87.1 Simultaneous Linear Equations Problem Set

SIMULTANEOUS LINEAR EQUATIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: Solve the following system of equations: 𝑥 + 2𝑦 − 𝑧 = 6 3𝑥 + 5𝑦 − 𝑧 = 2 −2𝑥 − 𝑦 − 2𝑧 = 4 A. 𝑥 = 10, 𝑦 = 4, and 𝑧 = 29 B. 𝑥 = −2, 𝑦 = −7, and 𝑧 = 3 C. 𝑥 = 9, 𝑦 = 4, and 𝑧 = −3 D. 𝑥 = 22, 𝑦 = −16, and 𝑧 = −16



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PROBLEM 2: Determine the currents within an electrical circuit represented by the following equations: 𝐼! + 𝐼! + 𝐼! = 0 2𝐼! − 5𝐼! = 6 5𝐼! − 𝐼! = −3 A. 𝐼! = 1.4𝐴, 𝐼! = 7𝐴, and 𝐼! = 2𝐴 B. 𝐼! = 1.236𝐴, 𝐼! = −.708𝐴, and 𝐼! = −.528𝐴 C. 𝐼! = 1.26𝐴, 𝐼! = −.41𝐴, and 𝐼! = 2 − .8𝐴 D. 𝐼! = 1.762𝐴, 𝐼! = −.998, and 𝐼! = .421𝐴



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PROBLEM 3: We want 10𝐿 of gas containing 2% additive and have drums of the following: • “𝑥” gasoline without additive • “𝑦” gasoline with 5% additive • “𝑧” gasoline with 6% additive We need to use 4 times as much pure gasoline as 5% additive gasoline. Determine how much of each is needed. A. 𝑥 = 6.4, 𝑦 = 1.6, and 𝑧 = 2 B. 𝑥 = 1, 𝑦 = 3, and 𝑧 = 7 C. 𝑥 = 7.5, 𝑦 = 4, and 𝑧 = 1.2 D. 𝑥 = 4.4, 𝑦 = 4, and 𝑧 = 9.3



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PROBLEM 4: Determine the values of 𝑥! and 𝑥! that satisfy the following linear system. The linear equations represented by this system are: 3𝑥! + 7𝑥! = 2 2𝑥! + 6𝑥! = 4 4 −2 2 B. −4 −2 C. 4 −4 D. 2 A.



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PROBLEM 5: Determine the values of “𝑥” and “𝑦” 3𝑥 + 2𝑦 = 19 𝑥+𝑦 =8 A. 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 2 B. 𝑥 = 3 𝑎𝑛𝑑 𝑦 = 7 C. 𝑥 = 6 𝑎𝑛𝑑 𝑦 = −9 D. 𝑥 = 3 𝑎𝑛𝑑 𝑦 = 5



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SIMULTANEOUS LINEAR EQUATIONS | SOLUTIONS SOLUTION 1: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook. Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES



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We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. In this problem, we will deploy our knowledge of MATRICES, solving it by hand, but illustrating how it is set up so you can HACK it using your CALCULATOR…which is always our go to approach. But for edification purposes, we will solidify an understanding of the process behind the scenes, so let’s get in to it. It’s possible to take any set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that: 𝐴𝑋 = 𝐵 Where each of these VARIABLES represent a UNIQUE MATRIX developed through the definition of each of our EQUATIONS, such that: • The matrix “A” is the COEFICIENT MATRIX of the system used to represent the coefficients of variables in an equation. These will be the numbers in front of the

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variables on the left side of the equal sign. • The matrix “X” is the MATRIX OF VARIABLES used to represent the variables in the equation, which are usually X, Y, and Z. These will be any variables on the left side of the equal sign. • The matrix “B” is the MATRIX OF CONSTANTS used to represent the constants that are on the other side of the equal signs for each equation. These are usually the numbers on the right side of the equal sign. In this problem, we are given: 𝑥 + 2𝑦 − 𝑧 = 6 3𝑥 + 5𝑦 − 𝑧 = 2 −2𝑥 − 𝑦 − 2𝑧 = 4 Our first step would be to always ensure that the equations are in STANDARD FORM, meaning all the VARIABLES are on the left side and all the CONSTANTS are on the right side of the equation. We are good, lets move forward. Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵



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We get: 1 3 −2

2 5 −1

−1 𝑥 6 −1 𝑦 = 2 −2 𝑧 4

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables -1, 2, -1, etc…all on the left side of the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are x, y, and z. • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, which are 6, 4, and 2. Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the MATRIX OF VARIABLES on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 It is important to remember that a matrix multiplied by the inverse of itself will

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produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B. The inverse of any matrix is defined as:

𝐴!! =

𝑎𝑑𝑗(𝐴) 𝐴

Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with their cofactors • |A| is the determinant of the matrix [A] The ADJOINT of a MATRIX is a complex term made extremely simple when viewing the GENERAL FORMULA expressed as: 𝑎𝑑𝑗(𝐴) = (𝐴! )!



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Where: • 𝐴! represents the COEFFICIENT MATRIX of MATRIX A • (𝐴! )! represents the TRANSPOSE of the COEFFICIENT MATRIX A To find the ELEMENTS of a COEFFICIENT MATRIX, you will go ELEMENT by ELEMENT of the ORIGINAL MATRIX, replacing the ELEMENT with a new ELEMENT representing the MATRIX OF MINORS. We will then take the MATRIX OF MINORS, and run through the ELEMENTS changing their sign per the standard TEMPLATE illustrated as: + − +

− + −

+ − +

Per this TEMPLATE, if the SECOND ELEMENT in the first ROW is a negative number, it will change to a positive…if the FIRST ELEMENT in the second ROW is a negative number, it will change to positive, and so forth. Let’s illustrate this with our current problem. We are given the MATRIX: 1 3 −2

2 5 −1

−1 −1 −2 Made with

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The COEFFICIENT MATRIX will be defined by first developing the MATRIX OF MINORS. We will do this by replacing the ELEMENT in the original MATRIX with the DETERMINANT formulated when considering that element. Let’s run through developing the first two ELEMENTS, first focusing on ELEMENT a11 of our MATRIX A: 1 3 −2

2 5 −1

−1 −1 −2

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are: 1 3 −2

2 5 −1

−1 −1 −2

The DETERMINANT can be written as: 5 −1



−1 = 5 −2 − −1 −1 −2

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Which gives us a value: 𝑎!! = −11 Placing this in to position in our MATRIX OF MINORS, we get: −11

Moving to the next ELEMENT a12 of our MATRIX A: 1 3 −2

2 5 −1

−1 −1 −2

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are: 1 3 −2

2 5 −1

−1 −1 −2

The DETERMINANT can be written as: 3 −2

−1 = 3 −2 − −1 −2 −2 Made with

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Which gives us a value: 𝑎!" = −8 Placing this in to position in our MATRIX OF MINORS, we get: −11

−8

And that’s the general process…hence, the reason we want you to HACK this work using your CALCULATOR. Filling out the remaining values in our MATRIX OF MINORS, we get: −11 −5 3

−8 −4 2

7 3 −1

Now we need to apply the proper sign convention established to ultimately define THE MATRIX OF COFACTORS, which again is: + − +



− + −

+ − +

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Applying these sign conventions ELEMENT by ELEMENT, we find that our MATRIX OF COFACTORS is: −11 𝐴! = 5 3

8 −4 −2

7 −3 −1

The ADJOINT MATRIX is the TRANSPOSE of the MATRIX OF COFACTORS, which is:

𝑎𝑑𝑗 𝐴 = 𝐴!

!

−11 = 8 7

5 −4 −3

3 −2 −1

Now we need to define the DETERMINANT of MATRIX A, which is: 𝐴 = 1 −11 − 2 −8 + (−1)(7) Or: 𝐴 = −2 This gives us what we need to determine the INVERSE MATRIX, which is:

!!

𝐴



1 −11 =− 8 2 7

5 −4 −3

3 5.5 −2 = −4 −1 −3.5

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−1.5 1 .5

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Or:

!!

𝐴

5.5 = −4 −3.5

−2.5 2 1.5

−1.5 1 .5

With everything defined to carry out our OPERATION, we can determine the SOLUTIONS to this set of SIMULTANEOUS LINEAR EQUATIONS but plugging in what we have up to this point, which gives us: 5.5 𝑋 = −4 −3.5

−2.5 2 1.5

−1.5 6 1 2 4 .5

Carrying out our standard MATRIX OPERATIONS, we get: 22 𝑋 = −16 −16 The SOLUTION to our set of SIMULTANEOUS LINEAR EQUATIONS is: 𝑥 = 22 𝑦 = −16 𝑧 = −16



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We can do a quick confirmation of these results by plugging them in to the original equations, such that: 22 + 2 −16 − −16 = 6 3(22) + 5 −16 − −16 = 2 −2 22 − −16 − 2 −16 = 4 The solutions check out. The correct answer choice is D. 𝒙 = 𝟐𝟐, 𝒚 = −𝟏𝟔, 𝐚𝐧𝐝 𝒛 = −𝟏𝟔

SOLUTION 2: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook. Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that:

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𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. In this problem, we will deploy our knowledge of MATRICES, solving it by hand, but illustrating how it is set up so you can HACK it using your CALCULATOR…which is always our go to approach. But for edification purposes, we will solidify an understanding of the process behind the scenes, so let’s get in to it.

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It’s possible to take any set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that: 𝐴𝑋 = 𝐵 Where each of these VARIABLES represent a UNIQUE MATRIX developed through the definition of each of our EQUATIONS, such that: • The matrix “A” is the COEFICIENT MATRIX of the system used to represent the coefficients of variables in an equation. These will be the numbers in front of the variables on the left side of the equal sign. • The matrix “X” is the MATRIX OF VARIABLES used to represent the variables in the equation, which are usually X, Y, and Z. These will be any variables on the left side of the equal sign. • The matrix “B” is the MATRIX OF CONSTANTS used to represent the constants that are on the other side of the equal signs for each equation. These are usually the numbers on the right side of the equal sign. In this problem, we are given: 𝐼! + 𝐼! + 𝐼! = 0 2𝐼! − 5𝐼! = 6 5𝐼! − 𝐼! = −3 Our first step would be to always ensure that the equations are in STANDARD FORM,

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meaning all the VARIABLES are on the left side and all the CONSTANTS are on the right side of the equation. We are good, lets move forward. Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵 We get: 1 2 0

1 −5 5

1 𝐼! 0 0 𝐼! = 6 −1 𝐼! −3

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables 1, 1, 1, 2, etc…all on the left side of the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are 𝐼! , 𝐼! , and 𝐼! . • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, which are 0, 6, and -3. Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we Made with by Prepineer | Prepineer.com

can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the MATRIX OF VARIABLES on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 It is important to remember that a matrix multiplied by the inverse of itself will produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B. The inverse of any matrix is defined as:

𝐴!! =

𝑎𝑑𝑗(𝐴) 𝐴

Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with

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their cofactors • |A| is the determinant of the matrix [A] The ADJOINT of a MATRIX is a complex term made extremely simple when viewing the GENERAL FORMULA expressed as: 𝑎𝑑𝑗(𝐴) = (𝐴! )! Where: • 𝐴! represents the COEFFICIENT MATRIX of MATRIX A • (𝐴! )! represents the TRANSPOSE of the COEFFICIENT MATRIX A To find the ELEMENTS of a COEFFICIENT MATRIX, you will go ELEMENT by ELEMENT of the ORIGINAL MATRIX, replacing the ELEMENT with a new ELEMENT representing the MATRIX OF MINORS. We will then take the MATRIX OF MINORS, and run through the ELEMENTS changing their sign per the standard TEMPLATE illustrated as: + − +

− + −

+ − +

Per this TEMPLATE, if the SECOND ELEMENT in the first ROW is a negative number, it will change to a positive…if the FIRST ELEMENT in the second ROW is a negative

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number, it will change to positive, and so forth. Let’s illustrate this with our current problem. We are given the MATRIX: 1 2 0

1 −5 5

1 0 −1

The COEFFICIENT MATRIX will be defined by first developing the MATRIX OF MINORS. We will do this by replacing the ELEMENT in the original MATRIX with the DETERMINANT formulated when considering that element. Let’s run through developing the first two ELEMENTS, first focusing on ELEMENT a11 of our MATRIX A: 1 2 0

1 −5 5

1 0 −1

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are:



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1 2 0

1 −5 5

1 0 −1

The DETERMINANT can be written as: −5 5

0 = −5 −1 − 0 5 −1

Which gives us a value: 𝑎!! = 5 Placing this in to position in our MATRIX OF MINORS, we get: 5

Moving to the next ELEMENT a12 of our MATRIX A: 1 2 0

1 −5 5

1 0 −1

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are:



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1 2 0

1 −5 5

1 0 −1

The DETERMINANT can be written as: 2 0

0 = 2 −1 − 0 0 −1

Which gives us a value: 𝑎!" = −2 Placing this in to position in our MATRIX OF MINORS, we get: 5

−2

And that’s the general process…hence, the reason we want you to HACK this work using your CALCULATOR. Filling out the remaining values in our MATRIX OF MINORS, we get: 5 −6 5

−2 −1 −2

10 5 −7

Now we need to apply the proper sign convention established to ultimately define THE MATRIX OF COFACTORS, which again is:

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+ − +

− + −

+ − +

Applying these sign conventions ELEMENT by ELEMENT, we find that our MATRIX OF COFACTORS is: 5 𝐴! = 6 5

2 −1 2

10 −5 −7

The ADJOINT MATRIX is the TRANSPOSE of the MATRIX OF COFACTORS, which is:

𝑎𝑑𝑗 𝐴 = 𝐴!

!

5 = 2 10

6 −1 −5

5 2 −7

Now we need to define the DETERMINANT of MATRIX A, which is: 𝐴 = 1 5 − 1 −2 + 1(10) Or: 𝐴 = 17



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This gives us what we need to determine the INVERSE MATRIX, which is:

!!

1 5 = 2 17 10

!!

. 294 = . 118 . 588

𝐴

6 −1 −5

5 2 −7

Or:

𝐴

. 353 −.059 −.294

. 294 . 118 −.412

With everything defined to carry out our OPERATION, we can determine the SOLUTIONS to this set of SIMULTANEOUS LINEAR EQUATIONS but plugging in what we have up to this point, which gives us: . 294 𝑋 = . 118 . 588

. 353 −.059 −.294

0 . 294 . 118 6 −.412 −3

Carrying out our standard MATRIX OPERATIONS, we get: 1.236 𝑥 = −.708 −.528 The SOLUTION to our set of SIMULTANEOUS LINEAR EQUATIONS is: 𝐼! = 1.236𝐴 𝐼! = −.708𝐴

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𝐼! = −.528𝐴 The correct answer choice is B. 𝑰𝑨 = 𝟏. 𝟐𝟑𝟔𝑨, 𝑰𝑩 = −. 𝟕𝟎𝟖𝑨, 𝐚𝐧𝐝 𝑰𝑪 = −. 𝟓𝟐𝟖𝑨

SOLUTION 3: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook. Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES Made with by Prepineer | Prepineer.com

We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. This is a word problem where we need to create our set of linear equations. Let’s revisit the problem statement. We want 10𝐿 of gas containing 2% additive and have drums of the following: • “𝑥” gasoline without additive • “𝑦” gasoline with 5% additive • “𝑧” gasoline with 6% additive We need to use 4 times as much pure gasoline as 5% additive gasoline. Determine how much of each is needed. Let us define the following variables: 𝑥 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑝𝑢𝑟𝑒 𝑔𝑎𝑠 𝑦 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 5% 𝑔𝑎𝑠 𝑧 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 6% 𝑔𝑎𝑠

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So the first sentence gives us the equation: 𝑥 + 𝑦 + 𝑧 = 10 The next information tells us that: • We get no additive from the first drum • We get 5% additive per liter from the second drum • We get 6% additive per liter from the third drum We want 2% total additive in 10𝐿 of gas so our second equation is: . 05𝑦 + .06𝑧 = .2 Multiplying through by 100 we get: 5𝑦 + 6𝑧 = 20 Finally, the last bit of information tells us that we need to use 4 times as much pure gasoline as 5% additive gasoline, so: 𝑥 = 4𝑦 Or 𝑥 − 4𝑦 = 0

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Now we have our set of linear equations and can run through the process of solving. The equations are: 𝑥 + 𝑦 + 𝑧 = 10 5𝑦 + 6𝑧 = 20 𝑥 − 4𝑦 = 0 Our first step would be to always ensure that the equations are in STANDARD FORM, meaning all the VARIABLES are on the left side and all the CONSTANTS are on the right side of the equation. We are good, lets move forward. Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵 We get: 1 0 1

1 5 −4

1 𝑥 10 6 𝑦 = 20 0 𝑧 0

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables 1, 1, 1, 0, etc…all on the left side of

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the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are 𝑥, 𝑦, and 𝑧. • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, which are 10, 20, and 0. Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the MATRIX OF VARIABLES on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 It is important to remember that a matrix multiplied by the inverse of itself will produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B.

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The inverse of any matrix is defined as:

𝐴!! =

𝑎𝑑𝑗(𝐴) 𝐴

Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with their cofactors • |A| is the determinant of the matrix [A] The ADJOINT of a MATRIX is a complex term made extremely simple when viewing the GENERAL FORMULA expressed as: 𝑎𝑑𝑗(𝐴) = (𝐴! )! Where: • 𝐴! represents the COEFFICIENT MATRIX of MATRIX A • (𝐴! )! represents the TRANSPOSE of the COEFFICIENT MATRIX A To find the ELEMENTS of a COEFFICIENT MATRIX, you will go ELEMENT by ELEMENT of the ORIGINAL MATRIX, replacing the ELEMENT with a new ELEMENT representing the MATRIX OF MINORS.



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We will then take the MATRIX OF MINORS, and run through the ELEMENTS changing their sign per the standard TEMPLATE illustrated as: + − +

− + −

+ − +

Per this TEMPLATE, if the SECOND ELEMENT in the first ROW is a negative number, it will change to a positive…if the FIRST ELEMENT in the second ROW is a negative number, it will change to positive, and so forth. Let’s illustrate this with our current problem. We are given the MATRIX: 1 0 1

1 5 −4

1 6 0

The COEFFICIENT MATRIX will be defined by first developing the MATRIX OF MINORS. We will do this by replacing the ELEMENT in the original MATRIX with the DETERMINANT formulated when considering that element.



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Let’s run through developing the first two ELEMENTS, first focusing on ELEMENT a11 of our MATRIX A: 1 0 1

1 5 −4

1 6 0

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are: 1 0 1

1 5 −4

1 6 0

The DETERMINANT can be written as: 5 −4

6 = 5 0 − 6 −4 0

Which gives us a value: 𝑎!! = 24 Placing this in to position in our MATRIX OF MINORS, we get: 24



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Moving to the next ELEMENT a12 of our MATRIX A: 1 0 1

1 5 −4

1 6 0

To find the MINOR COEFFICIENT that will replace this value, we will continue to ignore the remaining ELEMENTS in the ROW and COLUMN which our current ELEMENT resides, and define the DETERMINANT using the ELEMENTS remaining, which are: 1 0 1

1 5 −4

1 6 0

The DETERMINANT can be written as: 0 1

6 =0 0 − 6 1 0

Which gives us a value: 𝑎!" = −6 Placing this in to position in our MATRIX OF MINORS, we get: 24



−6

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And that’s the general process…hence, the reason we want you to HACK this work using your CALCULATOR. Filling out the remaining values in our MATRIX OF MINORS, we get: 24 4 −1

−6 −1 6

−5 −5 5

Now we need to apply the proper sign convention established to ultimately define THE MATRIX OF COFACTORS, which again is: + − +

− + −

+ − +

Applying these sign conventions ELEMENT by ELEMENT, we find that our MATRIX OF COFACTORS is: 24 𝐴! = −4 −1

6 −1 −6

−5 5 5

The ADJOINT MATRIX is the TRANSPOSE of the MATRIX OF COFACTORS, which is:

𝑎𝑑𝑗 𝐴 = 𝐴!



!

24 = 6 −5

−4 −1 5

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Now we need to define the DETERMINANT of MATRIX A, which is: 𝐴 = 1 24 − 1 −6 + 1(−5) Or: 𝐴 = 25 This gives us what we need to determine the INVERSE MATRIX, which is:

!!

1 24 = 6 25 −5

!!

. 96 = . 24 −.2

𝐴

−4 −1 5

−1 −6 5

Or:

𝐴

−.16 −.04 .2

−.04 −.24 .2

With everything defined to carry out our OPERATION, we can determine the SOLUTIONS to this set of SIMULTANEOUS LINEAR EQUATIONS but plugging in what we have up to this point, which gives us: . 96 𝑥 = . 24 −.2



−.16 −.04 .2

−.04 10 −.24 20 .2 0

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Carrying out our standard MATRIX OPERATIONS, we get: 6.4 𝑥 = 1.6 2 So if we want 10𝐿 of gas containing 2% additive and we need to use 4 times as much pure gasoline as 5% additive gasoline, then we need: •

6.4𝐿 of gasoline without additive

•

1.6𝐿 of gasoline with 5% additive

•

2𝐿 of gasoline with 6% additive

The correct answer choice is D. 𝒙 = 𝟔. 𝟒, 𝒚 = 𝟏. 𝟔, 𝐚𝐧𝐝 𝒛 = 𝟐

SOLUTION 4: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook.



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Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. In this problem, we will deploy our knowledge of MATRICES, solving it by hand, but illustrating how it is set up so you can HACK it using your CALCULATOR…which is always our go to approach.

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But for edification purposes, we will solidify an understanding of the process behind the scenes, so let’s get in to it. It’s possible to take any set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that: 𝐴𝑋 = 𝐵 Where each of these VARIABLES represent a UNIQUE MATRIX developed through the definition of each of our EQUATIONS, such that: • The matrix “A” is the COEFICIENT MATRIX of the system used to represent the coefficients of variables in an equation. These will be the numbers in front of the variables on the left side of the equal sign. • The matrix “X” is the MATRIX OF VARIABLES used to represent the variables in the equation, which are usually X, Y, and Z. These will be any variables on the left side of the equal sign. • The matrix “B” is the MATRIX OF CONSTANTS used to represent the constants that are on the other side of the equal signs for each equation. These are usually the numbers on the right side of the equal sign. In this problem, we are given: 3𝑥! + 7𝑥! = 2 2𝑥! + 6𝑥! = 4

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Our first step would be to always ensure that the equations are in STANDARD FORM, meaning all the VARIABLES are on the left side and all the CONSTANTS are on the right side of the equation. We are good, lets move forward. Laying this out in the MATRIX FORM: 𝐴𝑋 = 𝐵 We get: 3 2

7 𝑥! 2 = 𝑥 6 ! 4

Where: • The COEFICIENT MATRIX, A, would represent the coefficients of variables in the equations, which in this case, are the variables 3, 7, 2, and 6…all on the left side of the equation. • The MATRIX OF VARIABLES, X, would represent the variables in the equation, which are 𝑥! and 𝑥! . • The MATRIX OF CONSTANTS, B, would represent the constants on the right side of the equation, which are 2 and 4.



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Once we have written the SYSTEM OF LINEAR EQUATIONS in the form AX=B, we can now SOLVE for the MATRIX OF VARIABLES, X, by MULTIPLYING each side by the INVERSE OF MATRIX A. This will isolate the MATRIX OF VARIABLES on the left side, making the matrix operation: 𝐴!! 𝐴𝑋 = 𝐴!! 𝐵 It is important to remember that a matrix multiplied by the inverse of itself will produce the identity matrix, which gives us: 𝑋 = 𝐴!! 𝐵 This may be a simple definition problem on the exam, so remember that. Now that we know how to write an equation to solve for the MATRIX OF VARIABLES, we need to figure out how to calculate the INVERSE OF MATRIX A and then MULTIPLY it by MATRIX B. The inverse of any matrix is defined as:

𝐴!! =



𝑎𝑑𝑗(𝐴) 𝐴

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Where: • Adj(A) is the ADJOINT of MATRIX A obtaining by replacing 𝐴! elements with their cofactors • |A| is the determinant of the matrix [A] The ADJOINT of a MATRIX is a complex term made extremely simple when viewing the GENERAL FORMULA expressed as: 𝑎𝑑𝑗(𝐴) = (𝐴! )! Where: • 𝐴! represents the COEFFICIENT MATRIX of MATRIX A • (𝐴! )! represents the TRANSPOSE of the COEFFICIENT MATRIX A To find the ELEMENTS of a COEFFICIENT MATRIX, you will go ELEMENT by ELEMENT of the ORIGINAL MATRIX, replacing the ELEMENT with the new ELEMENT represented in the COEFFICIENT MATRIX. Let’s illustrate this with our current problem.



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We are given the MATRIX: 3 2

7 6

The ADJOINT of this MATRIX will then be is found by first swapping the leading diagonal of the 2 𝑥 2 matrix and switching the signs of the other two elements such that: 6 −2

−7 3

Now to define the DETERMINATE of “𝐴”, we proceed as we would with any COFACTOR, such that: 𝐴 = 6 3 − −7 −2 Or: 𝐴 =4 This gives us what we need to determine the INVERSE MATRIX, which is:

𝐴!! =



1 6 4 −2

−7 3

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Or:

𝐴!!

6 = 4 −2 4

−7 4 3 4

With everything defined to carry out our OPERATION, we can determine the SOLUTIONS to this set of SIMULTANEOUS LINEAR EQUATIONS but plugging in what we have up to this point, which gives us: 6 𝑋= 4 −2 4

−7 4 2 3 4 4

Carrying out our standard MATRIX OPERATIONS, we get:

𝑋=

−4 2

The SOLUTION to our set of SIMULTANEOUS LINEAR EQUATIONS is: 𝑥! = −4 𝑥! = 2

The correct answer choice is D.



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SOLUTION 5: The TOPIC of SIMULTANEOUS LINEAR EQUATIONS is not one that is directly referenced in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it’s a situation where knowing how to deal with SIMULTANEOUS LINEAR EQUATIONS can be a major contributor the success on any given problem. For this reason, we will break it down despite it being neglected from the NCEES Supplied Reference Handbook. Given any NUMBER of LINEAR EQUATIONS each having a unique set of UNKNOWN SHARED VARIABLES, such that 𝐴𝑥! + 𝐵𝑥! + 𝐶𝑥! + . . . +𝑍𝑥! = 𝑦! 𝐷𝑥! + 𝐸𝑥! + 𝐹𝑥! + . . . +𝑌𝑥! = 𝑦! 𝐺𝑥! + 𝐻𝑥! + 𝐼𝑥! + . . . +𝑋𝑥! = 𝑦! Where: • 𝑥! , 𝑥! , 𝑥! , … , 𝑥! are UNKNOWN SHARED VARIABLES • 𝐴, 𝐵, 𝐶, … , 𝑍 are KNOWN COEFFICIENTS and • 𝑦! , 𝑦! , 𝑎𝑛𝑑 𝑦! are KNOWN QUANTITIES We can solve this set of equations using one of a two approaches, either: 1. Solve them using the method of LINEAR ELIMINATION or LINEAR Made with by Prepineer | Prepineer.com

COMBINATION…this is a manual process done By hand, acceptable, but too time consuming for the more complex systems. 2. Solve using your NCEES APPROVED CALCULATOR…using MATRICES or the SYSTEM SOLVER, both a standard functionality of the calculator which will significantly decrease our time solving these problems. In all the problems up to this point, we moved forward developing and solving the set of MATRICES, as we would using our CALCULATOR. It’s possible to take any set of SIMULATRENOUS LINEAR EQUATIONS and place them in to a specific MATRIX FORMAT, such that: 𝐴𝑋 = 𝐵 In this problem, however, we will take a different approach, using LINEAR ELIMINATION. In this problem, we are given: 3𝑥 + 2𝑦 = 19 𝑥+𝑦 =8 Our first step would be to focus in on ELIMINATING one of the VARIABLES so that we have ONE EQUATION with ONE UNKNOWN. We will then work our way back up the CHAIN OF WORK, defining the other

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UNKNOWN VARIABLE. Let’s hone in on ELIMINATING the variable x. There are infinite routes to take here, but essentially, we want the two equations to have equal, yet opposite, COEFFICIENTS for the x variable. Our first equation as a COEFFICIENT of 3, therefore, let’s get our second equation to have that same COEFFICIENT, but opposite in sign. We can do this by multiplying our second equation by -3 and get: −3(𝑥 + 𝑦) = (−3)8 Or: −3𝑥 − 3𝑦 = −24 Now we can set these two EQUATIONS against one another, vertically, and simply ADD them, giving us: 3𝑥 + 2𝑦 = 19 −3𝑥 − 3𝑦 = −24 −3𝑥 − 𝑦 = −5 Which tells us that: 𝑦=5

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Taking this value and plugging it in to our original second equation, we get: 𝑥+5=8 Rearranging to isolate and solve for x, we get: 𝑥=3 The SOLUTION to our set of SIMULTANEOUS LINEAR EQUATIONS is: 𝑥=3 𝑦=5 The correct answer choice is D. 𝒙 = 𝟑 𝐚𝐧𝐝 𝒚 = 𝟓



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