**Unformatted text preview: **12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight [PRINT]
MAT-140-T2457 20EW2 Precalculus, 8-2 Problem Set: Module Eight
Dallas Weber, 12/20/20 at 2:36:59 PM EST Question1: Score 4/4 Solve the system of equations by any method. −2x + 5y = −21 7x + 2y = 15 Enter the exact answer as an ordered pair, (x, y) .
If there is no solution, enter NS. If there is an infinite number of solutions, enter the general
solution as an ordered pair in terms of x .
Include a multiplication sign between symbols. For example, a * x . Your response
(3,-3) Correct response
(3,-3) Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: Adding these equations as presented will not eliminate a variable. However, we see that the first
equation has −2x in it and the second equation has 7x. So if we multiply the first equation by 7
and the second equation by 2, the x -terms will add to zero. 1/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 7 (−2x + 5y) = 7 (−21) −14x + 35y = −147 2 (7x + 2y) = 2 (15) 14x + 4y = 30 Multiply both sides by 7.
Use the distributive property. Multiply both sides by 2.
Use the distributive property. Now, let’s add them. −14x + 35y = −147 14x + 4y = 30 39y = −117 y = −3 For the last step, we substitute y = −3 into one of the original equations and solve for x . −2x + 5y = −21 −2x + 5 (−3) = −21 −2x − 15 = −21 −2x = −6 x = 3 Our solution is the ordered pair (3, −3). Check the solution in the original second equation. 2/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 7x + 2y = 15 7 (3) + 2 (−3) = 15 True 21 − 6 = 15 Question2: Score 4/4 Solve the system of equations by any method. 6x + 11y x + 2y = 16 = 5 Enter the exact answer as an ordered pair, (x, y) .
If there is no solution, enter NS. If there is an infinite number of solutions, enter the general
solution as an ordered pair in terms of x .
Include a multiplication sign between symbols. For example, a * x . Your response
(-23,14) Correct response
(-23,14) Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: In this case we use substitution.
First, we will solve the second equation for x . 3/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight x + 2y x = 5 = −2y + 5 Now we can substitute the expression −2y + 5 for x in the first equation. 6x + 11y = 16 6 (−2y + 5) + 11y = 16 −12y + 30 + 11y = 16 −y = −14 y Now, we substitute y = 14 = 14 into the second equation and solve for x . x + 2 (14) = 5 x + 28 = 5 x = −23 Our solution is (−23, 14) .
Check the solution by substituting (−23, 14) into both equations. 4/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 6x + 11y = 16 6 (−23) + 11 (14) = 16 −138 + 154 = 16 x + 2y = 5 (−23) + 2 (14) = 5 −23 + 28 = 5 True True Question3: Score 4/4 Solve the system of equations by any method. −x + 2y = −1 3x − 6y = 4 Enter the exact answer as an ordered pair, (x, y) .
If there is no solution, enter NS. If there is an infinite number of solutions, enter the general
solution as an ordered pair in terms of x .
Include a multiplication sign between symbols. For example, a * x . Your response
NS Correct response
NS Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: 5/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Adding these equations as presented will not eliminate a variable. However, we see that the first
equation has −x in it and the second equation has 3x. So if we multiply the first equation by 3, the
x -terms will add to zero. −x + 2y = −1 3 (−x + 2y) = 3 (−1) −3x + 6y = −3 Multiply both sides by 3.
Use the distributive property. Now, let’s add them. −3x + 6y = −3 3x − 6y = 4 0 ≠ 1 This statement is a contradiction. Therefore, the system has no solution. Question4: Score 0/4 6/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Solve the system of equations by any method. −2x + 8y = 6 x − 4y = −3 Enter the exact answer as an ordered pair, (x, y) .
If there is no solution, enter NS. If there is an infinite number of solutions, enter the general
solution as an ordered pair in terms of x .
Include a multiplication sign between symbols. For example, a * x . Your response Correct response -3+4y (x, 1/4*x+3/4) Auto graded Grade: 0/1.0 Total grade: 0.0×1/1 = 0%
Feedback: With the addition method, we want to eliminate one of the variables by adding the equations. In this
case, let’s focus on eliminating x . If we multiply both sides of the second equation by 2, then we
will be able to eliminate the x -variable. x − 4y = −3 2 (x − 4y) = 2 (−3) 2x − 8y = −6 Multiply both sides by 2.
Use the distributive property. Now, add the equations. 7/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight −2x + 8y = 6 2x − 8y = −6 0 = 0 We can see that there will be an infinite number of solutions that satisfy both equations.
Solve the second equation for y . x − 4y = −3 −4y = −x − 3 y = So the general solution is (x, 1
4 x + 3
4 ) 1
4 x + 3
4 . Question5: Score 4/4 A fast-food restaurant has a cost of production C (x) = 11x + 110 and a revenue function
R (x) = 6x . When does the company start to turn a profit?
Enter the exact answer.
If there is no solution, enter NS. If there is an infinite number of solutions, enter IS. x = Your response
NS Correct response
NS Auto graded Grade: 1/1.0 8/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Total grade: 1.0×1/1 = 100%
Feedback: Write the system of equations using y to replace function notation. y = 11x + 110 y = 6x Substitute the expression 11x + 110 from the first equation into the second equation and solve for
x. 11x + 110 = 6x 5x = −110 x = −22 Since x is negative, we conclude the company never turns a profit. Question6: Score 4/4 Use a system of linear equations with two variables and two equations to solve.
A number is 10 more than another number. Twice the sum of the two numbers is 36 . Find the two
numbers.
Enter the numbers in increasing order. First Number:
Your response
4 Correct response
4 Auto graded Grade: 1/1.0 9/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Second Number:
Your response Correct response 14 14 Auto graded Grade: 1/1.0 Total grade: 1.0×1/2 + 1.0×1/2 = 50% + 50%
Feedback: Use the given information to create two equations with two variables.
Let x be one number, and y be the other. y = x + 10 2 (x + y) = 36 x + y = 18 x + (x + 10) = 18 (1) (2) Solve the system using substitution. 2x = 8 x = 4 For the last step, we substitute x = 4 into one of the original equations and solve for y . y = x + 10 y = 4 + 10 y = 14 Therefore, the two numbers are 4 and 14 .
10/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Question7: Score 4/4 Use a system of linear equations with two variables and two equations to solve.
students enrolled in a freshman-level chemistry class. By the end of the semester, 5 times the
number of students passed as failed. Find the number of students who passed, and the number of
students who failed.
228 Passing
Students:
Your response Correct response 190 190 Your response Correct response 38 38 Auto graded Grade: 1/1.0 Failing Students: Auto graded Grade: 1/1.0 Total grade: 1.0×1/2 + 1.0×1/2 = 50% + 50%
Feedback: Use the given information to create two equations with two variables.
Let x be the number of failing students, and y be the number of passing students. y x + y = 5x (1) = 228 (2) Solve the system using substitution. 11/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight x + (5x) = 228 6x = 228 x = 38 For the last step, we substitute x = 38 into one of the original equations and solve for y . y = 5x y = 5 (38) y = 190 Therefore, 190 students passed and 38 students failed. Question8: Score 4/4 Determine whether the given ordered triple is a solution to the system of equations. x − y = 0 x − z = 5 x − y + z Your response and (4, 4, −1) = −1 Correct response Yes, it is a solution.
Feedback:
Correct. Yes, it is a solution. Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: Check each equation by substituting in the values of the ordered triple (4, 4, −1) for x , y , and z.
12/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight x − y = 0 (4) − (4) = 0 True
x − z = 5 (4) − (−1) = 5 4 + 1 = 5 True
x − y + z = −1 (4) − (4) + (−1) = −1 True The ordered triple (4, 4, −1) satisfies all of the equations, so it is a solution to the system. Question9: Score 5/5 13/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Solve the system of equations by any method. = −19 (1) 2x + 4y + z = 20 (2) 2x + 3y + 5z = 23 (3) 3x − 4y + 2z Enter the exact answer as an ordered triple, (x, y, z).
Hint: There are multiple ways to solve this system of equations.
A strategy is to eliminate one of the variables and end up with 2 equations in 2 variables.
One way to do that is to begin with equation 2 to get z = 20 − 2 x − 4 y.
Then substitute 20 − 2 x − 4 y for z in equations 1 and 3.
You now have 2 equations (the modified equations 1 and 3) in 2 variables (x and y ).
Solve this smaller system for x and y and then use z = 20 − 2 x − 4 y to calculate z. Your response
(-1,5,2) Correct response
(-1,5,2) Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: First, we will solve the second equation for z. 2x + 4y + z = 20 z = −2x − 4y + 20 (2)
(4) Now we can substitute the expression −2x − 4y + 20 for z into the other equations. 14/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 3x − 4y + 2z = −19 3x − 4y + 2 (−2x − 4y + 20) = −19 3x − 4y − 4x − 8y + 40 = −19 (1) −x − 12y + 40 = −19 = −12y + 59 (5) 2x + 3y + 5z = 23 (3) 2x + 3y + 5 (−2x − 4y + 20) = 23 x 2x + 3y − 10x − 20y + 100 = 23 −8x − 17y + 100 = 23 8x + 17y = 77 (6) Now, we substitute equation (5) into equation (6) and solve for y . 8 (−12y + 59) + 17y = 77 −96y + 472 + 17y = 77 −79y y Now, we substitute y = 5 = −395 = 5 into equation (5) and solve for x . x = −12 (5) + 59 = −60 + 59 = −1 15/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Now, we substitute x = −1, y = 5 into equation (4) and solve for z. z = −2 (−1) − 4 (5) + 20 = 2 − 20 + 20 = 2 Therefore, our solution is (−1, 5, 2) .
Check the solution by substituting (−1, 5, 2) into all equations. 3x − 4y + 2z = −19 3 (−1) − 4 (5) + 2 (2) = −19 −3 − 20 + 4 = −19 2x + 4y + z True = 20 2 (−1) + 4 (5) + (2) = 20 −2 + 20 + 2 = 20 2x + 3y + 5z = 23 2 (−1) + 3 (5) + 5 (2) = 23 −2 + 15 + 10 = 23 True True Question10: Score 4/4 16/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight Use the matrices below to perform the indicated operation if possible. −11 8 B = [ −5 ,C 3 0 10 7 1 = [ If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it's next to the infinity symbol. Specify the appropriate number of rows (horizontal) and columns
(vertical). 2C + B = Your response
−11 Correct response 28 [ 9 −11 28 9 5 [ 5 Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: First perform the scalar multiplication for 2C . Multiply each entry in C by 2. 2C 0 10 7 1 = 2[ 0 20 14 2 = [ Now perform the addition for 2C + B . Add the corresponding entries. 17/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 0 20 2C + B = [ −11 8 −5 3 + [
14 2 −11 28 9 5 = [ Question11: Score 0/4 Use the matrices below to perform matrix multiplication. 2 7 3 −9 0 12 B = [ ,C ⎡ 4 = ⎢ −3
⎣ 5 10 ⎤ 6⎥
8 ⎦ If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it's next to the infinity symbol. Specify the appropriate number of rows (horizontal) and columns
(vertical). BC = Your response
2 84 24 6 [ Correct response
2 86 24 6 [ Auto graded Grade: 0/1.0 Total grade: 0.0×1/1 = 0%
Feedback: The dimensions of B are 2 × 3 and the dimensions of C are 3 × 2 . The inner dimensions match so
the product is defined and will be a 2 × 2 matrix. 18/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight 2 7 = [ BC ⎡ 3 4 10 ⎢ −3
−9 0 12 ⎣ ⎤ 6⎥ 5 8 ⎦ 2 (4) + 7 (−3) + 3 (5) 2 (10) + 7 (6) + 3 (8) −9 (4) + 0 (−3) + 12 (5) −9 (10) + 0 (6) + 12 (8) = [ 2 86 24 6 = [ Question12: Score 0/5 Use the matrices below to perform matrix multiplication. B = [ 3 7 4 −9 0 11 3 −4 , D = ⎢9 4 ⎡ ⎣ 0 8 11 ⎤ 1⎥
−10 ⎦ If the operation is undefined, enter NA. Hint: To create a matrix in Mobius, click on the icon that has three rows of three dots - it's next to the infinity symbol. Specify the appropriate number of rows (horizontal) and columns
(vertical). BD = Your response
66 48 Correct response −1 [ −27 124 −229 72 48 0 −27 124 −209 [ Auto graded Grade: 0/1.0 Total grade: 0.0×1/1 = 0%
Feedback: 19/20 12/20/2020 Southern New Hampshire University - 8-2 Problem Set: Module Eight The dimensions of B are 2 × 3 and the dimensions of D are 3 × 3 . The inner dimensions match so
the product is defined and will be a 2 × 3 matrix. 3 7 4 BD = [
−9 0 11 3 −4
⎢9 4 ⎡ ⎣ 0 8 11 ⎤ 1⎥
−10 ⎦ 3 (3) + 7 (9) + 4 (0) 3 (−4) + 7 (4) + 4 (8) 3 (11) + 7 (1) + 4 (−10) −9 (3) + 0 (9) + 11 (0) −9 (−4) + 0 (4) + 11 (8) −9 (11) + 0 (1) + 11 (−10) = [ 72 48 0 −27 124 −209 = [ 20/20 ...

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