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FIN401_Solving equations primer.doc

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SOC 202
Louis Pike

FIN401 – Solving Equations To solve linear equations we will make heavy use of the following facts. 1. we can add a number, c, to both sides of the equation and not change the equation. 2. we can subtract a number, c, from both sides of an equation. 3. we can multiply both sides of an equation by a number, c, without changing the equation. 4. We can divide both sides of an equation by a non-zero number, c, without changing the equation. These facts form the basis of almost all the solving techniques that we’ll be looking at so it’s very important that you know them and don’t forget about them. One way to think of these rules is the following. What we do to one side of an equation we have to do to the other side of the equation. In this section we will be solving linear equations and there is a nice simple process for solving linear equations. Let’s first summarize the process and then we will work some examples. Process for Solving Linear Equations If the equation contains any fractions use the least common denominator to clear the fractions. We will do this by multiplying both sides of the equation by the LCD. Alternatively, you can simply multiply both sides by the largest denominator. Simplify both sides of the equation. This means clearing out any parenthesis, and combining like terms. Use the first two facts above to get all terms with the variable in them on one side of the equations (combining into a single term of course) and all constants on the other side. If the coefficient of the variable is not a one we usually just divide both sides of the equation by the coefficient if it is an integer or multiply both sides of the equation by the reciprocal of the coefficient if it is a fraction. 1. VERIFY YOUR ANSWER! This is the final step and the most often skipped step, yet it is probably the most important step in the process. With this step you can know whether or not you got the correct answer long before your instructor ever looks at it. We verify the answer by plugging the results from the previous steps into the original equation. It is very important to plug into the original equation since you may have made a mistake in the very first step that lead you to an incorrect answer. Also, if there were fractions in the problem and there were values of the variable that give division by zero (recall the first step…) it is important to make sure that one of these values didn’t end up in the solution set. It is possible, as we’ll see in an example, to have these values show up in the solution set. Let’s take a look at some examples. Example 1 Solve each of the following equations. (a) (b) Solution In the following problems we will describe in detail the first problem and the leave most of the explanation out of the following problems. (a) For this problem there are no fractions so we don’t need to worry about the first step in the process. The next step tells to simplify both sides. So, we will clear out any parenthesis by multiplying the numbers through and then combine like terms. The next step is to get all the x’s on one side and all the numbers on the other side. Which side the x’s go on is up to you and will probably vary with the problem. As a rule of thumb we will usually put the variables on the side that will give a positive coefficient. This is done simply because it is often easy to lose track of the minus sign on the coefficient and so if we make sure it is positive we won’t need to worry about it. So, for our case this will mean adding 4x to both sides and subtracting 15 from both sides. Note as well that while we will actually put those operations in this time we normally do these operations in our head. The next step says to get a coefficient of 1 in front of the x. In this case we can do this by dividing both sides by a 7. Now, if we’ve done all of our work correct is the solution to the equation. The last and final step is to then check the solution. As pointed out in the process outline we need to check the solution in the original equation. This is important, because we may have made a mistake in the very first step and if we did and then checked the answer in the results from that step it may seem to indicate that the solution is correct when the reality will be that we don’t have the correct answer because of the mistake that we originally made. The problem of course is that, with this solution, that checking might be a little messy. Let’s do it anyway. So, we did our work correctly and the solution to the equation is, (b) Okay, with this one we won’t be putting quite as much explanation into the problem. In this case we have fractions so to make our life easier we will multiply both sides by the LCD, which is 21 in this case. After doing that the problem will be very similar to the previous problem. Note as well that the denominators are only numbers and so we won’t need to worry about division by zero issues. Let’s first multiply both sides by the LCD. Be careful to correctly distribute the 21 through the parenthesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At this point we’ve got a problem that is similar the previous problem and we won’t bother with all the explanation this time. Another set of general rules of equations: - If you do the same thing to both sides of an equation, the result will still be an equation, even if it’s a different equation. - Usually, you want to isolate the desired variable/unknown on the left hand side of the equation and have one number on the right hand side of the equation. - You can flip the two sides of an equation if you want to, because a+b = c+d is the same as c+d = a+b - You must have at least as many independent equations as you have unknowns. For the most part, this means that if you have one variable, you must have one equation. If you have two variables, you must have two equations, etc. Abbreviations: X … desired variable LHS., RHS…. left-hand side and right-hand side MBS… multiply both sides DBS… divide both sides Examples of use in FIN401: In class 2 of the course, we discuss the cost of capital that the firm needs to achieve on its projects in order to compensate its shareholders and bondholders based on their expectations. There are many examples of having to work with equations in order to work through this section of the course. A few examples follow: Calculating a variable in the Cost of Equity calculation If the cost of equity is 12%, the risk-free rate is 4%, and the expected return on the market is 14%, what is the Beta of the firm. In order to answer this question, you must manipulate the CAPM equation: re= r f B(r (m)– r)f The easiest way to do this question is probably to subtract rf from both sides in order to start isolating B. re- rf= B(r (m)– rf If we then DBS by (r – r), we can completely isolate B (m) f (re- rf /r (m)– rf = B If you flip sides, the solution is then B = (.12-.04) / (.14-.04) = .8 Calculating a variable in the WACC calculation If the WACC is 12%, the cost of debt is 6%, and the cost of equity is 16%, what is the percentage weight of debt as a percentage of the entire capital structure of the firm? Ignore taxes or other market imperfections. In order to answer this question, you must manipulate the WACC equation: WACC = w *r + we*r e d d At first glance, there are two variables (w and w ) aed one eqdation, but the required insight here is that
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