If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Solving tricky proportions

Sal solves the equation (x-9)/12=2/3. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user Lennie SkyWalker
    at , why did Sal add 7 to the 7x? (7x + 7)
    (4 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Alina Razi
    oh and for some people, at when he says what x equals to. You have to divide 51 by 3. Just saying in case someone gets confused
    (5 votes)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user David Buckalew
    The number of sparrows sitting on the top of the fence is inversely proportional to the square of the velocity of the wind. At noon there were 18 sparrows on the fence with the wind speed at 6 mph. At 2pm, there are 6 sparrows on the fence. What is the wind speed in mph rounded to the nearest tenth? (I have the answer--10.4 mph. How do you work it?)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user skofljica
      first you need to set equation for number of sparrows (N) in relation to velocity (v). they are inversely proportional so we can write N=C/(v^2) ; C-constant. Constant is needed because we don't know anything about relations except that they are inversely proportional.
      Then we can calculate constant from first condition: N=18 and v=6mph
      And finally, we can calculate wind speed.
      (6 votes)
  • blobby green style avatar for user Ayaalsamawi2007
    What do you mean "hairy proportions"?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user Learner2525
    Why do they even call this HAIRY proportions??
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user anavarro2025
    Thank you Sal it always help when I miss a day in algebra and I have no clue what im doing. My brother told me he uses you and when I first tried I saw its the same thing my old math teacher used to help teach us. It also helps since im in a high school course but im in middle school. Thanks again Sal your a big help!!
    (3 votes)
    Default Khan Academy avatar avatar for user
  • winston default style avatar for user Srishti
    How would I solve a proportion with variables in both denominators?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user StudyBuddy
      Same way: cross multiply.
      Say we had 2/x+1 = 3/3x-2.
      If we cross multiply we get 2(3x+2)=3(x+1)
      Using the distributive property...
      6x+4=3x+3
      Then, solve like a normal equation. subtract 4 from both sides to get
      6x=3x-1
      Subtract 3x from both sides to get
      3x=-1
      And then divide both sides by 3 to get
      x=-1/3
      Hope this helps!
      (2 votes)
  • aqualine seedling style avatar for user deanna
    51 into 17 how did you do it could you possibly explain I think I might have to pause the video and put it back at that part huh?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Chirag
      Good question. At , Sal writes:

      3x = 51

      The next step would be:

      x = 51/3

      And finally since 17 times 3 is equal to 51:

      x = 17

      Hope this helped. Here's a great tip for you - next time you ask something specific with regard to the video, use a timestamp by mentioning the time like I did above referring to the point in the video at 48 seconds.
      (2 votes)
  • male robot hal style avatar for user lkcapc1
    how do you know what step to take first when there is a new problem?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Jermaine Race
      I always start by defining the information I have, then defining a variable for the information I want. Then you describe the known relationships between the information mathematically. For example, lets say I have a triangle with sides 1, 2, 3 and a similar triangle with the smallest side of 2 and I have been asked to find the size of the largest side of the larger triangle. The first thing I would do is draw two similar triangles, both with sides of three sizes, and one larger, but in proportion to the smaller. Then I would label the sides of them, as described, including variables for the unknown sides. Since these are similar triangles there is a proportional relationship between them, so I would write a proportion describing the relationship including the side of the unknown triangle that I know, the one I want to find out (lets say c/2), and the analogous sides of the smaller triangle (since we have the smallest and we want to find the largest sides of the large triangle, we use the numbers from the largest and smallest sides of the smaller triangle, that is 3/1). Take those two expressions and relate them to each other. Since they are similar triangles, the proportions are equal, which looks like this c/2 : 3/1 (when I did proportions, my math teachers were particular to use a colon for some reason, something about proportions not being proper equations, idk). After this point you just have to follow the rules you've been taught to solve proportions. You'll notice that you didn't use the length of the medium side of the smaller triangle. That's because you don't need to use it. A lot of people get hung up on extra information, but if you know what you need to come up with the answer, you'll know what information you can ignore.

      I believe that this process of 1-define what you have, 2-define what you need, 3-relate it together, 4-solve the resulting equation, is what proportions are really trying to introduce you to. I've never used cross-multiplication since learning the more algebraic methods of solving proportions, because they are easier once you understand them, but they take more practice.
      (1 vote)
  • aqualine ultimate style avatar for user Red Yu
    Are there any other ways to do proportional fractions?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] We have the proportion x minus nine over 12 is equal to two over three, and we wanna solve for the x that satisfies this proportion. Now, there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they wanna cross-multiply. They wanna say, hey, three times x minus nine is going to be equal to two times 12. And that's completely legitimate. You would get, let me write that down. So three times x minus nine, three times x minus nine is equal to two times 12. So it would be equal to two times 12. And then you can distribute the three. You'd get three x minus 27 is equal to 24. And then you could add 27 to both sides, and you would get, let me actually do that. So let me add 27 to both sides, and we are left with three x is equal to, is equal to, let's see, 51. And then x would be equal to 17. X would be equal to 17. And you can verify that this works. 17 minus nine is eight. 8/12 is the same thing as 2/3. So this checks out. Another way you could do that, instead of just straight up doing the cross-multiplication, you could say look, I wanna get rid of this 12 in the denominator right over here, let's multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus nine. And on your right-hand side, 2/3 times 12, well, 2/3 of 12 is just eight. And you could do the actual multiplication, 2/3 times 12 over one. 12, 12 and three, so 12 divided by three is four. Three divided by three is one. So it becomes two times four over one, which is just eight. And then you add nine to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. And you can also, you can multiply both sides by 12 and both sides by three, and then that would be functionally equivalent to cross-multiplying. Let's do one more. So here, another proportion, and this time the x is in the denominator. But just like before, if we want, we can cross-multiply. And just to see where cross-multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this eight right over here on the left-hand side. If we wanna get rid of this eight on the left-hand side in the denominator, we can multiply the left-hand side by eight. But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, similar-larly, if (laughs) I, if I wanna get this x plus one out of the denominator, I could multiply by x plus one right over here. But I have to do that on both sides if I want my equality to hold true. And notice, when you do what we just did, this is going to be equivalent to cross-multiplying. Because these eights cancel out, and this x plus one cancels with that x plus one right over there. And you are left with, you are left with x plus one, x plus one times seven, and I could write it as seven times x plus one, is equal to five times eight, is equal to five times eight. Notice, this is exactly what you have done if you would've cross-multiplied. Cross-multiplication is just a shortcut of multiplying both sides by both the denominators. We have seven times x plus one is equal to five times eight. And now we can go and solve the algebra. So distributing the seven, we get seven x plus seven is equal to 40. And then subtracting seven from both sides, so let's subtract seven from both sides, we are left with seven x is equal to 33. Dividing both sides by seven, we are left with x is equal to 33 over seven. And if we wanna write that as a mixed number, this is the same thing, let's see, this is the same thing as 4 5/7, And we're done.