002. ********************************************* Question: `q001. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All the c’s not having a variable makes it easy to eliminate them from the equations. If we subtract the 1st equation from the 2nd it will give us a new equation 1 and if we eliminate the 1st equation from the 3rd it will give us a new equation 2. From these we can fill in the answers to one of the original problem to find the answer for c. 60-2 = 58 a and 5-3= 2b and 90-128 =-38 58+2b= -38. 200-2=198 a and 10-5= 5b and 0-128=-128 198a+7b= -128 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: The easiest variable to eliminate would be b. To eliminate the b’s we must first multiply the first equation by -7 and the 2nd by -2. Which gives us the equations -406-14b=266 -396-14b=256 as you can see our b are eliminated which leaves us with -10a=10 then we divide by -10. a=-1 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128) Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: if we plug in 58 (-1)+2b= -38 -58+2b= -38 we add 58 to both sides 2b=20 divide by 2 b=10 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: To find c I plugged in a and b into the first equation 2(-1)+3 (10)=28 , with the answer being 128 on the equation I new c had to equal 100 but I plugged it into all of my equations. confidence rating #$&*6 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100. Substituting these values into the second equation, in order to check our solution, we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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 the form y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we plug in the point (1, -2) into the equation we get y = a x^2 + b x + c -2=a(1)^2+b(1)+c confidence rating #$&*5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: If we plug in the points (3, 5) and (7,8) in to the equation y = a x^2 + b x + c we get: 5 = a (3)^2 + b( 3) + c or 9 a + 3 b + c = 5. 8 = a (7)^2 + b ( 7) + c or 49 a + 7 b + c = 8. confidence rating #$&*7 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: The system consists of the three equations from the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c =8 confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different solution, you should show your solution. Start by indicating the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b. ADDITIONAL DETAILS ON SOLUTION OF SYSTEM You should have enough practice by now to be able to solve the system; however signs can trip us all up, and I've decided to append the following: The second equation minus the first gives us 8a + 2 b = 7. To avoid a common error in subtracting these questions, note that the right-hand sides of these equations are 5 and -2, and that 5 - (-2) = 5 + 2 = 7. It is very common for students (and the rest of us as well) to get a little careless and calculate the right-hand side as 5 - 2 = 3. The third equation minus the first gives us 48 a + 6 b = 10 (again the right-hand side can trip us up; 8 - (-2) = 10. I often see the incorrect calculation 8 - 2 = 6). Now we solve these two equations, 8 a + 2 b = 7 and 48 a + 6 b = 10: If you subtract 3 times the first from the second you will get 24 a = -11, so that a = -.45833. Substituting this into 8 a + 2 b = 7 and solving for b you get b = 5.33333. Substituting these values of a and b into any of the three original equations you get c = -6.875. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* 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: If we plug in the values for a, b and c into the given form we get the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we plug in 1 y = -.45833 (1)^2 + 5.33333 (1 )- 6.875 = -2. When we plug in 3 y = -.45833 (3)^2 + 5.33333( 3 )- 6.875 = 5. When we plug in 5 y = -.45833 (5)^2 + 5.33333(5)- 6.875 = 8.33333. When we plug in 7 y = -.45833 (7)^2 + 5.33333(7 )- 6.875 = 8. The points are (1, -2), (3, 5), (5,833333) and (7, 8). confidence rating #$&*4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). "