When we have an equation where the variable is in the denominator of a quotient, that's a rational equation. We can solve it by multiplying both sides by the denominator, but we have to look out for extraneous solutions in the process. Created by Sal Khan.
Want to join the conversation?
- Shouldn't you state the domain at the beginning?(20 votes)
- I think it's at least important to confirm that your answer doesn't result in an undefined expression, whether asked to state the domain or not.(13 votes)
- Is cross multiplication a standard method in mathematics or a wrong method of solving fraction.(19 votes)
- cross multiply is the style we use in Naija; however, I think this is a more standard method.(5 votes)
- What makes this a rational equation? Is it the fraction on the left side? Is this true for any equation of this form where you have two expressions on top of each other?(4 votes)
- What grade do you usually learn this in?(9 votes)
- How can I solve 1/4.8=1/4R+1/6R(3 votes)
- The trick in most rational equations is to get rid of the denominators, though you definitely want to write off to the side what they are so you remember. for instance 1/(4R) can't have R = 0 otherwise it would be 1/0 which is undefined. So just keep in mind the denoinators (with variables at least) are 4R and 6R, so in both cases R can't be 0.
Anyway, when you have your rational equation you want to multiply by all of the denominators. You could also simplify the right side so it s just one fraction, but I think it would be easier to mulitply first. so multiply both sides by 4.8*4R*6R
(4.8*4R*6R) 1/4.8 = [1/(4R) + 1/(6R) ]*4.8*4R*6R
4R*6R = 4.8*6R + 4.8*4R
24R^2 = 4.8(6R + 4R)
24R^2 = 4.8*10R
24R^2 = 48R
Can you handle it from here? A hint, you DO NOT want to divide by a variable.
if you had something like 4x = 5x you could divide both sides by x and get 4 = 5, which is not true. 4x = 5x if x = 0, so there is a solution.(9 votes)
- Is it possible to just multiply (9-x) and 3 to both sides from the start? This seems much simpler and yields the same answer. As follows:
- For this problem, corss multiplication works and works well as you have shown. You get the same answer. And, you have the added benefit of eliminating the fractions. I'm not sure why Sal didn't use that technique.
However, if the equation is more complicated where there are multiple factors in a denominator, you will want to find the LCM for all denominators and mulitply both sides by the LCM. As you multiply, you cancel out common factors and your equation will usually be simpler than what you would get by cross multiplying. Sal shows using an LCM in later videos in this section, though he is apt to do it one factor at a time.(6 votes)
- I was doing the practice problems for 'Find inverses of rational functions'. In one problem, it said to find the inverse for (5x-3)/(x-1). My answer was (x-3)/(x-5). I got it wrong, looked at the hints, and they said that the answer was (3-x)/(5-x). There is really no difference except that, basically, they just multiplied by negative one. It seemed totally arbitrary to me. When and why do they do this?(3 votes)
- My experience is that the KA exercises for rational expressions and rational equations don't always include all acceptable answers. I think your answer if better because the fraction is fully reduced. The KA answer is not fully reduced because both the numerator and denominator share a common factor of -1. If I were you I would report it as an error within the exercise (use "report problem" link).(8 votes)
- SMH I don’t understand this at all smh(5 votes)
- how do you check your answer for this problem, I'm having trouble.(5 votes)
- Replacing all of the x's in the original equation and solving both sides is how to check your answer. As long as both sides end up equaling the same number, your answer is correct.(1 vote)
- why divide it by 2 at the end ?(3 votes)
- To simplify it further and make it easier to reand and understand.
-20/38 is the same as -10/19.
If you want, you can try drawing it to see it visually.(3 votes)
- [Instructor] Let's say we wanna solve the following equation for x. We have x plus one over nine minus x is equal to 2/3. Pause this video and see if you can try this before we work through it together. All right now let's work through this together. Now, the first thing that we might wanna do, there's several ways that you could approach this, but the thing I like to do is get rid of this x here in the denominator. And the easiest way I can think of doing that, is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero 'cause clearly if somehow you do all this algebraic manipulation and you got x is equal to nine that still wouldn't be a valid solution 'cause if you were to substitute nine back into the original equation you'd be dividing by zero in the denominator. So, let's just put that right over here, x cannot be equal to nine. And so then, we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply, and divided by nine minus x they cancel out, and we'll just be left with an x plus one, and on the right-hand side, if you multiply 2/3 times nine minus x, we get 2/3 times nine is six and then 2/3 times negative x is negative 2/3 x and once again, let's remind ourselves, that x cannot be equal to nine. And then we can get all of our x's on the same side so let's out that on the left. So let's add 2/3 x to both sides. So plus 2/3, 2/3 x plus 2/3 x, and then, what do we have? Well, on the left-hand side we have one x which is the same thing as 3/3 x plus 2/3 x is going to give us 5/3 x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5/3 x, 5/3 x is equal to five. And then, last but not least, we can multiply both sides of this equation, times the reciprocal of 5/3 which is of course 3/5, and I'm doing that so I just have an x isolated on the left-hand side. So times 3/5, and we are left with 3/5 times 5/3 is of course equal to one. So we're left with x is equal to five times 3/5 is three. And so we're feeling pretty good about x equals three, but we have to make sure that that's consistent with our original expression. Well if we look up here, or if you substitute back x equals three, you don't get a zero in the denominator, x is not equal to nine. X equals three is consistent with that. So we should feel good about our solution. If we did all this algebraic manipulation and we get an x is equal to nine, then that still wouldn't be a valid solution because it would have made the original expression on the left be undefined.