A system of linear equations is a set of two or more linear equations.For example, a system of linear equations in two variables may be-2x - 3y = 13x - y = -7The goal is to find the values of the variables x and y which satisfy every equation in the system. This ordered pair is called the SOLUTION to the system of equations. When the solution issubstitued back into each equation in the system, the results are true statements.

For example, (-2,1) is a solution to the above system of equations because substituting x = -2 and y = 1, into -2x - 3y = 1gives the true statement -2(-2) - 3(1) = 1 1 = 1and substituting x = -2 and y = 1, into 3x - y = -7gives3(-2) - (1) = -7 -7 = -7Notice that (-5,3) is a solution to the first equation of the system, resulting in the true statement -2(-5) - 3(3) = 110 - 9 = 11 = 1However, if (-5,3) is substituted into the second equation, the result is:3(-5) - (3) = -7-15 - 3 = -7-18 = -7not a true statement. So (-5,3) is not a solution to the system.Maple can help us check solutions to equations. First store the appropriate values for x and y.x:=-2; y := 1;Then enter the equation to be evaluated.-2*x - 3*y;3*x - y;When finished be sure to use the "restart" command to clear the values stored for x and y.restart:NOTICE:When solving a system of linear equations, three possibilities may occur:1. Exactly one solution. (Independent System)2. No solution. (Dependent System)3. An infinite number of solutions. (Inconsistent System)

Check whether each of the following points are solutions to the system:x + y = 42x + 3y = 10a. (3,2)b. (1,3)c. (2,2)

We can represent a system of linear equations by graphing. The solution can bedetermined by locating the point or points of intersection of the equations in the system.For example, graph the following equations on the same coordinate plane:-2x - 3y = 13x - y = -7First put the equations into slope-intercept form:y = (-2/3)x - 1/3y = 3x+7Then use the "plot" command to graph them:plot([(-2/3)*x - 1/3, 3*x+7], x=-3..3, y=-3..3, color=[red,blue]);The solution to the system is the point of intersection, (-2,1).

Solve the following systems of linear equations graphically:a.3x + 2y = 12y - 2 = 1

(2,3)

b.y = -3x + 2y = -3x

No solution

c.x - y = -13x - 3y = -3

All real numbers

At times, using graphing to solve a system of linear equations can be imprecise. For example, consider the following system of equations:x - 2y = 52x + y = 3The equations in slope-intercept form are:y = (1/2)x - 5/2y = -2x + 3When graphed:plot([(1/2)*x - 5/2, -2*x+3], x=-3..3, y=-3..3, color=[red,blue]);The point of intersection can only be observed and therefore the solution must be estimated.The solution to this system is actually (11/5, -7/5).The following methods give us a more precise way of finding the solution to a system.

Recall that we often use the method of substitution for checking solutions. For example:y = 5x - 3x=2Substituting 2 for x in the first equation gives y = 5(2) - 3 = 10 - 3 = 7.So the solution is (2,7).This same method can be used for solving systems of linear equations.

Consider the following system:b = 3a - 15a = 2bSubtituting 2b in for a in the first equation gives:b = 3(2b) - 15 b = 6b - 15-5b = -15b = 3Then using "back subtitution" we get:a = 2(3) = 6The solution to the system is (6,3).Maple can help us to find the solution using substitution:First assign the value to be subsituted:a:=2*b;Next assign the equation for the substitution:eq := 3*a-15=b;Solve the equation:b:=solve(eq,b);Evaluate the other expression for the remaining variable:a:= a(b);Solution: (6,3)solution:=(a,b);restart:

Consider the following system:a.c + 2d = -283c + d = -9Solve one of the equations for a variable (say the first one):c = - 28 - 2dSubstitute (- 28 - 2y) for c i n the second equation and simplify:3(- 28 - 2d) + d = -9-84 - 6d + d = -9-84 - 5d = -9Solve for d:-5d = -9+84-5d = 75d = -15Using back subsitution, we get:c +2(-15) = -28c - 30 = -28c = 2The solution is (2,-15).c:=solve(c+ 2*d = -28,c):eq := 3*c + d = -9:d:=solve(eq,d):c:=c(d):solution:=(c,d):restart:

Solve the following systems using substitution:(Be sure to use the "restart" command to clear the values stored.)

4y = x + 45y = 2x + 11

(-8,-1)

y = -3x + 22y + 2x = 2 + y - x

all ordered pairs that satisfy y = -3x+2

3y = x + 6x - 3y = 9

No solution

Sometimes, the method of substitution can get to messy when we have systemsof equations in more than two variables.The method of elimination gives us a more consice way of solving. It allows us to "eliminate"one variable at a time. Once we have found the value of one variable, we can use backsubstitution to find the remaining variables.This method allows us to perform "elementary operations" to the equations in a system to help us eliminate variables. Any of the following elementary operations(or a combination of them) give an equivalent system: Interchange two equations in the system. Multiply an equation by a non-zero constant. Add two equations together.Recall that there are three possibilities for the results:a. One solutionb. No solutionc. Infinite solutionsUse the following steps to help solve a system of linear equations using the method of elimination: Put the equations into standard form (ax + by + cz + ... = real number constant, where a,b,c,...are also real numbers) If necessary, multiply one or more of the original equations by a real number constant so that when like terms are added, one of the variables "drops out." Add the equations in column form. Repeat the previous two steps until one variable remains. Solve for the remaining variable. Use back substitution on one of theresulting equations to find the value another variable. Repeat the back subsitution until all values for the variables have been found.

Solve the following systems using the method of elimination:x + 7y = 192y = x - 1First, the equations need to be written in standard form:x + 7y = 19-x + 2y = -1Notice that if we add the equations in column form, we can eliminate the x variable:x + 7y = 19+ -x + 2y = -1 9y = 18 Dividing by 9, we get:y = 2Use back substitution in either of the original equations to find the value of x:x + 7(2) = 19x + 14 = 19x = 5The solution (which is stated below) is x = 5, y = 2.To check the solution, substitute the values into the other original equation:2(2) = (5) - 14 = 5 - 14 = 4We get a true statement, as expected.

(5,2)

Solve the following systems using the method of elimination:5x + y = -243x - 2y = 9The equations are in standard form already.Suppose we want to eliminate the y variable. We need to multiply the firstequation by 2, so that when we add the equations together, the y's "drop out."2( 5x + y = -24)10x + 2y = -48So now our system looks like:10x + 2y = -483x - 2y = 9Adding the equations in column form, we get:13x = -39Dividing both sides by 13:x = -3Either of the original equations will work for back substitution. Suppose we try the first one:5(-3) + y = -24-15 + y = -24Adding 15 to both sides:y = -9The solution (as stated below) is x = -3, y = -9.To check, substitute the values into the second original equation:3(-3) - 2(-9) = 9-9 + 18 = 99 = 9We get a true statement.

(-3,-9)

Solve the following systems using the method of elimination:3x - 5y = 664x + 3y = 1The equations are already in standard form.Choose a variable to eliminate, x or y (see below):

Notice that if we mulitply the first equation by -4, and the second equation by 3,the results for the x terms will be -12x and 12x, respectively. When added together, the x variable will "drop out."-4( 3x - 5y = 66)3( 4x + 3y = 1)-12x + 20y = -26412x + 9y = 329y = -261Dividing by 29, we get:y = -9Using the first original equation for back subsitution:3x - 5(-9) = 663x + 45 = 66Subtracting 45 from both sides:3x = 21Dividing both sides by 3:x = 7The solution (as stated below) is x = 7, y = -9.Checking the solution in the second original equation:4(7) + 3(-9) = 128 - 27 = 11 = 1We get a true statement:

Notice that if we mulitply the first equation by 3, and the second equation by 5,the results for the y terms will be -15y and 15y, respectively. When added together, the y variable will "drop out."3( 3x - 5y = 66)5( 4x + 3y = 1)9x - 15y = 19820x + 15y = 5Adding the equations in column form:29x = 203Dividing both sides by 29:x = 7Using back subsitution on the first original equation:3(7) - 5y = 6621 - 5y = 66Subtracting 21 from both sides:-5y = 45Dividing both sides by -5 we get:y = -9The solution (as stated below) is x = 7, y = -9.Checking the solution in the second original equation:4(7) + 3(-9) = 128 - 27 = 11 = 1We get a true statement:

(7,-9)

Solve the following systems using the method of elimination:2x + 3y + z = 7x - 2y + z = -4-x + 3y - z = 6Notice how we can add the second and third equations to eliminate thex and z variables at the same time:x - 2y + z = -4-x + 3y - z = 6The resulting equation is:y = 2Now we need to use two of the original equations to eliminate the x or z variable,so that we can we can use back subsitution to find another value.Consider the first and third equations:2x + 3y + z = 7-x + 3y - z = 6The z variable can be eliminated simply by adding the two equations in column form:x + 6y = 13Using back subtitution, we have:x + 6(2) = 13x + 12 = 13Subtracting 12 from both sides, we get:x = 1We can now use back subsitution with x and y on any of the original equations to find z:Consider the second original equation:(1) - 2(2) + z = -41 - 4 + z = -4-3 + z = -4Adding 3 to both sides:z = -1The solution to the system (as stated below) is x = 1, y = 2, z = -1.Checking the solution in the first and third original equations:2x + 3y + z = 72(1) + 3(2) + (-1) = 72 + 6 - 1 = 77 =7and-x + 3y - z = 6-(1) + 3(2) - (-1) = 6-1 + 6 + 1 = 66 = 6We get a true statement for each.

(1,2,-1)

Solve each system of equations by using the method of elimination:a.5x + 40y = 775x + 15y = 17

(-19/5,12/5)

b.x - y + z = 14x + y + z = 8x + 2y - 5z = -37

(4,-3,7)

c.5x - 3y + 7z = 8x + 4y - 3z = 116x + y + 4z = 19

All real numbers

d.8a - 3b + 2c = -295a + b + 7c = 23a - 4b - 5c = 30

No solution

At times, the method of elimination is not the most efficient way to solve a system of linear equations. (Especially if we have more than three variables and/or more than three equations.) The following method allows us to more efficiently solve these types of systems.

Consider the following system of equations in standard form:3x +2y + 6z = 182x - 2y + 5z = 7x + 0y + 5z = 20We can abbreviate this system by removing the variables and creating a rectangular array ofnumbers, called an "augmented matrix." with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

A:=matrix(3,4,[3,2,6,18,2,-2,5,7,1,0,5,20]):

The rows of an augmented matrix can be treated just like the equations of the system. Just as we can manipulateequations, we can perform "row operations" on an augmented matrix to obtain an equivalent system.The row operations are as follows:1. Interchange any two rows. Consider interchanging row 1 and row 3.A1:=swaprow(A,1,3):2. Replace a row by the sum of that row and a nonzero mulitple of another row. Consider multiplying row 1 by -2 and adding it to row 2.A2:=addrow(A1,1,2,-2): Consider multiplying row 1 by -3 and adding it to row 3.A3:=addrow(A2,1,3,-3):3. Multiply a row by a non-zero consant. Consider mutiplying row 2 by (-1/2).A4:=mulrow(A3,2,-1/2);

A matrix is said to be in row echelon form when: Rows that have all 0s are at the bottom. The first entry in each row that is not a 0 is a 1 (called a leading 1). As the matrix is viewed from top to bottom, each leading 1 appears to the right of the leading 1 in any previous row.The following matrices are in row echelon form:B:=matrix(3,3,[1,2,3,0,1,4,0,0,1]):c:=matrix(4,4,[1,6,0,-3,0,1,1,-6,0,0,1,1/2,0,0,0,0]): The following matrices are NOT in row echelon form:E:=matrix(3,3,[2,1,4,-5,6,-7,0,1,3]):F:=matrix(4,4,[1,2,3,4,0,0,0,0,5,7,1,6,-1,4,-2,0]):

Recall that each row of the matrix represents an equation of the system.Consider the following matrix in row echelon form:G:=matrix(3,4,[1,-3,1.5,5,0,1,3.125,.25,0,0,1,2]):If we rewrite each row by replacing the variables, we have:1x - 3y + 1.5z = 51y + 3.125z = .25z = 2Since z = 2, use back substitution on the second equation to solve for y:1y + 3.125(2) = .25y + 6.25 = .25Subtracting 6.25 from both sides:y = 6Use back substitution once more on the first equation to find x:1x - 3(6) + 1.5(2) = 5x - 18 + 3 = 5x - 15 = 5Adding 15 to both sides:x = 20The solution is x = 20, y = 6, z = 2.

A matrix is said to be in reduced row echelon form when: The matrix is in echelon form Every column containing a leading 1 has 0s everywhere else.For example, the following matrices are in reduced row echelon form:H:=matrix(3,3,[1,0,0,0,1,0,0,0,1]):J:=matrix(3,4,[1,0,0,2,0,1,0,5,0,0,1,-3]):

Consider the above matrix in reduced row echelon form:J:=matrix(3,4,[1,0,0,2,0,1,0,5,0,0,1,-3]);

NiM+JSJKRy0lJ21hdHJpeEc2IzclNyYiIiIiIiFGKyIiIzcmRitGKkYrIiImNyZGK0YrRiohIiQ=

Recalling that each row is a linear equation, we can solve this system by replacingthe variables and observing the solution.x =2y = 5z = -3

We can use elementary row operations to change a matrix into row echelon form,or more preferrably, reduced row echelon form, to solve the system. Put the equations of the system into standard form. Rewrite the system into an equivalent augmented matrix by eliminating the variables. Use elementary row operations to transform the matrix into row echelon form or reduced row echelon form. (Hint: work column by column to get leading 1s and 0s underneath.) Replace the variables and solve the resulting equations if necessary.

So far, we have seen only one type of result from solving a system of linear equations,and that is "exactly one solution."Recall, there are three different types of answers:1. Exactly one solution.2. An infinite number of solutions.3. No solution.

with(linalg):F:=matrix(2,3,[1,-4,1,0,0,0]):Notice the bottom row reads 0 = 0. A true statement.Use the first row to solve for one of the variables:x - 4y = 1Solving for x:x = 4y + 1Then we express the solution as "all ordered pairs (x,y) that satisfy x = 4y + 1."ORx - 4y = 1Solving for y:y = (1/4)x - 1/4The solution would be "all ordered pairs (x,y) that satisfy y = (1/4)x - 1/4."Note that there are infinitely many solutions.

with(linalg):F:=matrix(3,5,[1,1,2,3,1,0,1,3,2,1,0,0,0,0,-3]);

NiM+JSJGRy0lJ21hdHJpeEc2IzclNyciIiJGKiIiIyIiJEYqNyciIiFGKkYsRitGKjcnRi5GLkYuRi4hIiQ=

The last row reads 0 = -3. Obviously a false statement...meaning "no solution."

Use Maple to write the augmented matrix of each of the following systems.

Do not solve the systems.a.x + 2y + 3z = 83x - y + 2z = 5-2x - 4y - 6z = 5with(linalg):A:=matrix(3,4,[1,2,3,8,3,-1,2,5,-2,-4,-6,5]);b.3x + y - z = 02x - y + 3z = -7B:=matrix(2,4,[3,1,-1,0,2,-1,3,-7]);c.x + y = 35x - y = 39x - 4y = 1C:=matrix(3,3,[1,1,3,5,-1,3,9,-4,1]);d.x + 2y = 0y - z = 2x + y + z = -2E:=matrix(3,4,[1,2,0,0,0,1,-1,2,1,1,1,-2]);

Solve:2x + y + z = 9-x - y + z = 13x - y + z = 9with(linalg):A:=matrix(3,4,[2,1,1,9,-1,-1,1,1,3,-1,1,9]):A1:=swaprow(A,2,1):A2:=mulrow(A1,1,-1):A3:=addrow(A2,1,2,-2):A4:=addrow(A3,1,3,-3):A5:=mulrow(A4,2,-1):A6:=addrow(A5,2,3,4):A7:=addrow(A6,2,1,-1):A8:=mulrow(A7,3,-1/8):The matrix is now in row echelon form. You may wish to replace the variablesand solve from here...or continue to reduced row echelon form, as follows:A9:=addrow(A8,3,2,3):A10:=addrow(A9,3,1,-2):The matrix is now in reduced row echelon form. The solution is:x = 2, y = 1, z = 4or (2,1,4).

Solve:x - 3y + 4z = 34x + y - z = 83x + 4y - 5z = 5B:=matrix(3,4,[1,-3,4,3,4,1,-1,8,3,4,-5,5]):B1:=addrow(B,1,2,-4):B2:=addrow(B1,1,3,-3):B3:=mulrow(B2,2,1/13):B4:=addrow(B3,2,1,3):B5:=addrow(B4,2,3,-13):Stop right here...we can see that the last row says 0 = 0..."an infinite number of solutions."OR "all ordered triples (x,y,z) that satisfy x + (1/13)y = 27/13 and y - (17/13)z = -4/13.Usually, when we have more than two variables, we simply say "an infinite number of solutions."

Solve:3x - 2y - 8z = 19x - 6y - 24z = -2x - y + z = 1C:=matrix(3,4,[3,-2,-8,1,9,-6,-24,-2,1,-1,1,1]):Use row operations to transform the matrix into reduced row echelon form:C1:=swaprow(C,1,3):C2:=addrow(C1,1,2,-9):C3:=addrow(C2,1,3,-3):C4:=swaprow(C3,2,3):C5:=addrow(C4,2,3,-3):Stop here...right away we can see that the last row says 0 = -5. No solution

Solve2x + y + z = 9-x - y + z = 13x - y + z = 9A:=matrix(3,4,[2,1,1,9,-1,-1,1,1,3,-1,1,9]):The "rref" command returns the matrix in reduced row echelon form.linalg[rref](A):Solution: (2,1,4).

Solvex - 3y + 4z = 34x + y - z = 83x + 4y - 5z = 5B:=matrix(3,4,[1,-3,4,3,4,1,-1,8,3,4,-5,5]):linalg[rref](B);

NiMtSSdtYXRyaXhHNiRJKnByb3RlY3RlZEdGJkkoX3N5c2xpYkc2IjYjNyU3JiIiIiIiISNGLCIjOCMiI0ZGLzcmRi1GLCMhIzxGLyMhIiVGLzcmRi1GLUYtRi0=

Solution: An infinite number of solutions.

Solve3x - 2y - 8z = 19x - 6y - 24z = -2x - y + z = 1C:=matrix(3,4,[3,-2,-8,1,9,-6,-24,-2,1,-1,1,1]):linalg[rref](C):Solution: No solution

Solvex + 2y = 65x - 11y = -54

(-2,4)

Solve3x + 6y = 12x + 2y = -5

No solution

Solve2x + 9y = 165y = 8 - x

(8,0)

Solvex + 2y + z = -1x - 2y - 2z = 5x + 6y + z = 2

(1/2,3/4,-3)

Solve3x - 2y + z = -4x - y - z = 32x - y + 2z = -7

An infinite number of solutions.