I determined what the least common multiple of 2 and 7 is, which is 14. So I multiplied the top equation by 7 and the bottom equation by -2. I multiplied by -2 so that the values for b will cancel out. Then I solved for a. 7 [58 a + 2 b = -38] -2 [198 a + 7 b = -128] 406a + 14b = -266 -396a - 14b = 256 406a = -266 -396a = 256 10a = -10 a = -1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128) Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To obtain a value for b, all I had to do was substitute -1 for a in either of the equations. 58(-1) + 2b = -38 -58 + 2b = -38 2b = 20 b = 10 Just to be sure, I solved for b in the other equation as well by using the same process. 198(-1) + 7b = -128 -198 + 7b = -128 7b = 70 b = 10 If both equation produce the same value for b (which they did) then the answer is surely b = 10. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started out with solving for c with the easiest equation, the first one. I plugged in the values for a and b and then solved. 2(-1) + 3(10) + c = 128 -2 + 30 + c = 128 28 + c = 128 c = 100 Just to be 200% sure the c = 100, I plugged in all of the values (a, b, and c) to see if I got the right answer. I did this with the last two equations. 60(-1) + 5(10) + 100 -60 + 50 + 100 -10 + 100 90 90 was the correct answer for the second equation 200(-1) + 10(10) + 100 -200 + 100 + 100 -200 + 200 0 0 was the correct answer for the third equation. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100. Substituting these values into the second equation, in order to check our solution, we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8). We are going to use simultaneous equations to obtain the equation of that parabola. A graph has a parabolic shape if its the equation of the graph is quadratic. The equation of a graph is quadratic if it has the form y = a x^2 + b x + c. y = a x^2 + b x + c is said to be a quadratic function of x. To find the precise quadratic function that fits our points, we need only determine the values of a, b and c. As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c. The first step is to obtain an equation using the first known point. What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into theform y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First I plugged in 1 for every x in the equation then plugged in -2 for y. So the equation should look like this: -2 = a(1)^2 + b(1) + c Then I solved as much as I could: -2 = a(1*1) b(1) + c -2 = a +b + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Again, I substituted x for 3 and y for 5, so: 5 = a(3)^2 + b(3) + c 5 = a(3*3) + b(3) + c 5 = 9a + 3b + c I did the same for the third coordinate. x = 7 and y = 8 8 = a(7)^2 + b(7) + c 8 = a(7*7) + b(7) + c 8 = 49a + 7b + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this. Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c. Solve the system to find the values of a, b and c. What is the solution of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you have to eliminate the easiest variable, which is c. FOR THE SECOND EQUATION: To get the c to cancel out, I multiplied the first equation by -1 and subtracted it from the second equation. -1 (a + b + c = -2) 9a + 3b + c = 5 -a - b - c = 2 The new second equation is: 8a + 2b = 7 FOR THE THIRD EQUATION: Again, to get the c to cancel out, I multiplied the first equation by -1 and subtracted it from the third equation. -1 (a + b + c = -2) 49a + 7b + c = 8 -a - b - c = 2 The new third equation is 48a + 6b = 10. I then simplified it by dividing by 2: 24a + 3b = 5. SOLVING FOR A: First I figured out what the least common multiple of 2b and 3b is. The LCM is 6, so I multiplied the equation containing 2b by 3 and the equation contained 3b by -2. I multiplied with a -2 so that the values for b will cancel out. Then I solved for a. 3 ( 8a + 2b = 7) -2 (24a + 2b = 5) The equations will look like this: 24a + 6b = 21 -48a - 6b = -10 24a = 21 -48a = -10 -24a = 11 a = -0.4583 SOLVING FOR B: Again I figured out the LCM of 8a and 24a which is 24. To get the 24a to cancel out, I multiplied the first equation by -3. Then I solved for b. -3 (8a + 2b = 7) So…. 24a + 3b = 5 -24a -6b = -21 -3b = -16 b = 5.3333 SOLVING FOR C: To solve for c, I simply substituted the values I got for a and b into one of the 3 original equations. a + b + c = -2 -0.4583 + 5.3333 + c = -2 4.875 + c = -2 c = -6.875 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different solution, you should show your solution. Start by indicating the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK QUESTION!: I noticed that the values you got for a and b had five (5) decimal places and the value you got for c only had three (3). The values I got for a, b, and c all have four (4) decimal places. I got the answer right even though I had four decimal places. How do you determine the number of decimal places to use for a value (or an answer)???
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Given Solution: Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function. What is your function? What y values do you get when you substitute x = 1, 3, 5 and 7 into this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By substituting the values I got for a, b, and c into y = a x^2 + bx + c, the specific function is: y = -0.4583 x^2 + 5.3333 x - 6.875 If x = 1 then y = -0.4583 (1*1) + 5.3333(1) - 6.875 = -2. If x = 3 then y = y = -0.4583 (3*3) + 5.3333 (3) - 6.875 = 5.0002 (round down to 5) If x = 5 then y = -0.4583 (5*5) + 5.3333 (5) - 6.875 = 8.334 If x = 7 then y = -0.4583 (7*7) + 5.3333 (7) - 6.875 = 8.0014 (round down to 8) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!