**Unformatted text preview: **6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight [PRINT]
MAT-140-H7324 21EW5 Precalculus, 8-2 Problem Set: Module Eight
Corvinna Curtis, 6/27/21 at 2:11:18 PM EDT Question1: Score 4/4 Solve the system of equations by any method. −2x + 5y = −42 7x + 2y = 30 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
(6,-6) Correct response
(6,-6) 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 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Multiply both sides by 7. Use the distributive property. Multiply both sides by 2. Use the distributive property. 7 (−2x + 5y) = 7 (−42) −14x + 35y = −294 2 (7x + 2y) = 2 (30) 14x + 4y = 60 Now, let’s add them. −14x + 35y = −294 14x + 4y = 60 39y = −234 y = −6 For the last step, we substitute y = −6 into one of the original equations and solve for x . −2x + 5y = −42 −2x + 5 (−6) = −42 −2x − 30 = −42 −2x = −12 x = 6 Our solution is the ordered pair (6, −6) . Check the solution in the original second equation. 2/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight True 7x + 2y = 30 7 (6) + 2 (−6) = 30 42 − 12 = 30 Question2: Score 4/4 Solve the system of equations by any method. 5x + 9y = 15 x + 2y = 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
(-15,10) Correct response
(-15,10) 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 6/27/2021 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. 5x + 9y = 15 5 (−2y + 5) + 9y = 15 −10y + 25 + 9y = 15 −y = −10 y = 10 Now, we substitute y = 10 into the second equation and solve for x . x + 2 (10) = 5 x + 20 = 5 x = −15 Our solution is (−15, 10) .
Check the solution by substituting (−15, 10) into both equations. 4/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight = 15 True = 5 = 5 True 5x + 9y = 15 5 (−15) + 9 (10) = 15 −75 + 90 x + 2y (−15) + 2 (10) = 5 −15 + 20 Question3: Score 4/4 Solve the system of equations by any method. −x + 2y = −1 8x − 16y = 9 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 6/27/2021 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 8x. So if we multiply the ﬁrst equation by 8, the
x -terms will add to zero. Multiply both sides by 8. Use the distributive property. −x + 2y = −1 8 (−x + 2y) = 8 (−1) −8x + 16y = −8 Now, let’s add them. −8x + 16y = −8 8x − 16y = 9 0 ≠ 1 This statement is a contradiction. Therefore, the system has no solution. Question4: Score 0/4 6/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Solve the system of equations by any method. −3x + 12y = 21 x − 4y = −7 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 (x, 4y-7) (x, 1/4*x+7/4) 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 3, then we
will be able to eliminate the x -variable. Multiply both sides by 3. Use the distributive property. x − 4y = −7 3 (x − 4y) = 3 (−7) 3x − 12y = −21 Now, add the equations. 7/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight −3x + 12y = 21 3x − 12y = −21 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 = −7 −4y = −x − 7 y = 1
4 x + 7
4 So the general solution is (x, 1
4 x + 7
4 ) . Question5: Score 4/4 A fast-food restaurant has a cost of production C (x) = 14x + 140 and a revenue function
R (x) = 7x . 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 Correct response NS NS Auto graded Grade: 1/1.0 8/20 6/27/2021 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 = 14x + 140 y = 7x Substitute the expression 14x + 140 from the ﬁrst equation into the second equation and solve for
x. 14x + 140 = 7x 7x = −140 x = −20 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
numbers. 9 more than another number. Twice the sum of the two numbers is 46 . Find the two Enter the numbers in increasing order. First Number: Your response Correct response 7 7 Auto graded Grade: 1/1.0 9/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Second Number: Your response
16 Correct response
16 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 + 9 (1) (2) 2 (x + y) = 46 x + y = 23 Solve the system using substitution. x + (x + 9) = 23 2x = 14 x = 7 For the last step, we substitute x = 7 into one of the original equations and solve for y. y = x + 9 y = 7 + 9 y = 16 Therefore, the two numbers are 7 and 16.
10/20 6/27/2021 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, 7 times the
number of students passed as failed. Find the number of students who passed, and the number of
students who failed.
312 Passing Students:
Your response
273 Correct response
273 Auto graded Grade: 1/1.0 Failing Students: Your response
39 Correct response
39 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 = 7x (1) (2) x + y = 312 Solve the system using substitution. 11/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight x + (7x) = 312 8x = 312 x = 39 For the last step, we substitute x = 39 into one of the original equations and solve for y. y = 7x y = 7 (39) y = 273 Therefore, 273 students passed and 39 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 = 9 x − y + z and (8, 8, −1) = −1 Your response Correct response Yes, it is a solution.
Yes, it is a solution.
Feedback:
Correct.
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 (8, 8, −1) for x , y, and z .
12/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight x − y = 0 (8) − (8) = 0 True x − z = 9 (8) − (−1) = 9 8 + 1 = 9 True x − y + z = −1 (8) − (8) + (−1) = −1 True The ordered triple (8, 8, −1) satisﬁes all of the equations, so it is a solution to the system. Question9: Score 5/5 13/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Solve the system of equations by any method. = −22 (1) 2x + 4y + z = 18 (2) = 21 (3) 3x − 4y + 2z 2x + 3y + 5z 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 = 18 − 2 x − 4 y.
Then substitute 18 − 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 = 18 − 2 x − 4 y to calculate z . Your response
(-2,5,2) Correct response
(-2,5,2) Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: First, we will solve the second equation for z . (2) = −2x − 4y + 18 (4) 2x + 4y + z = 18 z Now we can substitute the expression −2x − 4y + 18 for z into the other equations. 14/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight 3x − 4y + 2z = −22 (1) 3x − 4y + 2 (−2x − 4y + 18) = −22 3x − 4y − 4x − 8y + 36 = −22 −x − 12y + 36 = −22 = −12y + 58 (5) x 2x + 3y + 5z = 21 (3) 2x + 3y + 5 (−2x − 4y + 18) = 21 2x + 3y − 10x − 20y + 90 = 21 −8x − 17y + 90 = 21 8x + 17y = 69 (6) Now, we substitute equation (5) into equation (6) and solve for y. 8 (−12y + 58) + 17y = 69 −96y + 464 + 17y = 69 −79y = −395 y = 5 Now, we substitute y = 5 into equation (5) and solve for x . x = −12 (5) + 58 = −60 + 58 = −2 15/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Now, we substitute x = −2 , y = 5 into equation (4) and solve for z . z = −2 (−2) − 4 (5) + 18 = 4 − 20 + 18 = 2 Therefore, our solution is (−2, 5, 2) .
Check the solution by substituting (−2, 5, 2) into all equations. 3x − 4y + 2z = −22 3 (−2) − 4 (5) + 2 (2) = −22 −6 − 20 + 4 = −22 True −4 + 20 + 2 = 18 True 2x + 3y + 5z = 21 2 (−2) + 3 (5) + 5 (2) = 21 −4 + 15 + 10 = 21 True 2x + 4y + z = 18 2 (−2) + 4 (5) + (2) = 18 Question10: Score 4/4 16/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight Use the matrices below to perform the indicated operation if possible. −11 6 B = [ −4 , C 0 11 7 1 = [ 2 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 28 10 4 [ Correct response −11 28 10 4 [ 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 = 2 [ = [ 0 11 7 1 0 22 14 2 Now perform the addition for 2C + B . Add the corresponding entries. 17/20 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight 0 22 2C + B = [
14 = [ −11 6 −4 2 + [ 2 −11 28 10 4 Question11: Score 4/4 Use the matrices below to perform matrix multiplication. 2 6 ⎡ 3 B = [ −8 0 , C 5 = ⎢ −3 11 ⎣ 6 10 ⎤ 7⎥
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
10 86 26 8 [ Correct response
[ 10 86 26 8 Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
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 6/27/2021 Southern New Hampshire University - 8-2 Problem Set: Module Eight BC 2 = [ ⎡ 3 5 10 ⎢ −3
−8 0 11 ⎣ ⎤ 7⎥ 6 8 ⎦ 2 (5) + 6 (−3) + 3 (6) 2 (10) + 6 (7) + 3 (8) −8 (5) + 0 (−3) + 11 (6) −8 (10) + 0 (7) + 11 (8) = [ = [ 6 10 86 26 8 Question12: Score 5/5 Use the matrices below to perform matrix multiplication. B = [ 4
−9 7
0 5
12 3 −4 , D = ⎢ 8 4 ⎡ ⎣ 0 7 12 ⎤ 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
[ 68
−27 47
120 5 Correct response −228 [ 68 47 5 −27 120 −228 Auto graded Grade: 1/1.0 Total grade: 1.0×1/1 = 100%
Feedback: 19/20 6/27/2021 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. 4 7 5 BD = [
−9 =[ =[ 0 12 3 −4
⎢8 4 ⎡ ⎣ 0 7 12 ⎤ 1⎥
−10 ⎦ 4 (3) + 7 (8) + 5 (0) 4 (−4) + 7 (4) + 5 (7) 4 (12) + 7 (1) + 5 (−10) −9 (3) + 0 (8) + 12 (0) −9 (−4) + 0 (4) + 12 (7) −9 (12) + 0 (1) + 12 (−10) 68 47 5 −27 120 −228 20/20 ...

View
Full Document