8th grade (Illustrative Mathematics)
Manipulating formulas: temperature
Sal takes the formula for converting from Fahrenheit temperature to Celsius, and solves it for F so it serves to convert from Celsius to Fahrenheit. Created by Sal Khan and Monterey Institute for Technology and Education.
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- As some have pointed out, the problems initially seem to be much more difficult than Sal covers in this and any videos leading up to this point. Despite having been here for an hour just taking hints repeatedly to try to understand it that way, I still haven't grasped it. One of the most befuddling things to me is the "Combining of contant terms on right." For example:
Why 8, not 12? There has been no indication of subtraction? I would appreciate help!(30 votes)
- Oh my gosh! Thank you so much! It finally clicked -- at least, for now. Doubtless, I'll run into more trouble, later. Thank you, again!(5 votes)
- Why or when do you have to multiple the answer of an equation that is in fraction form by -1??? I know that isn't in this video, but it is in some of the problems in "practice this concept" for this video. And I don't see it explained in either video and it's driving me nuts!
-g-4h-4i+2 = -h+10i+4
The last step you have -1 as the denominator, and it say you have to make the denominator positive. Well, I swear I've seen answers with negatives in the denominator. ???
I found one!
8uv-6uw-9u+1 = 2v-7
The last step says to multiple top and bottom by -1. WHY!? You still are left with a negative in the denominator. The denominator goes from 8v-6w-9 to -8v+6w+9
Then ya got this!
4np+3nq-n-8 = 5p-6
Denominator is 4p+3q-1 but it doesn't say to change the denominator even though there is a negative number?? The only thing I can think is that the largest number in the denominator must be positive??
AND THIS!?!? What's going on?
5pq-5pr+p+6 = -q-5
You end up with the denominator being 5q-5r+1 but it say to change that to -5q+5r-1 WHY?(14 votes)
- THE ANSWER TO THE LAST STEP IS EQUIVALENT FRACTION
what is happening at the last step is not about maintaining the equality of the equation (2x = 2x ) is using a property of fractions that is called "Equivalent fractions" that means for example 30/60 = 3/6 if we divide the first fraction by 10 the same goes to the other direction 2/4 = 70/140 if we multiply the first fraction by 35, so if we multiply the fraction, and i mean both the numerator and denominator by the same number or divide the numerator and denominator by the same number we're going to end up with an equivalent fraction and by equivalent i mean that the end result will be the same 4/2 = 2 , 40/20 = 2 and with the negative term 4/-2 = -2, if we multiply the same fraction (both numerator and denominator) by -1 the fraction will be equal to -4/2 = -2 and will have the same answer -2.(3 votes)
- how do you isolate the variable for a question like 9n+6=60 because i now ycan look at it and solve it but I need to learn how to isolate it and verify.(4 votes)
- and you verify by plugging 6 into the place that n holds in the original question. so you get 9 times 6 plus 6 equals 60 (which is true) therefore n does equal 6.(2 votes)
- I have the equation: -10mp-2n+9p-6=-3n-p+9
How do I do these kind of problems with multiple unknowns? You can't combine them because a "p" is different than a "n" or "m". Please help. None of these videos cover this.(2 votes)
- usually, you might be given 2 equations where you can substitute 1 into the other. For example:
-3m+9p-3=0 and 7m-3p=47
you could do something like this:
divide both sides by3
multiply both sides by -1
add 3p on both sides
m=3p-1 and then you can use this in the other equation.(7m-3p=47)
since m=3p-1 you can just substitute it in the other one 7(3p-1)-3p=47
simplify further to get 21p-7-3p=47
simplify even further to get 18p-7=47
add 7 to both sides to get 18p=54
divide both sides by 18
then you insert this into either one of the equations to find m. I'm going to take -3m+9p-3=0 as my equation.
So, since we found p, we can do this: -3m+9*3-3=0
subtract 24 from each equation
multiply by -1 (this step is optional, but will be necessary at one point)
divide both sides by 3
and, if you want to check your answer, plug those numbers in to find your answer. For the sake of shortness, i will leave you to do it.(2 votes)
- wouldn't you also have to multiply the (F-32) by 9/5 too(3 votes)
- No, you don't have to because the f - 32 is in parentheses. The rule for carrying out a problem is:
3. multiplication and division
4. addition and subtraction
always go from left to right.(3 votes)
- I cannot enter my answer on mastery page https://www.khanacademy.org/mission/algebra2/task/5733135296888832
If I enter a value in the first field, the page go away. I then must reload and try to answer again, with the same result. What can I do after I report this and look in the community for faults and bugs?(3 votes)
- Try different tasks and see if it occurs. If it occurs, I would wait for a few days and see if it changed/works again. Although, if it doesn't change, try asking somebody else/post the question to get more help.(3 votes)
- Why Would it be a fraction on the problem? On the degree scale there are no fractions(2 votes)
- for precision(4 votes)
- when you divide something by something do you always have to flip it or not
does that matter or not
why would you flip it italics what would happen(2 votes)
- I don't exactly understand what you mean, but you have the expression: (1/2) divided by (5/3) that would be the same as (1/2) times (3/5). You'd basically flip fraction that is dividing, and multiply it instead of divide it. And really, yes you always do that, because take this example: "1 divided by 5" is the same thing as "(1/1) divided by (5/1)" which then it would be equal to "(1/1) times (1/5)" which is "1/5".
Hope this helps,
- ncochran2(4 votes)
- Hey, to anyone who could help, I have to solve the following equation for L.
L- L(cos θ) = v^2/2
I can't seem to find a way to solve for L without dividing L by itself which always results in a 1.
Any help is appreciated, Thanks!(1 vote)
- Prior answer has an error as the right side would be 1/L, not L.
So, here's the approach you should take:
1) Factor out L rather than dividing: L (1-cos θ) = v^2/2
2) Then, divide by (1-cos θ): L = v^2 / 2 (1-cos θ)
Hope this helps.(4 votes)
- Sal is just flipping the equation around. If you take division on one side of the equation and put it on the other side it becomes multiplication. For example 1x-3/7=6x+5/8.. It seems like he is just taking the denominator and swapping it over to the other side making it multiplication so that 1x-3/7=6x+5/8 now becomes 8* (1x-3)=7*(6x+5)
Please correct me if I am wrong.
Thank you in advance.(3 votes)
- You are wrong (except if you forgot parenthesis around 1x-3 and 6x+5).
In your example, 7 divides only 3, and 8 divides only 5. You're not allowed to take any denominator on one side and put it as a multiplicator on the other side. What you're allowed to do is to multiply (or divide) both sides of the equation by a same number, as Sal does. Because when you have a=b, you also have 3a = 3b, right? You're also allowed to add (or subtract) a same number to both sides: when you have a=b, you also have a-4 = b-4.
So about your example: first subtract 5/8 to both sides. You find 1x - 3/7 + 5/8 = 6x. Now subtract 1x from both sides. Result is -3/7 + 5/8 = 6x - 1x.
That is 11/56 = 5x. Divide both sides by 5: you get 11/(5x56) = x.
Always be aware of the operation you perform on the equation. There are no magic moves of taking something on one side and put it on the other side. The two rules that apply are: if a=b, then you can write ka=kb or also a+c=b+c.(0 votes)
We're told that the formula for converting a Celsius temperature into Fahrenheit is the Celsius temperature, or C, is equal to 5/9 times the Fahrenheit temperature minus 32. Rewrite the formula so it solves for Fahrenheit. So just like I mentioned, the C is Celsius, is the Celsius temperature. And the F is the Fahrenheit temperature. So right now, it's expressed in terms of, we've solved for C in terms of F. Now we want to solve for F in terms of C. So let's see how we're going to do that. So I'll just rewrite it. C is equal to 5/9 times F minus 32. And we want to solve for F. So the first useful thing to do might be to get rid of this 5/9 from both sides, or at least from the right side of the equation, so we just have an F minus 32. And the easiest way to do that is to multiply by the inverse of 5/9. So if we multiply both sides of this equation by 9/5-- and I could put it here, but it'll squeeze a little tight, let me try it-- 9/5 in there. If I do it to the right-hand side of the equation, I also have to do it to the left-hand side of the equation. And what do I get? The left-hand side becomes 9/5 Celsius, or the Celsius temperature, 9/5 times the Celsius temperature, is equal to-- the whole reason we wanted to multiply the right side by 9/5 is these 9's cancel out, this 5 cancels out, and this just becomes a 1. So 9/5 times the Celsius temperature is going to be equal to this expression: the Fahrenheit temperature minus 32. We don't need the parentheses anymore, because this is the only expression on the right-hand side. Now, we want to solve for the Fahrenheit temperature. So let's see if we can get rid of this 32 from the right-hand side. The easiest way I know how to do that is to add 32. Add 32 to both sides of this equation. Add 32 to both sides. The left-hand side now becomes 9/5 times the Celsius temperature, plus 32. And that's going to be equal to-- this negative 32 and positive 32 cancel out, that was the whole point behind adding 32, so it's just going to be equal to F. And we're done. We have rewritten this formula, up here, so it solves for Fahrenheit. This is the Fahrenheit right here. We've solved for Fahrenheit in terms of the Celsius temperature.