Chapter 11. Systems of Linear Equations
11.1. Objectives*
After completing this chapter, you should
Solutions by Graphing (Section 11.2)
be able to recognize a system of equations and a solution to it
be able to graphically interpret independent, inconsistent, and dependent systems
be able to solve a system of linear equations graphically
Elimination by Substitution (Section 11.3)
know when the substitution method works best
be able to use the substitution method to solve a system of linear equations
know what to expect when using substitution with a system that consists of parallel lines or coincident lines
Elimination by Addition (Section 11.4)
know the properties used in the addition method
be able to use the addition method to solve a system of linear equations
know what to expect when using the addition method with a system that consists of parallel or coincident lines
Applications (Section 11.5)
become more familiar with the five-step method for solving applied problems
be able to solve number problems
be able to solve value and rate problems
11.2. Solutions by Graphing*
Overview
Systems of Equations
Solution to A System of Equations
Graphs of Systems of Equations
Independent, Inconsistent, and Dependent Systems
The Method of Solving A System Graphically
Systems of Equations
A collection of two linear equations in two variables is called a system of linear equations in two variables, or more briefly, a system of equations. The pair of equations 
is a system of equations. The brace { is used to denote that the two equations occur together (simultaneously).
Solution to A System of Equations
We know that one of the infinitely many solutions to one linear equation in two variables is an ordered pair. An ordered pair that is a solution to both of the equations in a system is called a solution to the system of equations. For example, the ordered pair ( 3, 5 ) is a solution to the system 
since ( 3, 5 ) is a solution to both equations. 
Graphs of Systems of Equations
One method of solving a system of equations is by graphing. We know that the graph of a linear equation in two variables is a straight line. The graph of a system will consist of two straight lines. When two straight lines are graphed, one of three possibilities may result.
Example 11.1.
The lines intersect at the point ( a, b ) . The point ( a, b ) is the solution to the corresponding system. 
Example 11.2.
The lines are parallel. They do not intersect. The system has no solution. 
Example 11.3.
The lines are coincident (one on the other). They intersect at infinitely many points. The system has infinitely many solutions. 
Independent, Inconsistent, and Dependent Systems
Systems in which the lines intersect at precisely one point are called independent systems. In applications, independent systems can arise when the collected data are accurate and complete. For example, The sum of two numbers is 10 and the product of the two numbers is 21. Find the numbers. In this application, the data are accurate and complete. The solution is 7 and 3.
Systems in which the lines are parallel are called inconsistent systems. In applications, inconsistent systems can arise when the collected data are contradictory. For example, The sum of two even numbers is 30 and the difference of the same two numbers is 0. Find the numbers. The data are contradictory. There is no solution to this application.
Systems in which the lines are coincident are called dependent systems. In applications, dependent systems can arise when the collected data are incomplete. For example. The difference of two numbers is 9 and twice one number is 18 more than twice the other. The data are incomplete. There are infinitely many solutions.
The Method of Solving A System Graphically
To solve a system of equations graphically: Graph both equations.
If the lines intersect, the solution is the ordered pair that corresponds to the point of intersection. The system is independent.
If the lines are parallel, there is no solution. The system is inconsistent.
If the lines are coincident, there are infinitely many solutions. The system is dependent.
Sample Set A
Solve each of the following systems by graphing.
Example 11.4.
Write each equation in slope-intercept form. 
Graph each of these equations. 
The lines appear to intersect at the point ( − 1, 3 ) . The solution to this system is ( − 1, 3 ) , or 
Check: Substitute
x = − 1, y = 3 into each equation. 
Example 11.5.
Write each equation in slope-intercept form. 
Graph each of these equations. 
These lines are parallel. This system has no solution. We denote this fact by writing inconsistent. We are sure that these lines are parallel because we notice that they have the same slope,
m = 1 for both lines. The lines are not coincident because the
y
-intercepts are different.
Example 11.6.
Write each equation in slope-intercept form. 
Both equations are the same. This system has infinitely many solutions. We write dependent.
Practice Set A
Solve each of the following systems by graphing. Write the ordered pair solution or state that the system is inconsistent, or dependent.
Exercises
For the following problems, solve the systems by graphing. Write the ordered pair solution, or state that the system is inconsistent or dependent.
Exercise 11.2.6.
Exercise 11.2.8.
Exercise 11.2.10.
Exercise 11.2.12.
Exercise 11.2.14.
Exercise 11.2.16.
Exercises For Review
Exercise 11.2.19. (Go to Solution)
(Section 7.5) Supply the missing word. The __________ of a line is a measure of the steepness of the line.
Exercise 11.2.20.
(Section 10.2) Supply the missing word. An equation of the form a x 2 + b x + c = 0,a ≠ 0 , is called a __________ equation.
Exercise 11.2.21. (Go to Solution)
(Section 10.8) Construct the graph of the quadratic equation
y = x
2
− 3.
Solutions to Exercises
Solution to Exercise 11.2.11. (Return to Exercise)
These coordinates are hard to estimate. This problem illustrates that the graphical method is not always the most accurate.
(
− 6,6
)
11.3. Elimination by Substitution *
Overview
When Substitution Works Best
The Substitution Method
Substitution and Parallel Lines
Substitution and Coincident Lines
When Substitution Works Best
We know how to solve a linear equation in one variable. We shall now study a method for solving a system of two linear equations in two variables by transforming the two equations in two variables into one equation in one variable.To make this transformation, we need to eliminate one equation and one variable. We can make this elimination by substitution.
The substitution method works best when either of these conditions exists:
One of the variables has a coefficient of 1, or
One of the variables can be made to have a coefficient of 1 without introducing fractions.
The Substitution Method
To solve a system of two linear equations in two variables,
Solve one of the equations for one of the variables.
Substitute the expression for the variable chosen in step 1 into the other equation.
Solve the resulting equation in one variable.
Substitute the value obtained in step 3 into the equation obtained in step 1 and solve to obtain the value of the other variable.
Check the solution in both equations.
Write the solution as an ordered pair.
Sample Set A
Example 11.7.
Solve the system 
Step 1: Since the coefficient of
y
in equation 2 is 1, we will solve equation 2 for
y
.
y = − 3x + 7 Step 2: Substitute the expression − 3x + 7 for
y
in equation 1. 2x + 3( − 3x + 7 ) = 14 Step 3: Solve the equation obtained in step 2. 
Step 4: Substitute
x = 1 into the equation obtained in step 1, y = − 3x + 7.
We now have
x = 1 and
y = 4. Step 5: Substitute
x = 1,y = 4 into each of the original equations for a check.
Step 6: The solution is ( 1,4 ). The point ( 1,4 ) is the point of intersection of the two lines of the system.
Substitution And Parallel Lines
The following rule alerts us to the fact that the two lines of a system are parallel.
If computations eliminate all the variables and produce a contradiction, the two lines of a system are parallel, and the system is called inconsistent.
Sample Set B
Example 11.8.
Solve the system 
Step 1: Solve equation 1 for
y.
Step 2: Substitute the expression 2x − 1 for
y
into equation 2. 4x − 2( 2x − 1 ) = 4 Step 3: Solve the equation obtained in step 2. 
Computations have eliminated all the variables and produce a contradiction. These lines are parallel. 
This system is inconsistent.
Substitution And Coincident Lines
The following rule alerts us to the fact that the two lines of a system are coincident.
If computations eliminate all the variables and produce an identity, the two lines of a system are coincident and the system is called dependent.
Sample Set C
Example 11.9.
Solve the system 
Step 1: Divide equation 1 by 4 and solve for
x.
Step 2: Substitute the expression − 2y + 2 for
x
in equation 2. 3( − 2y + 2 ) + 6y = 6 Step 3: Solve the equation obtained in step 2. 
Computations have eliminated all the variables and produced an identity. These lines are coincident. 
This system is dependent.
Systems in which a coefficient of one of the variables is not 1 or cannot be made to be 1 without introducing fractions are not well suited for the substitution method. The problem in Sample Set D illustrates this “messy” situation.
Sample Set D
Example 11.10.
Solve the system 
Step 1: We will solve equation ( 1 ) for
y.
Step 2: Substitute the expression 
for
y
in equation ( 2 ).
Step 3: Solve the equation obtained in step 2. 
Step 4: Substitute 
into the equation obtained in step 
We now have 
and 
Step 5: Substitution will show that these values of
x
and
y
check.Step 6: The solution is 
Exercises
For the following problems, solve the systems by substitution.
Exercise 11.3.6.
Exercise 11.3.8.
Exercise 11.3.10.
Exercise 11.3.12.
Exercise 11.3.14.
Exercise 11.3.16.
Exercise 11.3.18.
Exercise 11.3.20.
Exercise 11.3.22.
Exercise 11.3.24.
Exercise 11.3.26.
Exercise 11.3.28.
Exercise 11.3.30.
Exercises For Review
Solutions to Exercises
Solution to Exercise 11.3.1. (Return to Exercise)
The point ( 2, − 1 ) is the point of intersection of the two lines.
Solution to Exercise 11.3.2. (Return to Exercise)
Substitution produces 4 ≠ 1, or 
, a contradiction. These lines are parallel and the system is inconsistent.
Solution to Exercise 11.3.3. (Return to Exercise)
Computations produce − 2 = − 2, an identity. These lines are coincident and the system is dependent.
11.4. Elimination by Addition*
Overview
The Properties Used in the Addition Method
The Addition Method
Addition and Parallel or Coincident Lines
The Properties Used in the Addition Method
Another method of solving a system of two linear equations in two variables is called the method of elimination by addition. It is similar to the method of elimination by substitution in that the process eliminates one equation and one variable. The method of elimination by addition makes use of the following two properties.
If A , B , and C are algebraic expressions such that
a x + ( − a x ) = 0
Property 1 states that if we add the left sides of two equations together and the right sides of the same two equations together, the resulting sums will be equal. We call this adding equations. Property 2 states that the sum of two opposites is zero.
The Addition Method
To solve a system of two linear equations in two variables by addition,
Write, if necessary, both equations in general form, a x + b y = c.
If necessary, multiply one or both equations by factors that will produce opposite coefficients for one of the variables.
Add the equations to eliminate one equation and one variable.
Solve the equation obtained in step 3.
Do one of the following: (a) Substitute the value obtained in step 4 into either of the original equations and solve to obtain the value of the other variable, or (b) Repeat steps 1-5 for the other variable.
Check the solutions in both equations.
Write the solution as an ordered pair.
The addition method works well when the coefficient of one of the variables is 1 or a number other than 1.
Sample Set A
Example 11.11.
Solve 
Step 1: Both equations appear in the proper form. Step 2: The coefficients of
y
are already opposites, 1 and
− 1,
so there is no need for a multiplication.
Step 3: Add the equations. 
Step 4: Solve the equation
4x = 16.
4x = 16
x = 4
The problem is not solved yet; we still need the value of
y
. Step 5: Substitute
x = 4
into either of the original equations. We will use equation 1. 
We now have
x = 4, y = 2.
Step 6: Substitute
x = 4
and
y = 2
into both the original equations for a check. 
Step 7: The solution is
(
4,2
).
The two lines of this system intersect at
(
4,2
).
Practice Set A
Solve each system by addition.
Sample Set B
Solve the following systems using the addition method.
Example 11.12.
Solve 
Step 1: The equations are already in the proper form,
a
x + b
y = c.
Step 2: If we multiply equation (2) by —3, the coefficients of
a
will be opposites and become 0 upon addition, thus eliminating
a
. 
Step 3: Add the equations. 
Step 4: Solve the equation
− 11b = 44.
− 11b = 44
b = − 4
Step 5: Substitute
b = − 4
into either of the original equations. We will use equation 2. 
We now have
a = − 1
and
b = − 4.
Step 6: Substitute
a = − 1
and
b = − 4
into both the original equations for a check. 
Step 7: The solution is
(
− 1, − 4
).
Example 11.13.
Solve 
Step 1: Rewrite the system in the proper form. 
Step 2: Since the coefficients of
y
already have opposite signs, we will eliminate
y
. Multiply equation (1) by 5, the coefficient of
y
in equation 2. Multiply equation (2) by 2, the coefficient of
y
in equation 1. 
Step 3: Add the equations. 
Step 4: Solve the equation 23x = 0
23x = 0
x = 0 Step 5: Substitute
x = 0 into either of the original equations. We will use equation 1. 
We now have
x = 0 and
y = − 2. Step 6: Substitution will show that these values check. Step 7: The solution is ( 0, − 2 ).
Practice Set B
Solve each of the following systems using the addition method.
Addition And Parallel Or Coincident Lines
When the lines of a system are parallel or coincident, the method of elimination produces results identical to that of the method of elimination by substitution.
If computations eliminate all variables and produce a contradiction, the two lines of the system are parallel and the system is called inconsistent.
If computations eliminate all variables and produce an identity, the two lines of the system are coincident and the system is called dependent.
Sample Set C
Example 11.14.
Solve 
Step 1: The equations are in the proper form. Step 2: We can eliminate
x
by multiplying equation (1) by –2. 
Step 3: Add the equations. 
This is false and is therefore a contradiction. The lines of this system are parallel. This system is inconsistent.
Example 11.15.
Solve 
Step 1: The equations are in the proper form. Step 2: We can eliminate
x
by multiplying equation (1) by –3 and equation (2) by 4. 
Step 3: Add the equations. 
This is true and is an identity. The lines of this system are coincident. This system is dependent.
Practice Set C
Solve each of the following systems using the addition method.
Exercises
For the following problems, solve the systems using elimination by addition.
Exercise 11.4.11.
Exercise 11.4.13.
Exercise 11.4.15.
Exercise 11.4.17.
Exercise 11.4.19.
Exercise 11.4.21.
Exercise 11.4.23.
Exercise 11.4.25.
Exercise 11.4.27.
Exercise 11.4.29.
Exercise 11.4.31.
Exercise 11.4.33.
Exercise 11.4.35.
Exercise 11.4.37.
Exercise 11.4.39.
Exercise 11.4.41.
Exercises For Review
Solutions to Exercises
11.5. Applications*
Overview
The Five-Step Method
Number Problems
Value and Rate Problems: Coin Problems Problems and Mixture Problems
The Five-Step Method
When solving practical problems, it is often more convenient to introduce two variables rather than only one. Two variables should be introduced only when two relationships can be found within the problem. Each relationship will produce an equation, and a system of two equations in two variables will result.We will use the five-step method to solve these problems.
Introduce two variables, one for each unknown quantity.
Look for two relationships within the problem. Translate the verbal phrases into mathematical expressions to form two equations.
Solve the resulting system of equations.
Check the solution.
Write a conclusion.
Sample Set A (Number Problems)
Example 11.16.
The sum of two numbers is 37. One number is 5 larger than the other. What are the numbers?
Step 2: There are two relationships. (a) The Sum is 37.
x + y = 37 (b) One is 5 larger than the other.
y = x + 5
Step 3: 
We can easily solve this system by substitution. Substitute
x + 5 for
y
in equation 1. 
Step 4: The Sum is 37. 
One is 5 larger than the other. 
Step 5: The two numbers are 16 and 21.
Practice Set A
The difference of two numbers is 9, and the sum of the same two numbers is 19. What are the two numbers?
The problems in Sample Sets B and C are value problems. They are referred to as value problems because one of the equations of the system used in solving them is generated by considering a value, or rate, or amount times a quantity.
Sample Set B (Coin Problems)
Example 11.17.
A parking meter contains 27 coins consisting only of dimes and quarters. If the meter contains $4.35, how many of each type of coin is there?
Step 2: There are two relationships. (a) There are 27 coins.
D + Q = 27. (b) Contribution due to dimes = 
Contribution due to quarters = 
Step 3: 
We can solve this system using elimination by addition. Multiply both sides of equation ( 1 ) by − 10 and add. 
Step 4: 16 dimes and 11 quarters is 27 coins. 
The solution checks.Step 5: There are 11 quarters and 16 dimes.
Practice Set B
Exercise 11.5.2. (Go to Solution)
A bag contains only nickels and dimes. The value of the collection is $2. If there are 26 coins in all, how many of each coin are there?
Sample Set C (Mixture Problems)
Example 11.18.
A chemistry student needs 40 milliliters (ml) of a 14% acid solution. She had two acid solutions, A and B, to mix together to form the 40 ml acid solution. Acid solution A is 10% acid and acid solution B is 20% acid. How much of each solution should be used?
Step 2: There are two relationships. (a) The sum of the number of ml of the two solutions is 40.
x + y = 40 (b) To determine the second equation, draw a picture of the situation. 
The equation follows directly from the drawing if we use the idea of amount times quantity.
Solve this system by addition. First, eliminate decimals in equation 2 by multiplying both sides by 100. 
Eliminate
x
by multiplying equation 1 by − 10 and then adding. 
Step 4: 24 ml and 16 ml to add to 40 ml. 
The solution checks.Step 5: The student should use 24 ml of acid solution A and 16 ml of acid solution B.
Practice Set C
Exercise 11.5.3. (Go to Solution)
A chemistry student needs 60 ml of a 26% salt solution. He has two salt solutions, A and B, to mix together to form the 60 ml solution. Salt solution A is 30% salt and salt solution B is 20% salt. How much of each solution should be used?
Exercises
Exercise 11.5.4. (Go to Solution)
The sum of two numbers is 22. One number is 6 more than the other. What are the numbers?
Exercise 11.5.5.
The sum of two numbers is 32. One number is 8 more than the other. What are the numbers?
Exercise 11.5.6. (Go to Solution)
The difference of two numbers is 12 and one number is three times as large as the other. Whatare the numbers?
Exercise 11.5.7.
The difference of two numbers is 9 and one number is 10 times larger than the other. Whatare the numbers?
Exercise 11.5.8. (Go to Solution)
Half the sum of two numbers is 14 and half the difference is 2. What are the numbers?
Exercise 11.5.9.
One third of the sum of two numbers is 6 and one fifth of the difference is 2. What are the numbers?
Exercise 11.5.10. (Go to Solution)
A 14 pound mixture of grapes sells for $3.10 . Type 1 grape sells for 25¢ a pound and type 2 grape sells for 20¢ a pound. How many pounds of each type of grape were used?
Exercise 11.5.11.
The cost of 80 liters of a blended cleaning solution is $28. Type 1 solution costs 20¢ a liter and type 2 solution costs 40¢ a liter. How many liters of each solution were used to form the blended solution?
Exercise 11.5.12. (Go to Solution)
The cost of 42 grams of a certain chemical compound is $14.40 . Type 1 chemical costs 45¢ a gram and type 2 chemical costs 30¢ a gram. How many grams of each chemical were used to form the compound?
Exercise 11.5.13.
A play was attended by 342 people, some adults and some children. Admission for adults was $1.50 and for children 75¢. How many adults and how many children attended the play?
Exercise 11.5.14. (Go to Solution)
200 tickets were sold to a college’s annual musical performance. Tickets for students were $2.50 and for nonstudents $3.50 . The total amount collected was $537. How many nonstudents purchased tickets for the performance?
Exercise 11.5.15.
A chemistry student needs 22 ml of a 38% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 40% acid and acid solution B is 30% acid. How much of each solution should be used?
Exercise 11.5.16. (Go to Solution)
A chemistry student needs 50 ml of a 72% salt solution. He has two salt solutions, A and B, to mix together to form the solution. Salt solution A is 60% salt and salt solution B is 80% salt. How much of each solution should be used?
Exercise 11.5.17.
A chemist needs 2 liters of an 18% acid solution. He has two solutions, A and B, to mix together to form the solution. Acid solution A is 10% acid and acid solution B is 15% acid. Can the chemist form the needed 18% acid solution? (Verify by calculation.) If the chemist locates a 20% acid solution, how much would have to be mixed with the 10% solution to obtain the needed 2-liter 18% solution?
Exercise 11.5.18. (Go to Solution)
A chemist needs 3 liters of a 12% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 14% acid and acid solution B is 20% acid. Can the chemist form the needed 12% solution? (Verify by calculation.) If the chemist locates a 4% acid solution, how much would have to be mixed with the 14% acid solution to obtain the needed 3-liter 12% solution?
Exercise 11.5.19.
A chemistry student needs 100 ml of a 16% acid solution. He has a bottle of 20% acid solution. How much pure water and how much of the 20% solution should be mixed to dilute the 20% acid solution to a 16% acid solution?
Exercise 11.5.20. (Go to Solution)
A chemistry student needs 1 liter of a 78% salt solution. She has a bottle of 80% salt solution. How much pure water and how much of the 80% salt solution should be mixed to dilute the 80% salt solution to a 78% salt solution?
Exercise 11.5.21.
A parking meter contains 42 coins. The total value of the coins is $8.40 . If the meter contains only dimes and quarters, how many of each type are there?
Exercise 11.5.22. (Go to Solution)
A child’s bank contains 78 coins. The coins are only pennies and nickels. If the value of the coins is $1.50 , how many of each coin are there?
Exercises For Review
Solutions to Exercises
Solution to Exercise 11.5.3. (Return to Exercise)
The student should use 36 ml of salt solution A and 24 ml of salt solution B.
Solution to Exercise 11.5.20. (Return to Exercise)
25 ml of pure water; 975 ml of 80% salt solution.
11.6. Summary of Key Concepts*
Summary Of Key Concepts
A collection of two linear equations in two variables is called a system of equations.
An ordered pair that is a solution to both equations in a system is called a solution to the system of equations. The values
x = 3,y = 1
are a solution to the system
Systems in which the lines intersect at precisely one point are independent systems. In applications, independent systems can arise when the collected data are accurate and complete.
Systems in which the lines are parallel are inconsistent systems. In applications, inconsistent systems can arise when the collected data are contradictory.
Systems in which the lines are coincident (one on the other) are dependent systems. In applications, dependent systems can arise when the collected data are incomplete.
To solve a system by graphing:
Graph each equation of the same set of axes.
If the lines intersect, the solution is the point of intersection.
To solve a system using substitution,
Solve one of the equations for one of the variables.
Substitute the expression for the variable chosen in step 1 into the other equation.
Solve the resulting equation in one variable.
Substitute the value obtained in step 3 into the equation obtained in step 1 and solve to obtain the value of the other variable.
Check the solution in both equations.
Write the solution as an ordered pair.
To solve a system using addition,
Write, if necessary, both equations in general form a x + b y = c
If necessary, multiply one or both equations by factors that will produce opposite coefficients for one of the variables.
Add the equations to eliminate one equation and one variable.
Solve the equation obtained in step 3.
Substitute the value obtained in step 4 into either of the original equations and solve to obtain the value of the other variable.
Check the solution in both equations.
Write the solution as an ordered pair.
If computations eliminate all variables and produce a contradiction, the two lines of the system are parallel and no solution exists. The system is inconsistent.
If computations eliminate all variables and produce an identity, the two lines of the system are coincident and the system has infinitely many solutions. The system is dependent.
The five-step method can be used to solve applied problems that involve linear systems that consist of two equations in two variables. The solutions of number problems, mixture problems, and value and rate problems are examined in this section. The rate problems have particular use in chemistry.
11.7. Exercise Supplement*
Exercise Supplement
Solutions by Graphing (Section 11.2) - Elimination by Addition (Section 11.4)
For the following problems, solve the systems of equations.
Exercise 11.7.2.
Exercise 11.7.4.
Exercise 11.7.6.
Exercise 11.7.8.
Exercise 11.7.10.
Exercise 11.7.12.
Exercise 11.7.14.
Exercise 11.7.16.
Exercise 11.7.18.
Exercise 11.7.20.
Applications (Section 11.5)
Exercise 11.7.21. (Go to Solution)
The sum of two numbers is 35. One number is 7 larger than the other. What are the numbers?
Exercise 11.7.22.
The difference of two numbers is 48. One number is three times larger than the other. What are the numbers?
Exercise 11.7.23. (Go to Solution)
A 35 pound mixture of two types of cardboard sells for $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?
Exercise 11.7.24.
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 11.7.25. (Go to Solution)
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?
Exercise 11.7.26.
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 11.7.27. (Go to Solution)
One third the sum of two numbers is 12 and half the difference is 14. What are the numbers?
Exercise 11.7.28.
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 11.7.29. (Go to Solution)
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?
Exercise 11.7.30.
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?
Solutions to Exercises
11.8. Proficiency Exam*
Proficiency Exam
Exercise 11.8.5. (Go to Solution)
(Section 11.3, Section 11.4) Solve using either substitution or addition: 
Exercise 11.8.6. (Go to Solution)
(Section 11.3, Section 11.4) Solve using either substitution or addition: 
Exercise 11.8.7. (Go to Solution)
(Section 11.5) The sum of two numbers is 43 and the difference of the same two numbers is 7. What are the numbers?
Exercise 11.8.8. (Go to Solution)
(Section 11.5) A chemist needs 80 ml of an 18% acid solution. She has two acid solutions, A and B, to mix together to form the 80-ml solution. Acid solution A is 15% acid and acid solution B is 20% acid. How much of each solution should be used?
Exercise 11.8.9. (Go to Solution)
(Section 11.5) A parking meter contains 32 coins. If the meter contains only nickels and quarters, and the total value of the coins is $4.60 , how many of each type of coin are there?
Exercise 11.8.10. (Go to Solution)
(Section 11.5) A person has $15,000 to invest. If he invests part at 8% and the rest at 12%, how much should he invest at each rate to produce the same return as if he had invested it all at 9%?
