Integrated math 3
- Adding & subtracting rational expressions: like denominators
- Intro to adding & subtracting rational expressions
- Add & subtract rational expressions: like denominators
- Intro to adding rational expressions with unlike denominators
- Adding rational expression: unlike denominators
- Subtracting rational expressions: unlike denominators
- Add & subtract rational expressions (basic)
Adding rational expression: unlike denominators
Sal rewrites (5x)/(2x-3)+(-4x²)/(3x+1) as (-8x³+27x²+5x)/(2x-3)(3x+1).
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- What if we did not distribute the numbers when adding them and leave them as factors. and then we can cancel out the numbers.(16 votes)
- You CAN leave sums of products as factors if they share a common factor. So 3(x-4) + 5(x-4) can be added to get 8(x-4). But you cannot add two different products of two different sums without distributing them first. .For example 3(x+4) + 5(2x-5) share no common factors. You cannot add them without distributing first and then combining like terms. You have to distribute to get 3x +12 +10x -25 which then simplifies to 13x-13 and could be re factored to 13(x-1). You cannot see those two factors of the original expression, 13 and (x-1), until you do the intervening arithmetic.(23 votes)
- Why wouldn't you multiply the denominators together at the end?(15 votes)
- Often you need to factor the equations to see if there are any things you can cancel out to simplify it. I don't know about you but my teacher tells me to leave it like that in case there are further calculations.(12 votes)
- none of this makes sense.(13 votes)
- They all give numbers, you're just multiplying the expressions while keeping them the same, so they will give numbers back(2 votes)
- Why didn't he factor by grouping to see if something cancels out?(12 votes)
- Why would you not factor out the x in the numerator and leave it like x(-8x²+27x+5) over (2x-3)(3x+1)?(8 votes)
- Can't you just combine the like terms on the bottom to get 5x - 3 and then use that as a common denominator?(4 votes)
- Sorry, but no.
Each of the denominators represents a single quantity [ (2x-3) and (3x+1) ]---it's the same as if we were adding 4/7 + 3/5. Their common denominator wouldn't be (7 + 5), it would have to have FACTORS of 7 and 5 --- in other words 7 * 5.
That's why the common denominator turns out to be (2x-3)(3x+1).(6 votes)
- in x/2 - x/4 = 6 why do we multiply both sides by 4?(4 votes)
- Equations must be kept in balance. Visualize a scale that's in balance. If both sides have 2 pounds on them and you only multiply one side by 4, that side now has 8 pounds and the other still has 2. So, the scale is not in balance.
Unless you are simplifying one side (just reorganizing, not changing the value), then you must do the operation to both sides of the equation.
Hope this helps.(6 votes)
- I'm practicing SAT Test on Khan Academy and I do not get why I keep getting different answers. The question of the problem is 7/x-1 + 6/x^2-12x+11. To the answer of the problem, it shows that you can factor x^2-12x+11 denominator but when I multiply the denominators x-1 and x^2-12x+11 I would get different answer.
(7/x-1)x^2-12x+11 + (6/x^2-12x+11)x-1. The answer I keep getting is
7x^2-78x+76/(x-1)(x^2-12x+11). And the actual answer of the problem is
7x-71/(x-1)(x-11). Don't get why I'm getting this answer. Thank you.(3 votes)
- To find a common denominator, you need to first factor denominators where possible. You didn't factor the x^2-12x+11 so you aren't finding the smallest common denominator. The factors of x^2-12x+11 are (x-1)(x-11). So, the LCD = (x-1)(x-11) rather than your version of (x-1)(x^2-12x+11).
Since you didn't use the smallest LCD, your answer is coming out as a fraction that is not fully reduced. You need to factor your numerator and the denominator to fully reduce the fraction.
You also have an error in your numerator
7(x^2-12x+11) = 7x^2-84x+77
6(x-1) = 6x-6
7x^2-84x+77+6x-6 = 7x^2-84x+6x+77-6 = 7x^2-78x+71
You have 76 rather than 71 as your last terms.
Hope this helps.(5 votes)
- I also need major help on factoring anything...can anyone help?(2 votes)
- To deal with rational expressions, you really need to understand all the factoring techniques as every problem involves some level of factoring. I would recommend that you go back to that section and review the lessons until you really understand them. Ask specific questions when you don't understand. here's a link to that section: https://www.khanacademy.org/math/algebra-home/algebra/polynomial-factorization(6 votes)
- (3:41) Anyone else hear Sal's stomach growling?(4 votes)
- [Voiceover] Pause the video and try to add these two rational expressions. Okay, I'm assuming you've had a go at it. Now we can work through this together. So the first thing that you might have hit when you tried to do it, is you realized that they have different denominators and it's hard to add fractions when they have different denominators. You need to rewrite them so that you have a common denominator. And the easiest way to get a common denominator is you can just multiply the two denominators, especially in case like this where they don't seem to share any factors. Both of these are about as factor as you can get and they don't share anything in common. And so let's set up a common denominator. So this is going to be equal to it's going to be equal to something let's see, it's going to be equal to something over our common denominator. Let's make it let's make it 2x, I'm going to do this in another color. So we're going to make it 2x-3 times 3x+1 times 3x+1 and then plus plus something else over 2x-3 2x-3 times 3x+1. Times 3x+1. And so to go from 2x, to go from just a 2x-3 here the denominator to a (2x-3)(3x+1) we multiply the denominator by 3x+1. So if we do that to the denominator, we don't want to change the value of the rational expression. We'd also have to do that to the numerator. So the original numerator was 5x. I'll do that in blue color. So the original numerator was 5x and now we're going to multiply it by the 3x+1, so times 3x+1. Notice I didn't change the value of this expression. I multiplied it by 3x+1 over 3x+1, which is 1 as long as 3x+1 does not equal zero. So let's do the same thing over here. Over here I have a denominator of 3x+1, I multiplied it by 2x-3, so I would take my numerator, which is -4x² and I would also multiply it by 2x-3. 2x-3. Let me put parentheses around this so it doesn't look like I'm subtracting 4x². And so then I can rewrite all of this business as being equal to, well, in the numerator, in the numerator I'm going to have 5x times 3x which is 15x² 5x times 1, which is + 5x, and then over here, let me do this in green, let's see, I could do -4x times 2x which would be -8x² and then -4x times -3 which is +12x². Did I do that right? Negative... Oh, let me be very careful. - 4, my spider sense could tell that I did something shady. In fact, if you want to pause the video you could see, try to figure out what I just did that's wrong. So -4x² times 2x is -8x to the third power. - 8x³ and then -4x² times -3 is 12x² and then our entire denominator our entire denominator we have a common denominator now so we were able to just add everything is 2x-3 2x-3 times 3x+1 times 3x+1 and let's see, how can we simplify this? So this is all going to be equal to, let me draw, make sure we recognize it's a rational expression, and so let's see, we can look at, we can, our highest degree term here is the -8x³ so it's -8 - 8x³ and then we have a 15x² and we also have a 12x². We could add those two together to get a 27x² so we've already taken care of this, we've taken care, let me do it in that green color, so we've taken of this, we've taken care of those two and we're just left with a 5x, so + 5x and then all of that is over 2x-3 times 3x+1 3x+1 and we are and we are all done. It doesn't seem like there's any easy way to simplify this further. You could factor out an x out of the numerator but that's not going to cancel out with anything in the denominator and it looks like we are all done.