Setting up an EquationDate: 05/25/99 at 12:49:05 From: El Shaimaa Ghozzi Subject: Re: Equations I have a question about equations. When I add 4 to a number (x) I get the same result as halving the number and adding 10. What is the value of x? And another one: Paul has x toy cars. He buys 6 more, and then has twice as many. What is the value of x? Date: 05/25/99 at 17:00:58 From: Doctor Peterson Subject: Re: Equations Hi, Shaimaa. Let's write the first question as an equation: When I add 4 to a number I get the same result as 4 + x = halving the number and adding 10. x/2 + 10 That is, 4 + x = x/2 + 10 When we have both unknowns and known values on both sides of an equation, we need to separate them. We can first move the 4 to the right by subtracting 4 from both sides, and then move the x/2 to the left by subtracting that from both sides: 4 + x - 4 = x/2 + 10 - 4 x = x/2 + 6 x - x/2 = x/2 + 6 - x/2 x/2 = 6 Now we just have to undo the division by 2, which we can do by multiplying both sides by 2: x/2 * 2 = 6 * 2 x = 12 Finally, we have to check our answer, by putting x = 12 in the original equation and seeing whether it is correct: 4 + x = x/2 + 10 4 + 12 =? 12/2 + 10 16 =? 6 + 10 Yes, it works! Now I'll start the second problem for you, and you can finish it: Paul has x toy cars. He buys 6 more, and then has twice as many. x + 6 = 2x The equation is x + 6 = 2x. Your mission is to solve for x, by gathering the x's on one side. Let me know if you have trouble. (It's easier than the first one.) If there are any steps in my explanation that were too quick for you to follow, please write back and I'll explain a little more fully. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/26/99 at 14:59:50 From: El Shaimaa Ghozzi Subject: Re: Equations I understood the second one and a bit of the first one but what I don't understand in the first one is the 2/ thing. If you would kindly explain that to me, I'd be most greatful to you! Thanks again, Shaimaa Date: 05/26/99 at 16:52:17 From: Doctor Peterson Subject: Re: Equations Hi, Shaimaa. I'm glad you were able to finish what I started for you. Now let's work on the harder part. I hope I understand which part you need more help with; I'll comment on a couple points where I worked with /2 . I assume you understand that x/2 means half of x, or x divided by 2. As I said before: When we have both unknowns and known values on both sides of an equation, we need to separate them. We can first move the 4 to the right by subtracting 4 from both sides, and then move the x/2 to the left by subtracting that from both sides: 4 + x - 4 = x/2 + 10 - 4 x = x/2 + 6 x - x/2 = x/2 + 6 - x/2 x/2 = 6 Here we have subtracted half of x from x, leaving the other half of x. If you're not comfortable doing this, you can write it out more carefully: x x x - --- = 1 * x - 1/2 * x = (1 - 1/2) * x = 1/2 * x = --- 2 2 What I've done is to think of x as one times x, so that both x's have a coefficient. Then I can use the distributive property to combine the two coefficients, 1 and 1/2. I'm also using the distributive property in subtraction form; remember, the distributive property says that (a + b)*c = a*c + b*c and if we subtract rather than add, it looks like this: (a - b)*c = (a + -b)*c = a*c + -b*c = a*c - b*c That's the rule I just used. Then I said: Now we just have to undo the division by 2, which we can do by multiplying both sides by 2: x/2 * 2 = 6 * 2 x = 12 If x divided by 2 is 6, that's the same as saying that 6 is the number you multiply by 2 to get x; that's how division is defined. So if we multiply 6 by 2, we will get x. Here's another way to see this: x/2 = 6 1/2 * x = 6 rewriting division as mult by a fraction 2 * (1/2 * x) = 2 * 6 multiplying both sides by 2 (2 * 1/2) * x = 12 associative property 1 * x = 12 a number times its reciprocal is one x = 12 Both of these steps may be a little twisted from the way you have seen them before, so I can understand if you had trouble. In the first, we are dealing with a "phantom coefficient" of 1 that you have to get used to using when you need it; and we were also distributing with a negative sign; in the second, we are multiplying to undo a division, rather than dividing to undo a multiplication as we more often do. Think through these steps carefully, and if you still need help, let me know which part is the problem and I can work on a different problem for you. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/