0-5 Systems of Linear Equations and Inequalities
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1 21. Solve each system of equations. Eliminate one variable in two pairs of the system. Add the first equation and second equations to eliminate x. Multiply the first equation by 3 and add it to the second to eliminate x. Solve this system of two equations by multiplying the first by 1 and then adding the two equations together to eliminate the variable y. Substitute these two values into one of the original equations to find z. The solution is (7, 1, 1). esolutions Manual - Powered by Cognero Page 1
2 23. Eliminate one variable in two pairs of the system. Multiply the first equation by 2 and add it to the second equation to eliminate z. Multiply the first equation by 3 and add it to the third to eliminate z. Solve this system of two equations by multiplying the first by 4, the second by 7, and then adding the two equations together to eliminate the variable x. Substitute these two values into one of the original equations to find z. The solution is ( 6, 4, 7). esolutions Manual - Powered by Cognero Page 2
3 25. Eliminate the variable y in the system. Multiply the second equation by 2 and add it to the third equation to eliminate y. Solve the system of two equations formed by the above equation with the first equation in the original system. To do this, multiply the first equation by 3 and add it to the equation 3x 2z = 28 to eliminate the variable x. Substitute these two values into one of the original equations to find y. The solution is (10, 7, 1). esolutions Manual - Powered by Cognero Page 3
4 27. Eliminate one variable in two pairs of the system. Multiply the second equation by 2 and add it to the first equation to eliminate y. Multiply the second equation by 3 and add it to the third to eliminate y. Solve this system of two equations by adding the two equations together to eliminate the variable z. Substitute these two values into one of the original equations to find y. The solution is (5, 8, 9). esolutions Manual - Powered by Cognero Page 4
5 29. Solve each system of equations. State whether the system is consistent and independent, consistent and dependent, or inconsistent. Solve the first equation in the system for x. Substitute this value for x into the second equation and solve for y. Because 4 = 10 is not a true statement, this system has no solutions. Therefore, the system is inconsistent. 31. Eliminate the variable x in the system by multiplying the second equation by 3 and then adding the two equations. Because 0 = 0 is always true, there are an infinite number of solutions. Therefore, the system is consistent and dependent. esolutions Manual - Powered by Cognero Page 5
6 33. Eliminate the variable y in the system by multiplying the first equation by 5, the second equation by 3, and then adding the two equations. Substitute this value for x into one of the original equations and solve for y. The solution is. The system is consistent and independent because it has exactly one solution. esolutions Manual - Powered by Cognero Page 6
7 35. Eliminate one variable in two pairs of the system. Multiply the first equation by 1 and then add it to the second equation. Multiply the first equation by 3 and then add it to the third equation. Substitute the value for x into this equation and solve for y. Substitute these two values into one of the original equations to find z. The solution is (4, 8, 5). The system is consistent and independent because it has exactly one solution. esolutions Manual - Powered by Cognero Page 7
8 37. Solve each system of inequalities. If the system has no solution, state no solution. Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. >: above the line <: below the line y is greater that or equal to, so graph a solid line along and shade above the line. The solution of y x 3 is Regions 1 and 2. The solution of y 2x + 1 is Regions 2 and 3. Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system. 39. esolutions Manual - Powered by Cognero Page 8
9 Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. >: above the line <: below the line y is greater than or equal to, so graph a solid line along and shade above the line. The solution of is Regions 1 and 2. The solution of is Regions 2 and 3. Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system. esolutions Manual - Powered by Cognero Page 9
10 41. Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. >: above the line <: below the line y is less than, so graph a dashed line along y is greater than, so graph a dashed line along and shade above the line. The solution of is Regions 1 and 2. The solution of is Regions 2 and 3. Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system. esolutions Manual - Powered by Cognero Page 10
11 43. Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. >: above the line <: below the line y is greater than or equal to, so graph a solid line along and shade above the line.. The solution of is Region 1. The solution of is Region 2. Since these solutions do not contain points that are common to both inequalities, this system of inequalities has no solution. esolutions Manual - Powered by Cognero Page 11
12 45. Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. >: above the line <: below the line The solution of is Regions 1 and 2. The solution of is Regions 2 and 3. Region 2 contains points that are solutions to both inequalities, so this region is the solution to the system. 47. Rewrite each inequality in slope-intercept form. Then determine how to shade each graph. esolutions Manual - Powered by Cognero Page 12
13 >: above the line <: below the line y is greater than or equal to, so graph a solid line along and shade above the line. The solution of is Regions 1, 2, 3, and 4. The solution of is Regions 1, 4, 5, and 6. The solution of is Regions 1, 2, and 6. Region 1 contains points that are solutions to all three inequalities, so this region is the solution to the system. esolutions Manual - Powered by Cognero Page 13
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3. Evaluate the objective function at each vertex. Put the vertices into a table: Vertex P=3x+2y (0, 0) 0 min (0, 5) 10 (15, 0) 45 (12, 2) 40 Max
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MATH2210 Notebook 1 Fall Semester 2016/2017. 1 MATH2210 Notebook 1 3. 1.1 Solving Systems of Linear Equations... 3
MATH0 Notebook Fall Semester 06/07 prepared by Professor Jenny Baglivo c Copyright 009 07 by Jenny A. Baglivo. All Rights Reserved. Contents MATH0 Notebook 3. Solving Systems of Linear Equations........................
Dr Brian Beaudrie pg. 1
Multiplication of Decimals Name: Multiplication of a decimal by a whole number can be represented by the repeated addition model. For example, 3 0.14 means add 0.14 three times, regroup, and simplify,
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