Algebra (all content)
- Rational equations intro
- Rational equations intro
- Equations with one rational expression (advanced)
- Rational equations (advanced)
- Equations with rational expressions
- Equations with rational expressions (example 2)
- Rational equations
- Equation with two rational expressions (old example)
- Equation with two rational expressions (old example 2)
- Equation with two rational expressions (old example 3)
Equation with two rational expressions (old example 2)
Sal solves the equation 5/(2x)-4/(3x)=7/18 by first finding the LCM (least common multiple) of 2x and 3x. Created by Sal Khan and Monterey Institute for Technology and Education.
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- How can I find out what x cannot be?(13 votes)
- It's a mathematical rule that you can't have a 0 in the denominator because any number divided by 0 is undefined.
In general that's the logic you have to follow, " is there any number in this equation that would make it undefined (and thus impossible to solve) ?
"if yes, what would x had to be to get that number?"
And that is your "X that can't be"
Example given in the previous video:
You know that you can't have 0 in the denominator.
In the first fraction you have the denominator x-1. What would x had to be for the denominator be 0?
it had to be 1; 1-1=0. So X can't be 1.
By the same logic:
what number would make x+3=0.
-3 so -3+3=0
You can also rule out -3.
For this example you already know that X cannot be 1 or -3.(41 votes)
- what happens if there is a variable in the numerator?
ex. x-3/(3x+2)=1/5(4 votes)
- At2:06, 18 is divided by the denominator (2) and then the quotient (9) is multiplied by the numerator.
I'm pretty fuzzy on the logic behind this. If anyone can explain why this works I think I'd retain this a lot better than I am now.(3 votes)
- The key for this problem (and many problems involving fractions) is to ELIMINATE the fractions:) If you can find a number into which all the denominators are divisible, all you need to do is multiply that number by each term, and then when you simply the fractions the denominators will cancel out.
The specific issue you seem to have is the order in which Sal is simplifying. It would be just as correct to multiply the 18x by 5 (the numerator) to get 90x, and then divide that by 2x (the denominator) to get 45. See? Same result. Sal did it in smaller steps to avoid having to multiply 18x by 5, which some people can't do in their heads. Remember the 18x is really 18x/1 so you are just multiplying fractions; Sal chose to do some cancellation/simplification before he multiplied. Did this make sense or did I make it worse?(4 votes)
- I want to know the answer and know how to check it. Here's my problem: 3/3x+4=2/5x-6(3 votes)
- Since you are adding fractions, the first thing you should do is convert everything to a common denominator. You didn't use parentheses so I can't tell if you mean
'3/(3x+4) = 2/(5x-6)' or '3/(3x) + 4 = 2/(5x) - 6'.
My guess is you meant '3/(3x+4) = 2/(5x-6)'. So multiple the left side by (5x-6)/(5x-6), and the right side by (3x+4)/(3x+4). You're really just multiplying both sides by 1, so this is perfectly valid. That gives you a numerator of 3*(5x-6) on the left side and 2*(3x+4) on the right side. The denominator on both sides is (3x+4)(5x-6). Now you only have to compare the numerators to solve for x.
3*(5x-6) = 2*(3x+4) ... 15x - 18 = 6x + 8 ... 9x = 26 ... x = 26/9. To check if this is correct, substitute the solution back into the original expression and verify that both sides are equal. Doing that I get 9/38 = 9/38.(2 votes)
- How do you solve a quadratic equation with multiple different denominators?(2 votes)
- Well first of how did you end up with a quadratic equation with multiple denominators? was a=1/2 or c=1/2? considering the denominator consists of 2ac?(3 votes)
- I was lost after the first 1 minute and 12 seconds. I'm in college remedial math, and I still don't understand equations.(3 votes)
- Its like saying 4=4. We know that is true... but there is a bunch of ways ways we can write that same thing....
ex. 2x2=4, 2+2=4,6-2=4 etc..... correct?
now, if I say, some number 'x' + 2 gives me 3, then we know that 'x' must be 1..... correct?
So writing that in a bit more mathematical fashion we can say
x + 2 = 3.
now if you go back to the 4=4 equation, if i subtract 1 from the right side, we would get 4=3... that's not true. but if i subtracted 1 from both sides, i get 3=3 which is true. so in general the equality is maintained as long as we do the same operation on both sides of the equation. now back to x + 2 = 3... subtract both sides by 2 you get x = 1..... its the same with division and multiplication.... if you have doubts mail me or add me as your coach... my email id is email@example.com(2 votes)
- how do I solve the rational equation 3x=12+15/x(2 votes)
- It is essentially the same as any other equation. If a number does not have a denominator, it by definition technically has a denominator of one, so simply multiply all numbers by x, cancel out the denominator, then solve from there.(3 votes)
- 2a/3 + 1/5 = 7/6
How should I begin to solve this I'm confused.(2 votes)
- Here what you got to do:
2a/3 = 7/6 - 1/5 ---> 20a/30 = 35/30 - 6/30 ---> 20a = 29 ---> a = 29/20
If you want the decimal form, you divide 29 by 20, then you will have:
- How do you solve x/3 -1/2 = x/4.(1 vote)
- Multiply the equation by the LCD which in this case is 12. Then isolate 𝑥.(2 votes)
- what is the lcd of 13/90 and 11/90(1 vote)
- You already have a common denominator. Since the fractions can not be reduced, the LCD is your common denominator = 90.(2 votes)
Solve the equation, 5 over 2x minus 4 over 3x is equal to 7 over 18. And they tell us that x can't to be equal to 0, because that would make these two expressions here undefined. Hopefully the answer here is not 0, and then this becomes-- this is kind of extra, unnecessary information. So let's figure out how to solve this. So a good place to start-- I don't like having x's in my denominators. So let's multiply-- and in fact, in general, I don't like having fractions in my equations. So let's see if we can multiply both sides of this equation by some things that will get rid of the fractions. So let me just rewrite it so we have some space. 5 over 2x, and then we have minus 4 over 3x is equal to 7 over 18. Now, if we want to get rid of the 2 in the denominator here, we could multiply everything by 2. If we want to get rid of this 3 in the denominator, we could multiply everything by 3. If we want to get rid of this 18 in the denominator, we could multiply everything by 18. And 18 also includes a 2 and a 3. The prime factorization of 18 is 2 times 9, which is 3 times 3. So when you're multiplying both sides of the equation by 18, you're actually multiplying it by a 2 and a 3 and another 3. So let's just multiply both sides of this equation by 18. So I'll multiply this term right here by 18. And then, this term right here by 18. That'll get rid of all of these numbers in the denominator. 18 divided by 3 is 6. 18 divided by 2 is 9. But we don't just have numbers in the denominator, we also have these x's in the denominator. So let's also multiply both sides of the equation by x, so that we get rid of these. So we're essentially going to multiply both sides of the equation by 18x. We're taking, essentially, the least common multiple of 2x, 3x and 18. This is the smallest number that is divisible by all three of these characters. Now, when we do that our denominators will disappear. x divided by x is 1. 18 divided by 2 is 9. So this term becomes 9 times 5, which is 45. And then this term right here, x divided by x is 1. 18 divided by 3 is 6. So you have 6 times 4 is 24, but you have a subtraction sign here so minus 24 is equal to-- let me do that in that yellow-- is equal to-- and then you have this term, 7 over 18 times 18x. Well, the 18's cancel out and you're just left with 7 times x is equal to 7x. And now this becomes a much, much simpler equation. Now, what's 45 minus 24? Let's see, 45 minus 20 would be would be 25. Then you subtract 4 more, it's 21. So you get 21 is equal to 7x. Divide both sides by 7 and you get x is equal to 3. And let's verify that that works. So we have 5 over 2x. So that's the same thing as 5 over 2 times 3, minus 4 over 3 times 3. So this is 5/6 minus 4 over 18-- sorry, 4 over 9. We want to find a common denominator. 18 is the least common multiple of 6 and 9, so let's put it over 18. 5/6 is the same thing as 15 over 18. Multiply the numerator and denominator by 3. 4/9 is the same thing as 8 over 18. Multiply the numerator and denominator by 2. 15 minus 8. So this becomes 15 minus 8 over 18, which is equal to 7 over 18. So it works out. 5 over 2x when x is equal to 3, minus 4 over 3x when x is equal to 3 is indeed equal to 7/18. So we're done.