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Intro to adding rational expressions with unlike denominators

Sal rewrites a/b+c/d as a single rational expression.

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• is there a video anywhere where sal does fractional addition like 1/x+2/x+1? need help!
• Well, the problem you gave have the same concept the video above described. We need to have a common denominators first in order to add the two fractions. Since x is anything, the only multiple of (x+2) and (x+1) is (x+1)(x+2). You multiply the first fraction by (x+1) and will get (x+1)/(x+2)(x+1). The second fraction can be multiplied by x+2. You will get (x+2)/(x+1)(x+2).
Adding the fractions together, you will get x+1+x+2/(x+2)(x+1). Which is just 2x+3/x^2+3x+3. This is as simplified as possible. Hope that helped!
• in the question 1+x/y/x/y why does it not simplify to 1?
Why is y+x/x the answer ? does the x not cancel?
• Since dividing a number by a fraction is the same as multiplying the number by the reciprocal of the fraction 1+x/y/x/y = (1+x)/y * y/x. The y's cancel leaving (1+x)/x. Now, consider this example: (1+2)/2. According to the logic you suggest in your question, the 2's cancel out and the answer is 1, but that is incorrect. You need to do the 1+2 first, giving 3/2.

Remember, if you have multiple terms in the numerator that are being added and or subtracted, then you can not cancel them out with a single term in the denominator, the expression with additions/subtractions must be taken as a whole, therefore 1+x has nothing in common with x. If you had (1+x)/(1+x), then yes, the terms cancel and the result is 1.

You may want to review this:
http://www.purplemath.com/modules/rtnldefs2.htm
• I don't get what he means at ?
• he has to multiply that side by "b" so that the denominators are equal.
• Couldn't you just do this: bd(a/b + c/d) = ad + bc, where the bs and ds in the denominator cancel away? Really, is there any algebraic rule that this operation is violating?
• You are multiplying a basic property of math: the identify property of multiplication. It says we can multiply any number by 1 (or something that = 1) and we have an equivalent value.
"bd" does not = 1. You would have to multiply by "bd/bd" to be multiplying by a value = 1.

You may be trying to use a property of equations. Equations and inequalities have their own properties that allow us to multiply by any value as long as we multiple the entire equation by that value. But, `(ad+bc)/bd` is not an equation. So, we can't apply a property of equations.
• Came across this problem on another math site:

The difference between the numerator and the denominator of a fraction is 5. If 5 is added to the denominator, the fraction is decreased by 5/4
Find the fraction.

So the fraction equals something like n/n+5 and I set up an equation like this: n/n+10 = n/n+5 - 5/4.
The problem I have is I only know the difference between the numerator and the denominator is 5, so it could be n+5/n, n-5/n, n/n+5 or n/n-5, right? But when I get the answer to the question it claims that the only possible answer is n+5/n.

Anyone know why this might be?
• I don't quite understand what you mean, but i can walk you through the problem. This is what you do:

The problem says that "the difference between the numerator and denominator is 5," or, in other words:
n - d = 5
n = numerator, and d = denominator. If we rearrange it, we get:
n = 5 + d
So, our fraction, instead of being n/d, we have:
(5 + d)/d
since n and (5 + d) are equal. Now, the next part of the problem says that "if 5 is added to the denominator, the fraction is decrease by 5/4." So, we just add 5 to the bottom of the fraction, and set it equal to the original fraction minus 5/4. This gives us:
(5 + d)/(d + 5) = (5 + d)/d - 5/4
The two (5 + d)s cancel, and we are left with 1 on the right side of the equation. Now, we just get d by itself on one side:
1 = (5 + d)/d - 5/4
1 + 5/4 = (5 + d)/d
9/4 = (5 + d)/d
9/4d = 5 + d
9/4d - d = 5
5/4d = 5
d = 4
So, d equals 4. Now we just replace d with 4 in our equation we made above (n = 5 + d), and we get:
n = 5 + 4
n = 9
Finally, we put our two numbers in the fraction and we get:
n/d = 9/4
Hope this helps! :)
• It is not an equation, it is an expression because it is not equal to something, so it simplifies to
2/x+5/x^3 = (2x^3+5x)/x^4 = x(2x^2+5)/x^4 = (2x^2+5)/x^3
• I was just wandering if you can cancel down the bd in the denominator with the b and d in the numerator since divison and multiplication are commutative operations. So that you are left with a + c in the end?
• No, we can't cancel out the "bd" in denominator with the "b" and "d" in the numerator.
When we reduce fractions, we cancel out factors (things being multiplied).
In the answer: (db - da) / (abc), the db and the da in the numerator are not factors. They are terms (things being added/subtracted) since they are being subtracted with each other.

If you back up a step in the video, Sal had: d(b-a) / (abc)
This is the factored form. The numerator now has 2 factors: the "d" and the "(b-a)". There is no "d" in the denominator, so that can't be cancelled. And, there is no factor "(b-a)" in the denominator, so that also can't be cancelled. Thus, the fraction is fully reduced in it's current form.
Hope this helps.
• Sal is correct in saying you need to multiply the unlike denominators but for the numerators can he just say a+c?
(1 vote)
• No, that won't work. To create an equivalent fraction, you must multiply both the numerator & denominator by the same value.
For example: 1/2 = 1/2(5/5) = 5/10
If you take your approach, you would say that 1/2 = 1/10. They are not equal fractions.

Hope this helps.
• can you teach me how to add algebraic expressions
• Hello hope you all are doing great. What is the Parenthesis for? Sorry to ask this I am just bad in math.

Take care everyone and God bless. :)
• Parenthesis in math is a bracket uses for grouping a part of the equation. I could give you some examples below:
2x^2 & (2x)^2
They're both different because without the brackets, you're just taking the power of x and then multiply it by 2. On the other hand, you're taking 2x to the power of 2, which would be 4x^2.

Another example is that to use brackets to specify negative coefficients or constants. Ex:
2x - (-9w) = ...

The last kind of parenthesis usage I can think of is to use it when distributing a certain number into many other numbers in the parenthesis (only in one side of the equation!)

Ex: 3(92x+7u-8w)=3

I hope you understand my poor explanation.
(1 vote)

Video transcript

- What I want to do in this video is really make sure that we feel comfortable manipulating algebraic expressions that involve fractions. So we'll start with some fairly straightforward ones. So let's say that I had, let's say I had A over B plus C over D, and if I actually wanted to add these things, so it is just one fraction, how would I do that? Well, what we could do is, we could find a common denominator. Well, over here, we don't know what B is, we don't know what D is, but we know a common denominator is just going to be B times D. That is going to be a common multiple of B and D. So we could rewrite this as two fractions, with the common denominator BD, so, plus, BD, actually, let me color code it a little bit. So A over B is going to be the same thing as what over BD? Well, to get BD, I multiplied the denominator by D, so let me multiply the numerator by D as well, then I haven't changed the value of the fraction, I'm just multiplying by D over D. So this is going to be A times D over B times D. Notice I could divide the numerator and the denominator by D, and I'm going to get back to A over B. And then we can look at the second fraction, C over D, to go from D to BD, we multiplied by B and so, if I multiply the denominator by B, if I don't want to change the value of the fraction, I have to multiply the numerator by B as well. So let's multiply the numerator by B as well, and it's going to be BC, BC. BC over BD. This is C over D. So what I have here in magenta, this fraction is equivalent to this fraction. I just multiplied it by D over D, which we can assume is one, if we assume that D is not equal to zero, and then, if we just multiply C over D times one, which is the same thing as B over B, if we assume that B is not equal to zero, then this fraction and this fraction are equivalent. Now, why did I go through all of this trouble? Well, now, I have a common denominator, so I can add these two fractions. So what's this going to be? Well, common denominator is BD, so let me just, so the common denominator is BD, and I could just add the numerators, just like you would've done if these were numbers, if this wasn't an algebraic expression. So this is going to be, this is going to be AD plus BC, all of that over BD.