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Systems of Linear Equations: Exercise Supplement

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Beginning with the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent, and dependent systems and examples to illustrate the applications for these systems. The substitution method and the addition method of solving a system by elimination are explained, noting when to use each method. The five-step method is again used to illustrate the solutions of value and rate problems (coin and mixture problems), using drawings that correspond to the actual situation. This module contains the exercise supplement for the chapter "Systems of Linear Equations".

Exercise Supplement

Solutions by Graphing ((Reference)) - Elimination by Addition ((Reference))

For the following problems, solve the systems of equations.

Exercise 1

{ 4x+y=5 2x+3y=13 { 4x+y=5 2x+3y=13

Solution

( 2,3 ) ( 2,3 )

Exercise 2

{ 5x+2y=5 x+7y=1 { 5x+2y=5 x+7y=1

Exercise 3

{ x3y=17 8x+2y=46 { x3y=17 8x+2y=46

Solution

( 86 13 , 45 13 ) ( 86 13 , 45 13 )

Exercise 4

{ 6m+5n=9 2m4n=14 { 6m+5n=9 2m4n=14

Exercise 5

{ 3x9y=5 x+3y=0 { 3x9y=5 x+3y=0

Solution

No solution.

Exercise 6

{ y=2x5 8x75=5 { y=2x5 8x75=5

Exercise 7

{ x=8 9y=5x76 { x=8 9y=5x76

Solution

( 8,4 ) ( 8,4 )

Exercise 8

{ 7x2y=4 14x+4y=8 { 7x2y=4 14x+4y=8

Exercise 9

{ y=x7 x=y5 { y=x7 x=y5

Solution

( 6,1 ) ( 6,1 )

Exercise 10

{ 20x+15y=13 5x20y=13 { 20x+15y=13 5x20y=13

Exercise 11

{ x6y=12 4x+6y=18 { x6y=12 4x+6y=18

Solution

( 6,1 ) ( 6,1 )

Exercise 12

{ 8x+9y=0 4x+3y=0 { 8x+9y=0 4x+3y=0

Exercise 13

{ 5x+2y=1 10x4y=2 { 5x+2y=1 10x4y=2

Solution

Dependent (same line)

Exercise 14

{ 2x5y=3 5x+2y=7 { 2x5y=3 5x+2y=7

Exercise 15

{ 6x+5y=14 4x8y=32 { 6x+5y=14 4x8y=32

Solution

( 4,2 ) ( 4,2 )

Exercise 16

{ 5x7y=4 10x14y=1 { 5x7y=4 10x14y=1

Exercise 17

{ 2m+10n=0 4m20n=6 { 2m+10n=0 4m20n=6

Solution

Inconsistent (parallel lines)

Exercise 18

{ 7r2s=6 3r+5s=15 { 7r2s=6 3r+5s=15

Exercise 19

{ 28a21b=19 21a+7b=15 { 28a21b=19 21a+7b=15

Solution

( 2 7 , 9 7 ) ( 2 7 , 9 7 )

Exercise 20

{ 72x108y=21 18x+36y=25 { 72x108y=21 18x+36y=25

Applications ((Reference))

Exercise 21

The sum of two numbers is 35. One number is 7 larger than the other. What are the numbers?

Solution

The numbers are 14 and 21.

Exercise 22

The difference of two numbers is 48. One number is three times larger than the other. What are the numbers?

Exercise 23

A 35 pound mixture of two types of cardboard sells for $30.15 $30.15 . Type I cardboard sells for 90¢ a pound and type II cardboard sells for 75¢ a pound. How many pounds of each type of cardboard were used?

Solution

26 pounds at 90¢;  9 pounds at 75¢ 26 pounds at 90¢;  9 pounds at 75¢

Exercise 24

The cost of 34 calculators of two different types is $1139. Type I calculator sells for $35 each and type II sells for $32 each. How many of each type of calculators were used?

Exercise 25

A chemistry student needs 46 ml of a 15% salt solution. She has two salt solutions, A and B, to mix together to form the needed 46 ml solution. Salt solution A is 12% salt and salt solution B is 20% salt. How much of each solution should be used?

Solution

28 3 4  ml of solution A;  17 3 4  ml of solution B. 28 3 4  ml of solution A;  17 3 4  ml of solution B.

Exercise 26

A chemist needs 100 ml of a 78% acid solution. He has two acid solutions to mix together to form the needed 100-ml solution. One solution is 50% acid and the other solution is 90% acid. How much of each solution should be used?

Exercise 27

One third the sum of two numbers is 12 and half the difference is 14. What are the numbers?

Solution

x=32,y=4 x=32,y=4

Exercise 28

Two angles are said to be complementary if their measures add to 90°. If one angle measures 8 more than four times the measure of its complement, find the measure of each of the angles.

Exercise 29

A chemist needs 4 liters of a 20% acid solution. She has two solutions to mix together to form the 20% solution. One solution is 30% acid and the other solution is 24% acid. Can the chemist form the needed 20% acid solution? If the chemist locates a 14% acid solution, how much would have to be mixed with the 24% acid solution to obtain the needed 20% solution?

Solution

(a) No solution
(b) 1.6 liters (1600 ml) of the 14% solution; 2.4 liters (2400 ml) of the 24% solution. (b) 1.6 liters (1600 ml) of the 14% solution; 2.4 liters (2400 ml) of the 24% solution.

Exercise 30

A chemist needs 80 ml of a 56% salt solution. She has a bottle of 60% salt solution. How much pure water and how much of the 60% salt solution should be mixed to dilute the 60% salt solution to a 56% salt solution?

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