Using Systems of Equations

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1 Learning to set up word problems involving systems of equations is a matter of learning to read carefully, to extract data from the given situation and to apply a mathematical system to the data in order to obtain a desired answer. It is just a matter of following a system; much like making cookies is a matter of following a recipe. So far, all of the homework in this section has involved x's and y's and two equations. That is how we are going to solve these problems. Each of these problems is a story about two things, so every one of these is going to have an x and a y. In some problems, it s helpful to use different letters, to help keep straight what the variables stand for. For example, let L = the length of the rectangle and W = the width. The biggest advantage to this method is that when you have found that w = 3 you are more likely to notice that you still haven t answered the question, What is the length of the rectangle? Here are the steps to the solution process: Step 1: Figure out from the story what those two things are. one of these will be x the other will be y Step 2: Begin your solution with "Let x = " Step 3: The second sentence of your solution will be "Let y = " Step 4: Each story gives two different relationships between the two things. Use one of those relationships to write your first equation. Use the second relationship to write the second equation. Step 5: Now solve the system of two equations using substitution elimination

2 Number Problems: One number is blah, blah, blah......blah, blah, blah triple the second number The first number is triple the second Let x = the first number Let y = the second number x = 3y The sum of the numbers is 24 x + y = 24 For this one, I'd use substitution to solve the system of equations. 3y + y = 24 Geometry Problems: Formulas express standard relationships between measurements of things in the real world and are probably the mathematical tools that are used most frequently in real-life situations. As a member of modern society, it is assumed that you know certain common formulas such as the area of a square or the perimeter of a rectangle. If you are unsure about a formula, just Google it. Chances are excellent it will be in one of the first few hits. The following problem involves the perimeter of a rectangle P = 2L + 2W. The length of a rectangle blah, blah, blah blah, blah, blah twice the width Let L = the length of the rectangle Let W = the width of the rectangle The length is 8 inches less than twice the width. Notice that less than makes the 6 run around to the end of the expression. The perimeter of the rectangle is 64 L = 2W - 8 2L + 2W =64 For this problem, I'd use substitution to solve the system of equations: 2(2W 8) + 2W = 64

3 Coin Problems: Coin problems ask you something about the number of coins. The number of coins that you have is usually different from the value of those coins. If you have 5 quarters, they are not worth 5, they are worth 5 *.25 = $1.25. Use the variables to represent the number of coins and use the value of each coin times the variable to represent the value of those coins in the problem. Multiply all the values by 100 to kill the decimals! blah, blah, blah...dimes and quarters Let d = # of dimes Let q = # of quarters # of coins There are 54 dimes and quarters. d + q = 54.10d +.25q = Value of coins The purse contains $ Multiply everything by d + 25q = 1260 You could use either elimination -10d 10q = d + 25q = 1260 or substitution 10(54 q) + 25q = 1260

4 Mixture Problems: Like coin problems, mixture problems involve different products that have different unit prices. They will have to tell you something about the number of items or the price of each item. Use the variables to represent the unknown. Multiply the number of items times the unit price to get the value of that product in the problem. Example: Blah, blah, blah bought 4 cokes... Let c = the price of a coke... blah, blah, blah 4 hot dogs Let h = the price of a hot dog 4 cokes plus 4 hot dogs cost c + 4h = cokes plus 2 hot dogs cost c + 2h = 5.50 For this problem, I'd use elimination to solve the system of equations. Multiply the second equation by negative 2 and combine them. Investment Problems: These are a form of mixture problems but what you are mixing are monies invested at one rate and monies invested at another rate. Some of the money is invested at 8% Let x = the amount invested at 8% Some of the money is invested at 6% Let y = the amount invested at 6% A total of 10,000 is invested. x + y = 10,000 The investments earned x +.06y = For this one, I'd use elimination to solve the system of equations. Multiply the first equation by 6 and the second by 100 to clear the decimals.

5 Chemical Mixture Problems: These always involve a quantity of solution (5 liters of a 2% hydrochloric acid solution, 2 gallons of antifreeze, etc.) being mixed with a quantity of another solution. Sometimes these are easier to set up in tables: Example: A chemist needs 150 milliliters of a 50% saline solution but has only 18% and 78% solutions available. Find how many milliliters of each should be mixed to get the desired solution. That means that she needs 150(.05) or 75 ml of salt in her solution. Quantity % Salt 18% solution x *.18.18x 78% solution y *.78.78y Total 150 * First equation comes from the quantity column: x + y = 150 Second equation comes from the last column:.18x +.78y = 75 For this one, I'd use elimination to solve the system of equations. Multiply the first equation by 18 and the second by 100 to clear the decimals. -18x 18y = x + 78y = 7500

6 Uniform Motion Problems: Many students find motion problems to be difficult. However, since they are all basically set up the same way, if you can learn how to set them up, you can solve them! Motion problems usually involve either planes or boats. These are easy if you can remember one thing The boat has to go faster than the current or it will go backwards The plane has to fly faster than the wind or else! Setting these up, always put the speed of the plane or boat first. Planes: Let P = the speed of the plane in still air Let W = the speed of the wind flying with the wind (tailwind), rate = P + W flying against the wind (headwind), rate = P - W Boats: Let B = the speed of the boat in still water Let C = the speed of the current going with the current (downstream), rate = B + C going against the current (upstream), rate = B C Example: A plane can fly 510 miles in 3 hours with a tailwind. It can only fly 390 miles in 3 hours against a headwind. Rate x Time = Distance Tailwind: (P + W) * 3 = 510 both sides by 3 P + W = 170 Headwind: (P W) * 3 = 390 both sides by 3 P W = 130 Solve by elimination.