Solving for One Uknown among Many: How and WhyDate: 09/20/2010 at 05:19:44 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers I have no problem solving for unknowns in equations that look like this: 5(6 + x) = 9 + 1x But my math book gave me problems to solve for x that look like this: cdx + e = rd xy - b = q The book says that all the rules that I have learned work no matter what symbols are used. But how can you solve for x when none of the numbers is known? I have checked my work once I was done doing these problems, but I found them incorrect and didn't understand it. For instance, on the second problem, here are my attempts to solve it (they were incorrect): xy - b = q -y - ...? I can't subtract y from the other side because there is no y. So that's as far as I got. Thank you for your help! Date: 09/20/2010 at 08:40:05 From: Doctor Ian Subject: Re: Solving for the Unknown with Unknowns Instead of Numbers Hi Emma, It's true that all the rules that you have learned work no matter what symbols are used. They even work when no symbols are used at all. This is because a symbol like 'x' just represents a quantity, like any other quantity. To see what I mean, consider something like 3 + 5 ----- * 7 = 28 2 You can evaluate the left side, and see that this is true. Suppose I'd like to 'solve for 3.' I can do the same sorts of transformations you've been learning in algebra class. For example, I can divide both sides by 7 to get 3 + 5 ----- = 28/7 2 I can multiply both sides by 2 to get 3 + 5 = 2(28/7) And I can add -5 to both sides to get 3 = 2(28/7) + -5 You can verify that each intermediate step is true, and so is the final equation. Now, what did I do here? More importantly, what did I NOT do? I didn't evaluate every expression just because I could. Instead of simplifying 28/7 to 4, for example, I kept it around. I left everything implied. In this case, I did it to make a point. But when you do it with variables, you do it because you don't know yet what values you have, so you really have no choice. But it's EXACTLY the same idea, carried out exactly the same way. That is, suppose I started out with an equation like my example above, but this time with only one known quantity and the rest unknown: 3 + a ----- * c = d b To 'solve for 3,' I could do the same steps: 3 + a ----- * c = d b 3 + a ----- = d/c b 3 + a = b(d/c) 3 = b(d/c) + -a If you substitute the original values for a, b, c, and d, you find that it all still works. Now, I don't have to solve for 3. I could solve for the variable a instead: 3 + a ----- * c = d b 3 + a ----- = d/c b 3 + a = b(d/c) a = b(d/c) + -3 Or I could solve for b: 3 + a ----- * c = d b 3 + a ----- = d/c b 3 + a ----- = b d/c Or I could solve for c, or for d. I'll leave those for you to try. Remember to check your result by substituting the original values, and seeing that you get a true statement. The key idea, though, is that you learn to look at anything -- whether it's a number like '3,' or a variable like 'a,' or an expression like (x - 2)^2, or a function like f(x) -- as 'some quantity.' And then you can apply the rules you know, which keep an equation balanced, namely: You can add 'some quantity' to both sides of an equation. You can multiply both sides of an equation by 'some quantity.' You don't need special rules for variables, or parameters, or expressions, or functions. They're the same as the rules for numbers. Because they ARE numbers -- they're just numbers the values of which we don't yet know, or haven't yet chosen. So when you go on to wonder 'how can you solve for x,' you point up that we actually use the word 'solve' in a lot of different ways. At first, you learn that 'solve' means 'find a particular value,' like x + 4 = 7 x = 3 But what you're learning now is that you can 'solve for' one variable 'in terms of' others, like x + a = b x = b - a Here, we've 'solved for x in terms of a and b.' In this form, it means we can CHOOSE values for a and b, and once those choices are made, the value of x is determined. For example, if a is 3 and b is 5, then x must be 2. Or, if a is 4 and b is 2, then x must be -2. If we were in a situation where we'd normally know x and b, and want to find a, we could 'solve for a in terms of x and b': x + a = b a = b - x This doesn't contain any new information. It just puts it in a form that's easier to use. But -- and here's the part that can cause confusion -- often we just say we are 'solving for x,' leaving the rest implied. And we use 'solve' in other ways, too. For example, given an equation like ... x^2 = 4 ... there are TWO values that make this equation true. It's true when x is 2, and when x is -2. So we say that there is a SET of solutions, or a 'solution set,' which is {2, -2} In fact, an equation like ... x + 4 = 7 ... also has a solution set; that set happens to have one element: {3} And some equations have an empty solution set, e.g., this equation ... x = x + 1 ... has no solutions, so its solution set is {} And an equation like ... y = 3x + 4 ... has infinitely many solutions in its solution set. However, each solution is a PAIR of values, rather than a single value. For example, if x is 1, y is 7. So (x = 1,y = 7) is in the solution set. By convention we put the x value first, so we usually just write (1,7). Also, when x is 2.5, y is 11.5. So (2.5,11.5) is also in the set. Since we can choose ANY value for x, and find a corresponding value for y, we can't write down all the solutions. However, if we plot the (x,y) pairs on a graph, we find that in the case of ... y = 3x + 4 ... all of the corresponding points lie on a straight line. So the line is a kind of 'snapshot' of all the solutions to the equation at once. Other kinds of equations give other kinds of shapes, but this is one of the reasons we make a big deal out of graphing -- sometimes, it's much easier to see what's going on by looking at a picture of the solutions rather than by looking directly at the equations. For example, at some point you'll be asked to solve 'systems' of equations that look like this: y = 3x + 4 y = 2x - 5 When we write a system like this, what we're really saying is: 1) The equation y = 3x + 4 has some set of solutions. 2) The equation y = 2x - 5 has some set of solutions. 3) What solutions are in BOTH sets? That is, we're being asked to find the intersection of the solution sets. (If you haven't learned about sets yet, you can find a two-minute introduction here: Intersection, Difference, Union http://mathforum.org/library/drmath/view/61902.html It's worth knowing a little about them, since many ideas that are complicated to explain without sets are much easier to explain in terms of sets.) And it turns out that finding the intersection of the solution sets is the same as finding the intersection of the lines! (If a solution is in both sets, it will be on both lines, right? And to be on both lines, the lines have to intersect at that point.) So often, the easiest way to solve a system is to graph the equations, and look for the points of intersection. But again, the point is that 'solving a system' is a little different than 'solving an equation,' even though we keep using the same word ('solve') for all these different situations. Don't worry too much about understanding all this in detail right now. The main point I want to make here is that you need to pay attention to how the word 'solve' is being used, because the meaning can change on you as you go through your math classes, and often no one will alert you that its meaning has been expanded. Anyway, does the idea of solving your equations make more sense now? If so, try some of your equations, and let me know what you come up with. If you're still stuck, try assigning numbers to the variables you DON'T want to solve for, like ... x*2 - 3 = 4 y = 2, b = 3, q = 4 ... and solve for that -- again, leaving operations implied ... x*2 - 3 = 4 x*2 = 4 + 3 x = (4 + 3)/2 ... so you can undo the substitutions later: x = (q + b)/y It helps if each variable gets its own number! So you don't want to use, for example, 2 to stand in for both b and q, because you want to be able to tell them apart when you're done. I hope this helps! Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 09/22/2010 at 05:30:52 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers As I was reading through the advice you gave me, I encountered a side question I need to clarify before I can continue. What do you mean by 'solve for 3'? Three's value is already known -- it's 3! Thanks! Emma Date: 09/22/2010 at 07:27:46 From: Doctor Ian Subject: Re: Solving for the Unknown with Unknowns Instead of Numbers Hi Emma, Good question! This gets to the heart of what it means to 'solve for [one quantity] in terms of [other quantities].' All that really means is to rearrange an equation so that one of the quantities appears by itself -- isolated -- on one side and nowhere on the other side. So when we start with an equation like ... 3x + 4 = 5x - 6 ... and we 'solve for x,' ... 3x + 4 = 5x - 6 4 = 2x - 6 10 = 2x 5 = x ... we're solving for x 'in terms of' some constant -- and we want to find out what that constant is. But if we start with an equation like ... ax + 4 = bx - 6 ... then when we 'solve for x' ... ax + 4 = bx - 6 4 = (b - a)x - 6 10 = (b - a)x 10/(b - a) = x ... we don't get a unique answer. What we get is an equation that will let us choose values for a and b, and then evaluate an expression to get a corresponding value for x. This is what we mean when we say we're solving FOR x IN TERMS OF a and b. Think about a formula for something like the volume of a cylinder: V = pi * r^2 * h Here, V is the volume, r is the radius of either end, and h is the height. This is already in a good form if we have measurements for the radius and height, and want to calculate the volume. For example, if the radius is 10 cm, and the height is 8 cm, we can just substitute: V = pi * (10 cm)^2 * (8 cm) That is, the formula is already 'solved for V in terms of r and h.' But suppose we're told that the volume is 1200 cm^3, and that the radius is 10 cm. What we don't know is the height. We could still substitute those values to get an equation like ... 1200 cm^3 = pi * (10 cm^2) * h ... but it's not very convenient for finding h. What we can do is 'solve for h in terms of V and r': V = pi * r^2 * h V/pi = r^2 * h V/(pi * r^2) = h Now we can substitute our values for V and r, and calculate h directly. In the same way, if we expect to know V and h, we can 'solve for r in terms of V and h.' I'll leave that for you to do, if you're interested. Anyway, it IS a little strange to 'solve for 3' in an equation, but I was trying to make the point that the operations you do are the same, whether you're working with numbers or variables or any other kinds of quantities. The value of 'solving for 3,' though, is that you can, at each step, evaluate both sides to convince yourself that the equation is still correct. (3 + 5)/2 = 4 True 3 + 5 = 2(4) Still true 3 = 2(5) - 5 Still true That's not always so easy to see when working with variables: (x + 5)/2 = 4 True for some value(s) of x x + 5 = 2(4) Still true for the same values x = 2(5) - 5 Still true for the same values Does this answer your question? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 09/22/2010 at 12:07:11 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers Yes! Thank you. I just can't think about it too hard, because I'll start wondering WHY it works, WHY I have to do it, etc. I'll read the rest of your notes and get back to you if I have any other questions. Thanks so much, again. Emma Date: 09/22/2010 at 14:25:44 From: Doctor Ian Subject: Re: Solving for the Unknown with Unknowns Instead of Numbers Hi Emma, These are exactly the questions you SHOULD be thinking about. The basic ideas are ones you've been learning about as far back as you can remember, although often they're not spelled out as explicitly as they should be. For example, you learned long ago that you can multiply 3 and 4 to get 12, and you can divide 12 by 3 to get 4. But in that second case, what you REALLY have is a multiplication ... 3 * ? = 12 ... that's not written in a convenient form. If we write it like this instead, ... ? = 12 / 3 ... then we can use division to 'solve for the unknown factor' in a multiplication where we know the factor and a product, instead of the two factors. In trying to use mathematics to model situations in the world, we often find that there are certain relationships, like force = mass * acceleration or ___________________________ | | length of pendulum period = 2 * pi * | -------------------------- \| gravitational acceleration As a rule, if there are N quantities that might change, and you know N - 1 of them, that will force a particular value for the remaining quantity. Sometimes, as I showed in the example of the cylinder, we know all the values on one side of the equation, and we can just substitute and evaluate to find the value on the other side. But sometimes we have a missing quantity 'in the middle' of one side, and we'd like to 'solve for' that quantity in order to calculate it easily. So that's why we do this sort of thing. Why does it work? The whole idea of 'solving' equations is this. Write down any equation with variables, like these three: 2x + 4 = 10 x^2 - 1 = 8 y = 3x + 4 In each case, we have a set (called the 'solution set') of values, or collections of values, that can be assigned to the variables in order to give a true statement. In the first equation above, there is just one value that works: x = 3. So we can say that the solution set is the set {3} In the second equation, the equation is true when x is either 3 or -3. So now we have two elements in our solution set: {3, -3} That is, we can grab any element of this set, and substitute it for x, and we will get a true statement. If we try substituting anything NOT in this set, we will get a false statement. In the third equation, we need matched pairs of values to make the equation true. For example, when x is 1, y must be 7. So the pair (x = 1,y = 7) or just (1,7) is in the solution set. {(1,7), ...} So is the pair (2,10): {(1,7), (2,10), ...} In this case, there are infinitely many solutions in that set, so we can't write them all down. We can, however, make a graph, which is a sort of 'picture' of all the solutions at once. But here is the important point. When you do certain transformations on equations, like adding the same thing to each side, you are guaranteed that the solution set of the old equation is the same as the solution set of the new equation. In other words, you can change the appearance of an equation, without changing the conditions under which it's true. So we can start with something complicated, like ... (x + 4) ------- + 5x = 18 (x - 2) ... and we can keep transforming it, without changing the solution set, until we get something simple like this: x = 5 And now it's trivial to see what the solution set is! It's just {5} So that must be the solution set of the original equation, too. That means the main thing you're learning in algebra is WHICH transformations will let you do this kind of thing -- and which other ones can also be used, with certain caveats. And then you spend a lot of time practicing combining them -- much of which has to do with the larger issue of learning to solve problems: How to Motivate Students to Learn Math? http://mathforum.org/library/drmath/view/71952.html I hope this helps! Let me know if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 09/27/2010 at 10:10:06 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers I have finished reading your notes, and they helped SO much! Thank you very much! I just have a question about assigning numbers to the unknowns and then finishing the problem. For instance, with cdx, would I assign a number like two hundred thirty-four (234), or does the book mean something like 2(3 + 4)? Thanks again! Emma Date: 09/27/2010 at 13:08:18 From: Doctor Ian Subject: Re: Solving for the Unknown with Unknowns Instead of Numbers Hi Emma, As a rule of thumb, juxtaposing quantities indicates multiplication, so 'cdx' means 'c times d times x, whatever the values of c, d, and x turn out to be.' It's only a rule of thumb, because obviously it doesn't work for numbers like '2' and '15.' If we wrote '215' for '2 times 15,' we wouldn't be able to distinguish later between '2*15' and '21*5.' So you could always expand that to ... c*d*x ... or, if you like, (c)(d)(x) The latter has a somewhat more algebraic 'feel' to it, but they're completely equivalent. Then, if you wanted to work with concrete values, you would assign particular values to each quantity, and track those through your subsequent operations. In the early stages of learning algebra, it can actually be helpful to make everything explicit, rather than relying on abbreviations that are common. A simple example would be taking something like this ... x + 3x ... and writing it as this: 1*x + 3*x That re-written, equivalent form makes it much clearer that the distributive property can be used to combine terms: 1*x + 3*x = (1 + 3)*x = 4*x Then, as you become more comfortable with 'seeing' the implied quantities, you can cut down on actually writing them. But this way, it happens at YOUR pace, not at one that's imposed on you by someone else. If you're still having a little trouble getting comfortable with the idea of variables, and the notation used with them, you might find it helpful to look at this: Variables, Explained http://mathforum.org/library/drmath/view/62905.html Let me know if you'd like to talk more about this, or anything else. Also: do you think it would be worth adding our discussion to the public archive (with all identifying information about you removed) so that others with the same confusion could learn from it? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 09/28/2010 at 08:57:12 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers I think it definitely would be worth adding this to the archive. I was wondering if this might be a common problem for people, because it did seem kind of hard to understand. Thank you so much for all your help! You guys are awesome! If I need any more help on this subject, should I just write on this thread? Date: 09/28/2010 at 11:34:46 From: Doctor Ian Subject: Re: Solving for the Unknown with Unknowns Instead of Numbers Hi, Emma. If it's still about this particular subject, then it would be okay to continue this thread. If it's about something new, it would be better to start with a new question submission at the main page. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 09/28/2010 at 13:11:46 From: Emma Subject: Solving for the Unknown with Unknowns Instead of Numbers Okay. Thanks again! Emma |
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