Ch. 8 NotesNote: The portion of this chapter that we will study deals with sets of linear equations. Recall that a linear equation in two variables is an equation that describes a line and is of the form, ax + by = c. We will be solving the systems (sets of equations considered together) using 3 methods. The solution to a system of linear equations is always an ordered pair that solves both (all) equations at the same time. As with solving any equation, systems can be consistent and dependent (the same line, thus infinite solutions), inconsistent and independent (parallel lines, thus no solution), or consistent and independent (two lines that meet in a single point, thus one solution). We learn 3 methods because no one method is the best; each has its benefits and drawbacks. Here is a summary of those benefits and drawbacks:

SS8.1 Graphing Pros- This method is very easy because we already know how to

graph. Cons- The method is inexact and does not work well when the

solution is off the graph or not an ordered pair of whole numbers.

SS8.2 Substitution Pros- We know how to substitute a given value into an equation

already. Cons- When fractions are involved this method can become messy.

SS8.3 Addition Method (Also called elimination) Pros- Factions are easier to deal with. Cons- The method is entirely new, and a little harder to put

together.

SS 8.1 Solving Systems of Linear Equations by Graphing

Steps to Solving using Graphing1. Graph each line using methods from chapter three2. Locate the point where the lines meet, and determine and label its ordered pair.

This is the solution.

Example: y = 3x 4y = x + 2

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Example: y = x + 4y = -x + 2

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Example: y + 2x = 34x = 2 2y

Note: This is an inconsistent, independent system. It is inconsistent because there is not solution and independent because the equations each describe different lines.

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Example: 3x + y = 02y = -6x

Note: This is a dependent, consistent system. Since the equations describe the same line they are considered dependent and because there is a solution (infinite solutions) it is considered consistent.

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Example: 6x 3y = 5x + 2y = 0

Note: This is the reason that graphing is not always the most appropriate method for solving a system of equations!

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SS 8.2 Solving Systems of Linear Equations by Substitution

Steps in Substitution1. Solve one equation for either x or y2. Evaluate the second equation using the solution in part 1, and solve for the

remaining variable. (If you solved for y in part one, then you will be solving for x in this part, and this will give you your x coordinate.)

3. Plug the solution from part 2 into one of the original equations and solve for the remaining variable. (If you solve for x in part 2, this will yield the y coordinate.)

4. Write the answer as an ordered pair.

Note: If you have mastered the skill of putting lines into slope-intercept form, it is a good idea to put both (all) equations in this form before beginning so that you may see from the outset if you will have a solution, no solution or infinite solutions. If you have not mastered the skill of putting an equation into slope-intercept form this may be the time to practice the skill. I think that you will see its usefulness once we solve a couple of problems.

Example: x + y = 20 x = 3y

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Example: 3y = x + 64x + 12y = 0

Example: 4x + 2y = 52x + y = -4

Note: The untrue statement tells us that there is no solution.

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Example: 1/4x 2y = 1 x 8y = 4

Note: The true statement means that there are infinite solutions.

Example: x/3 + y/6 = 1x/2 y/4 = 0

Note: The fractions make this messy, and we would prefer another method.

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SS 8.3 Solving Systems of Linear Equations by Addition (Elimination)

Steps for Addition (Elimination)1. Put the equations in the same form! (y=mx + b; ax + by = c; etc.)2. Add the two equations in such a way that one variable is eliminated (goes away;

is subtracted out) – this may require multiplication of one or both equations by a constant.

3. Solve the new equation from step 2 for the remaining variable (for instance if you eliminated y in step 2, you will be solving for x).

4. Put the solution from step 3 into an original equation and solve for the remaining variable (if you solved for x in step 3, then you are solving for y now).

5. Write the answer as an ordered pair.

Example: 4x + y = 132x y = 15

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Example: 2x + 3y = 04x + 6y = 3

Note: One of the steps yields a false statement and this how we know that there is no solution to this problem.

Example: -x + 5y = -13x 15y = 3

Note: One of the steps yields a true statement, which tells us that this equation has infinite solutions.

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Example: 8x = -11y 162x + 3y = -4

Note: We can’t solve this problem until the equations are in the same form!

Example: 2y = x + 63x 2y = -6

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Example: x + 1/3y = 5/12

8x + 3y = -16

Note: The simplest way to deal with this system if you do not like to deal with fractions is to clear each equation of fractions by multiplying by the LCD. After that is done you follow the normal steps.

Your Turn

Example: x y/3 = -1-x/2 + y/8 = 1/4

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Example: 3x + y = 49x + 3y = 6

Example: 2x + y = 64x + 2y = 12

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SS 8.4 Systems of Linear Equations and Problem Solving

This section is about solving word problems with two equations and two unknowns. Most word problems that we have already worked on can be solved using this method, but up until this time we did not have the skill for solving such problems, although unwittingly we were using the substitution method in our solutions! At this point, most students find that word problems get much easier because they see the two equations quite simply and can think in terms of two unknowns, rather than one unknown and putting the other unknown in terms of the first (that, by the way, is where we were unwittingly doing substitution!).

Just as with every other section in which we have encountered word problems, the word problems all follow patterns. This chapter contains 3 of the same patterns that we have been dealing with since the end of chapter 2 and one that is new, but familiar. They are the number problems, mixture/interest problems, D=R*T problems and the new, but familiar, value problems (they are just number problems in disguise).

Number ProblemsThese problems deal with a simple translation problem that talks about two number, whether they are a large and small number, two consecutive integers, odd/even integers, or even amounts of money people have, it doesn’t matter, the pattern remains the same. The pattern has been that we have a statement about 2 numbers (giving us the information needed to define the numbers), and then a relationship between those two numbers (giving us the equation to plug into to solve the problem). Now, we will have two relationships between the two numbers, which will yield two equations, and we will define each of the numbers as different variables!

Example: Two numbers total 83 and have a difference of 17. Find the 2 numbers.

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Example: The sum of two numbers is 16. Three times the larger number decreased by the smaller number is 72. Find the two numbers.

Example: John has four times as much invested in his savings account as in his checking account. The total amount invested is $2300.

Example: The cost of the speakers for Corine's new stereo system was $45 more than three times the cost of the receiver. What did her speakers cost if the total system cost $929?

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Value Problems We have actually seen this type of problem before, but it really has never been stated as its own pattern. There have been coin value problems, car rental problems, buying items in the store, and others like this. This type of problem is a modified number problem. The only catch is that you must multiply the value of the item by the number of that item to get the value that you possess, purchase, etc. The scenario goes like this. You are told the total number of things that you have or are purchasing and this gives you your first equation. Then you are told the total value of the things that you possess or are purchasing and this allows you to form a second equation based upon the knowledge that the number that you possess or purchase multiplied by the value of each item when summed will give you the total value. The value of the items will either be given to you or they will be assumed known from common knowledge. We define the different things that we possess or are purchasing in terms of two variables.

Example: Ron has 14 coins with a total value of $2.30. The coins are nickels and quarters. How many of each coin does he have?

Example: Kelly sells 62 shares of stock she owns for a total of $433. If the stock was in two different companies, one selling at $6.50 a share and the other at $7.25 a share, how many of each did she sell?

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Example: A gambler finishes a session of blackjack with $5 chips and $25 chips. If he has 45 chips in all with a total value of $465, how many of each kind of chip does the gambler have?

Example: For a Saturday matinee, adult tickets cost $5.50 while kids under 12 pay only $4. If 70 tickets are sold for a total of $310, how many of the tickets were adult tickets and how many were sold to kids under 12?

*Problems from Elementary Algebra, edition 5 by Charles P. McKeague.

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Mixture/Interest ProblemsWe have already seen these problems time and again. We will still be solving

them using a chart, but instead of the sometimes-difficult step of putting one unknown in terms of the other we will use two unknowns and create a second equation in that column that adds down! Our other equation will come from the sum of the last column still. Recall that for a mixture problem we have three columns: the percentage of the solution/mixture, the total amount of the mixture (this is the column that adds to give us our first equation), and the amount of the pure thing (this is the column that we used before to get our only equation, and we will still use it to get our second equation). In the interest problems we have 4 columns, one of which (the time column) is rarely nothing more than 1. In the interest problems the first column is the principle (this column adds to give us our first equation), the second is our percentage rate, the third is the time, and the fourth is the interest (this is the column that yielded our only equation before and still yields an equation for us this time, but it is our second). In both mixture and interest problems the rows that are not totals multiply to yield the value in the last column. In both, you must remember that percentages must be converted to decimals in order to be multiplied. In both, the last column yields the second equation.

Example: Mr. Wilson invested money in two accounts. His total investment was $20,000. If one account pays 6% in interest and the other pays 8%, how much does he have in each account6 if he earned a total of $1,380 in one year.

P R T I

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Example: A woman invested four times as much at 5% as she did at 6%. The total amount of interest she earns in one year from both accounts is $520. How much did she invest at each rate?

P R T I

Note: This is a variation of the typical problem, where the first column will sum. Instead, we use the information about the relationship between the amounts to form a second equation. When we have only one variable and one equation we define one variable and then put the other amount in terms of that variable according to the relationship.

Example: How many liters of a 50% alcohol solution and a 20% solution must be mixed to obtain 18 liters of a 30% alcohol solution?

Percent of Alcohol Total Amount of Solution Amount of Pure Alcohol

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Example: A grocer wishes to mix peanuts worth $2.52 per pound with Brazil nuts worth $3.80 per pound to make 480 pounds of a mixture worth $3.44 per pound. How much of each should be used?

Amount of Nuts Cost per Pound Total Cost

Note: This type of problem can also be classed as a value problem. See if you can see the similarities with it and the value problems.

Distance ProblemsWe have already discussed these problems in several chapters. Our most recent

experience was in chapter 6 where the distance and time or rate and time were unknowns, but it was known that the times were equal, and the problem yielded an equation involving rational expressions. The problems that we will now see will not be where times are equivalent, they will be like those that we first encountered in chapter 2 where the distances are equivalent or add or subtract. Recall that in order to work these type of problems we must know the formula D=R*T. In one type of problem, two variables will come into the problems due to the speed of a current or wind. In defining the rate of travel for each item in the problem the rate will involve the sum or the difference of the rate that the item can travel and the rate that the wind or current is traveling. Because of this, we must make sure that outside the table we define what we are referring to as our 2 variables (in these problems before, all defining was taken care of by the table alone, so be careful). One equation will come from the exact same information that yielded our equation before – the fact that the distances are equal, sum or subtract. The other equation will come from one row of the chart, and more than likely, either will do nicely. In the other type of problem, the two variables will be used to define the two rates of the items, and one equation will come from the translation of the relationship between the two speeds, and the other will come from the sum, difference or equality of the distances (just as with the problems that we had solved like this prior to this section).

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Example: Alan can row 18 miles down the Delaware River in 2 hours, but the return trip took him 4 ½ hours. Find the rate Alan could row in still water, and find the rate of the current.

D R T

Example: Rich and Pat are frequent flyers with Delta Airlines. They often fly form Philadelphia to Chicago, a distance of 780 miles. On one particular flight they fly into the wind, and the flight takes 2 hours. The return trip, with the wind behind them, it only takes 1 ½ hours. Find the speed of the wind and the speed of the plane in still air.

D R T

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Example: How long will it take a bus traveling at 60 mph to overtake a care traveling at 40 mph if the car had a 1.5-hour head start?

D R T

Note: In this case the times are the unknown factors, instead of the rates, which is more typical, but there is a known relationship between the times, namely a difference of when the car left versus when the bus left, that allows us to create two equations.

Example: A bus traveled on a level road for 3 hours at an average speed 20 mph faster than it traveled on a winding road. The time spent on the winding road was 4 hours. Find the average speed on the level road if the entire trip was 305 miles.

D R T

Note: This problem involves the adding of the distances instead of the equality. The rates are where the unknowns occur, and the second equation comes from knowing that the difference in the rates is 20 mph.

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SS 8.5 Systems of Linear Inequalities (Optional)

Solving systems of linear inequalities relies solely upon graphing them. If you are weak on graphing a linear inequality, you should return to that section from chapter 3 and review their graphing.

Steps for Solving Systems of Linear Inequalities1. Graph both linear inequalities on the same graph. It is helpful to either use

different colors for the graphs or to use differing hash marks.2. The solution to the system is where both hash marks or both colors lie and the

solid lines that create the boundaries. Dotted boundary lines are still boundaries, but they are not part of the solution.

Example: y 2x + 1y x + 2

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Example: y + 2x 05x + 3y -2

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Example: y ½ x + 2y ½ x 3

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