Alex s Guide to Word Problems and Linear Equations Following Glencoe Algebra 1 What is a linear equation? It sounds fancy, but linear equation means the same thing as a line. In other words, it s an equation of a line. We will talk more about the line itself later, when we do graphing. What are linear equations useful for? They are useful for quite a lot. We will first look at how to translate between sentences (in English) and equations (in math). Translating between sentences and equations Three of the most important words in math are is, of, and per. Here s what they mean in math: Here are some examples: Words Word Math is = of multiply per divide more than add less than subtract Five s the number a is equal to three s the sum b plus c. Math 5a = 3(b + c) Nine s the number y subtracted from 95 is 37. 95 9y = 37 25 miles per hour is half of the speed limit. Lisa needs 7 cups of sugar in a recipe; she has already added 2 cups. How many more does she need to add? 25 miles 1 hour = 1 (speed limit) 2 2 + c = 7 The most important thing to do is use your intuition. Think about the problem. Visualize it in your head; make sure what you re writing in the equation makes sense. Here are two things you need to do first: 1. Define (choose) your variables. 2. Pull information out of the problem. (What is important, and what isn t?) Solving equations with variables on both sides We have already started to learn about solving equations in Alex s Guide to Expressions and Equations. We are going to look more at that here. Remember the three rules: 1. Isolate the variable using inverse (opposite) operations. 2. Work backwards (SADME) 3. Whatever you do to one side, you must do to the other. When you have variables on both sides of an equation, you still use the same three rules. Just follow the concept of combining like terms (CLT) to combine all terms with the same variable. You are essentially moving terms to the other side of the equal side if you need it over there. This is accomplished, as rule 1 states, by using opposite operations. Here is an example: 2 + 10x = 8x 1 Notice there are two variable terms, one on each side of the equal sign. We want to move them both to the same sign of the equal sign, so that we can CLT. The side you choose doesn t matter, but I usually choose to get the variable on the left-hand side. That means, the 8x on the right-hand side needs to be moved over to
the left-hand side. In order to do this, we follow the same three steps and subtract 8x from both sides, since that is the opposite operation from the positive 8x that is already there: 2 + 10x = 8x 1 8x When we balance the equation now, we see that our goal was accomplished, and we only have variables on one side of the equal sign (the arbitrarily-chosen left-hand side): 2 + 10x 8x = 1 Now it s just business as usual we CLT and use the 3 steps to solve for the variable: 2 + 2x = 1 +2 +2 2x = 1 2 2 8x x = 1 2 Ratios and proportions A ratio is a comparison of two numbers, and a proportion is a comparison of two ratios. A ratio is a fraction, and a proportion is an equation. Here s an example of a ratio: 1 Here s an example of a proportion: 1 = 3 2 2 6 Notice that the largest and smallest numbers (6 and 1) are diagonally across from each other, and the two middle numbers (3 and 2) are also diagonally across from each other. This will always be the case. The numbers in the biggest/smallest pair are called the extremes, and the numbers in the middle pair are called the means. Let s consider what happens when we cross-multiply each of the denominators: (6)(1) = (3)(2) 6 = 6 This can always be done for a proportion. If you cross-multiply the means and extremes and don t get the same thing on both sides, then the proportion has been set up wrong. The good news about this, is that we can use this technique to solve equations, since it always works. Consider this example: This is easy to figure out by using algebra. Since you have two parts of the data (4-hour-related and 6-hourrelated), you can set up a proportion to represent the information in the problem Remember that the word per means divided by create a fraction with that number. Therefore, 30 miles per 4 hours, and unknown miles per 6 hours, can both be written like this: 30 miles 4 hours =? miles 6 hours I ve written the units in just to help you see what I m doing here. Notice that they re directly across from each other is the same unit. This is explained below in the Dimensional Analysis section. Units are not required for the calculation, but they are essential to making sure you ve set everything up correctly. Get comfortable using the units! We can use our algebra skills to solve this. I m going to use the variable x for the number of miles I can ride in 6 hours: 30 4 = x 6
Ask yourself, What is being done to the x? It s being divided by 6. What are we going to do? Multiply by 6. Where? Both sides. Doing this gives us: (6) ( 30 4 ) = (x 6 ) (6) And simplifying gives us: And, finally, solving gives us: 180 4 = x 45 = x So, I can ride 45 miles in 6 hours, if I keep the same pace. Percent of change This topic is often made too hard for students to understand, so I m going to attempt a much simpler way. The concept is to look at how something changes over. The mantra is: Percent of change = change original Change means the amount the value changed before-and-after. Original means what the original value was. This works the same for both increases (where the number goes up) and decreases (where the number goes down). Commit this to memory, because it always works. Let s look into what it means. Imagine there s a shirt I want to buy, and it usually costs $38. The store is having a sale this week, and the shirt is now $24.70. How much of a discount (i.e.: what percent of change) am I getting? Here s how it would look: Percent of change = change $38 $24.70 = = 13.30 original $38 38 = 0.35 Convert 0.35 to a percent, and we find that the percent of change is a decrease of 35%. Obviously, whether it s an increase or decrease can be determined by just looking at the data. Dimensional analysis This sounds like something from Doctor Who, but actually it s real and straightforward. It also marks a very important skill in developing your mathematical understanding. When I say dimensions, I m actually talking about units. When you re doing a word problem, there are two steps (see page 1 of this Guide) to do first. Pick your variables, and pull information out of the problem. Dimensional analysis is a part of doing the latter. Imagine setting all your information up as fractions based on the units associated with each number. Here s an example: Let s do our initial two steps: define variables and pull information. I want to use the variable m for the number of miles I can drive. Now let s pull information from the problem: My fuel economy = 30 miles 1 gallon
That s the only piece of information I have, so now we can put all of this together in an equation: 30 miles m miles = 1 gallon 9.5 gallons Notice again that the same units are directly across from each other. This is part of dimensional analysis. It is saying that the values are comparable. A counterexample would be trying to say that 30 miles per gallon is the same as 4 people per car. They aren t the same units; they aren t comparable values. Using our algebra skills, we can solve for m: And finishing it up gives: (9.5) ( 30 1 ) = ( m 9.5 ) (9.5) 285 = m Therefore, I can drive 285 miles on 9.5 gallons of gas. Dimensional analysis helped me make sure I was setting up the problem s information correctly. This was a pretty simple problem; it s much more useful in a complex example, as we will see next. Dimensional analysis with added or subtracted units Here s an example of dimensional analysis where the units are added: v = v 0 + at v v 0 a t Before your pull your hair out and start screaming, WHAT?! just stay with me here. My point in using this example is to show you that you can still understand something if you haven t seen it before. (Hint: I know you haven t seen this before.) The first thing to do when trying to understand something new is look at the units (a.k.a. the dimensions). Since you haven t seen this equation before, I ll tell you what all the units are: v is velocity (also called speed), and its units are always some version of distance. v 0 is the initial velocity, which is usually zero, if something has started moving from a standstill. Since it s the same type of quantity as the regular velocity, it has the same units. a is acceleration, and its units are always some version of distance 2. t is, and is some unit of, like seconds, minutes, hours, etc. Using dimensional analysis, let s put all this together and see it the equation makes sense: v = v 0 + at It s common practice to use square brackets when doing a dimensional analysis. Your goal is to make sure your units match up, and the operations make sense. You want to make sure the equal sign is equating the same types of numbers: ] + [distance 2 ] [ 1 ]
Let s simplify this. We have no numbers in our dimensional analysis, but treat the units just like numbers. See what cancels. See what can be simplified. Remember that anything squared means s itself : ] + ] [ 1 ] One of the units in the denominator of the acceleration cancels with the unit of the t variable: ] + ] [ 1 ] Simplifying this gives us: ] + [distance ] Notice that they re all three the same unit now. Just like when adding fractions, you have to have the same denominator. This applies to units as well. Important: the same thing doesn t apply to units that are multiplied. There, just like with multiplying fractions, you go straight across the top and straight across the bottom, and usually some parts of the units will cancel with others, just like the part in this example. Dimensional analysis with multiplied/divided units Here s another example that illustrates dimensional analysis with units that are being multiplied: Obviously, you cannot compare these speeds as they are. They need to be in the same units to compare them. In order to do this, you will need to convert one of them into the other s units. It doesn t matter which one you choose, because you ll get the same answer either way. I m going to choose miles per hour. To start, you need to find all the relevant conversion factors in a book or online. I ll provide them for you here. Remember that per means divide: 60 sec 1 min 60 min 1 hr 5280 ft 1 mi 12 in 1 ft 2.54 cm 1 in 100 cm 1 m Now that we have all the necessary conversion factors, set up the conversion so that the unwanted units cancel out and you re left with only the units you want. Since I chose miles per hour, we don t need to do anything to the first speed, but I need to convert the second speed from meters per second to miles per hour. I m choosing x as my variable. Utilize the conversion factors to do what you want them to. Manipulate them until you have everything arranged so that stuff cancels out. Think about the units you don t want, and the ones you do want in the answer. Keep going until everything is canceled out and you re left with what you want. The mantra here is multiplication sign, division bar multiplication sign, division bar : 40 m cm [ ] [100 1 sec 1 m ] [ 1 in 2.54 cm ] [ 1 ft 12 in ] [ 1 mi sec min mi ] [60 ] [60 ] = [x 5280 ft 1 min 1 hr hr ] Look at how this turns out. Everything cancels except miles per hour, which is what I wanted: 40 m cm [ ] [100 1 sec 1 m ] [ 1 in 2.54 cm ] [ 1 ft 12 in ] [ 1 mi sec min mi ] [60 ] [60 ] = [x 5280 ft 1 min 1 hr hr ] Now, it s just straight across the top, straight across the bottom since we re multiplying fractions here: 40 100 1 1 1 60 60 mi [ ] = [x ] = 89.48 miles per hour 1 1 2.54 12 5280 1 1 hr